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Class 11

RBSE Solutions for Class 11 Physical Geography Chapter 9 Denudation

May 27, 2022 by Safia Leave a Comment

Rajasthan Board RBSE Class 11 Physical Geography Chapter 9 Denudation

RBSE Class 11 Physical Geography Chapter 9 Text Book Questions

RBSE Class 11 Physical Geography Chapter 9 Multiple Choice Questions

Question 1.
Spatial disintegration and decomposition of rocks is called:
(a) Denudation
(b) Erosion
(c) Weathering
(d) Solution
Answer:
(c) Weathering

Question 2.
What is Denudation?
(a) Erosion and Transportation
(b) Erosion and Deposition
(c) Erosion, Weathering and Mass Translocation
(d) Erosion and Dissolution
Answer:
(c) Erosion, Weathering and Mass Translocation

Question 3.
The process of Exfolation takes place in the regions where:
(a) Annual range of temperature is high
(b) Temperature is high
(c) Temperature is low
(d) Daily range of temperature is high
Answer:
(d) Daily range of temperature is high

Question 4.
In which region is the process of chemical weathering more active?
(a) Tropical and Arid
(b) Polar region
(c) Tropical and Humid
(d) Cold and Humid
Answer:
(c) Tropical and Humid

Question 5.
Translocation of massive rock debris along the slope by gravitational force is called:
(a) Weathering
(b) Erosion
(c) Mass Translocation
(d) Transportation
Answer:
(c) Mass Translocation

Weathering the Storm in Ersama MCQ Questions for Class 9 English Chapter 6 with Answers ·

RBSE Class 11 Physical Geography Chapter 9 Very Short Answer Type Questions

Question 1.
Which type of weathering is oxidation?
Answer:
Oxidation is a type of chemical weathering in which atmospheric oxygen dissolves in water to change rock minerals into oxides.

Question 2.
What does erosion mean?
Answer:
Erosion is a dynamic process in which rocks keep on disintegrating by being rubbed, eroded, trans – located and transported through glaciers, underground water, waves, wind and rivers.

Question 3.
Where does attrition take place, in weathering or in erosion?
Answer:
The process of attrition takes place in erosion. In this process, the rock fragments or particles being carried by wind, river or sea waves smash together and break into smaller particles.

Question 4.
Which type of weathering is block disintegration?
Answer:
Block Disintegration is a type of physical weathering, in which cracks occur in the rocks and they disintegrate due to extreme daily range of temperature.

Question 5.
Which type of weathering is Carbonation?
Answer:
Carbonation is a type of chemical weathering in which the atmospheric carbon – dioxide gas dissolves in water to form carbonic acid, which in turn erodes lime rocks rapidly.

RBSE Class 11 Physical Geography Chapter 9 Short Answer Type Questions

Question 1.
Explain the meaning of denudation in brief.
Answer:
The process by which the cover of rocks of the crust is eroded, is called denudation. This process enables the leveling of different land forms created by endogenetic forces. In this, after the fragmentation of uplifted landmass, its transportation, erosion, attrition and mass translocation take place. Eventually, by these processes, the uplifted terrain transforms into the lower terrain.

Question 2.
Write the types of weathering.
Answer:
Weathering is a spatial process, in which the rocks disintegrate and decompose at their own place. Mainly, there are three types of weathering:

1. Physical Weathering:
It is further classified into Block Disintegration, Exfolation, Frost Weathering and Pressure Release.

2. Chemical Weathering:
This type of weathering is further classified into Oxidation, Carbonation, Desilication, Hydration and Dissolution.

3. Biological Weathering:
This type of weathering is classified into weathering by vegetation, weathering by animals and weathering by humans.

Question 3.
What is Plucking?
Answer:
Water of glaciers uproots the rock fragments coming in their way and carries them along, this process is called Plucking. The process of plucking takes place in glacial regions.

Question 4.
What do you understand by Corrosion?
Answer:
By the chemical action of water, when the minerals of rocks dissolve in water and flow away, it is called Corrosion. This process of corrosion promotes the process of erosion.

Question 5.
Explain Physical Weathering.
Answer:
Physical Weathering:
Process of disintegration in the rocks by insolation, frost, water and wind is called Physical Weathering. The process of physical weathering is classified into the following parts:

1. Block Disintegration:
In this type of process, due to extreme daily range of temperature, cracks occur in the rocks and they disintegrate into large pieces.

2. Exfolation:
Breaking of rocks in the form of shells when the upper layer of rocks remain heated and the inner layers cool down.

3. Frost Weathering:
In extreme cold regions, the breaking down of rocks due to freezing and melting of water in the cracks of rocks on regular basis.

4. Pressure Release:
When the removal of upper rocks reduces the pressure on the lower rocks, then cracks start appearing in the rocks. This process is called Pressure Release.

RBSE Class 11 Physical Geography Chapter 9 Essay Type Questions

Question 1.
Explain the meaning of weathering and the main types of weathering in detail.
Answer:
Weathering is a spatial process in which the rocks break down by disintegration and decomposition at their own place.

Types of Weathering:
The process of weathering takes place in various forms. Considering the various components as the basis, weathering is classified into the following types:
RBSE Solutions for Class 11 Physical Geography Chapter 9 Denudation img-1
1. Physical Weathering:
Process of disintegration in the rocks by insolation, frost, water and wind is called Physical Weathering. It has the following types:

(a) Block Disintegration:
In hot desert regions, cracks occur in the rocks due to extreme daily range of temperature. Over time/these rocks disintegrate into large pieces.

(b) Exfolation:
Breaking of rocks in the form of shells when the upper layer of rocks remain heated and the inner layers cool down.

(c) Frost Weathering:
In extreme cold regions, the breaking down of rocks due to the freezing and melting of water in the cracks of rocks on regular basis.

(d) Pressure Release:
When the removal of upper rocks reduces the pressure on the lower rocks, then the cracks start appearing in the rocks.

2. Chemical Weathering:
When the rocks break, dissolve, decay and change into new compounds by the action of water and gas in a chemical process , it is called Chemical Weathering. It is classified into the following parts:

(a) Oxidation:
The atmospheric oxygen dissolves in water to change rock minerals into oxides by which decomposition of the rocks takes place.

(b) Carbonation:
The atmospheric carbon – dioxide gas dissolves in water to form carbonic acid by which the lime-rocks dissolve.

(c) Desilication:
Process of separation of silica from rocks.

(d) Hydration:
Absorption of water in rock minerals is called Hydration. Rocks scatter by the process of absorption of water.

(e) Dissolution:
Rainwater dissolves many types of acids and carbonic elements in the rock-materials and creates new chemical compounds. This reaction is called Dissolution.

3. Biological Weathering:
Biological weathering is the weakening and subsequent disintegration of rock by vegetation and living organisms. It has the following types:

(a) Weathering by vegetation:
Roots of trees enter into the rocks and disintegrate them.

(b) Weathering by animals:
Disintegration of rocks by earthworm, termite, rats, etc.

(c) Weathering by humans:
Weathering taking place due to agricultural, mining and construction activities done by human beings.

Question 2.
Explain the meaning of denudation and the main types of denudation in detail.
Answer:
The process by which the cover of rocks of the crust gets removed is called Denudation. This process of denudation completes the action of leveling by the medium of exogenetic forces. Denudation is regarded as the combined result of weathering, erosion and mass translocation. On this basis, it is classified into the following parts:
RBSE Solutions for Class 11 Physical Geography Chapter 9 Denudation img-2
Weathering:
Weathering is a spatial process in which the rocks break down by disintegration and decomposition at their own place. The process of weathering completes mainly in the form of physical, chemical and biological weathering. This process of weathering is affected by the structure and organization of rocks, slope of land, diversity in climate and vegetation. In this, the amount and periodicity of insolation, nature of glacier, structure of rocks and the effect of water and gases on them are included.

Erosion:
The word Erosion is made from the Latin word ‘Erodere’, the meaning of which is to rub or to nibble. Erosion is a dynamic process in which rocks keep on disintegrating by being rubbed, eroded, translocated or transported through glaciers, underground water, waves, air and rivers. The process of erosion takes place due to corrasion, attrition, hydraulic action, corrosion, deflation, cavitation and plucking.

Mass Translocation:
Movement and transfer of rock debris in massive amount along a slope due to gravitational force is called Mass Translocation. Gravitational force is responsible in rolling down of unorganized rock debris. Creeping down the slopes, the rock particles accumulate in the foothills. This accumulation or pile of rocks – flour is called Talus. On the basis of the amount of rolling or creeping of unorganized loose materials, the mass translocation is classified in the form of gradual flow, rapid flow and extreme rapid flow.

Question 3.
Explain in detail the concept of the Cycle of Erosion.
Answer:
American Geomorphologist, William Morris Davis propounded the concept of Cycle of Erosion in 1899. He explained, “During the period of Erosion Cycle, the uplifted terrain on being eroded, transforms into shape – less pedeplain.” For the presentation of Cycle of Erosion, Davis considered the three elements important: structure, process and stage. These three are known by the name of Trikoot of Davis. Their brief description is as follows:

1. Structure:
First, rock structure develops on a terrain. After that, various landscapes form there.

2. Process:
In the development of landscapes or land forms, any of the transformational processes like river, wind, waves, glacier, underground water, etc. plays a vital role.

3. Stage:
Like human life, there is a certain period of time for the development of landscapes. In the Erosion Cycle, Davis considered the direct connection between uplift and erosion. According to him, one of the two processes of uplift and erosion takes place at a time. When there is uplift, there is no erosion; and when there is erosion, there is no uplift. Erosion begins only when the uplift ends. The Erosion Cycle of Davis is classified into the following three stages:

  • Youth Stage
  • Mature Stage
  • Old Stage.

Youth Stage:
In this stage, erosion begins after the ending of uplift. In this stage, the river flows along an uneven surface and there is intensive bottom erosion, the gradients are steep and the erosion is rapid. The rapid deepening of the channel leads to the formation of V-shaped valleys.

Mature Stage:
In this stage, lateral erosion begins whereby the valleys begin to widen rather than deepen. In this stage, the velocity of rivers reduces due to gentle slope, and this decreases their transportation capacity. Most of the rivers get levelled according to the base surface of the erosion.

Old Stage:
In this stage, unevenness of the earth’s crust reduces. The entire area transforms into a pedeplain. Absolute and relative relief both are minimized. The valleys become shallow and extremely broad, and their lateral slope is concave in shape.

The Cycle of Erosion of Davis is being presented by the following diagram:
RBSE Solutions for Class 11 Physical Geography Chapter 9 Denudation img-3

RBSE Class 11 Physical Geography Chapter 9 Other Important Questions

RBSE Class 11 Physical Geography Chapter 9 Multiple Choice Questions

Question 1.
Weathering is a:
(a) Spatial process
(b) Dynamic process
(c) Process of movement
(d) None of these
Answer:
(a) Spatial process

Question 2.
Breaking of rocks in the form of shells is called:
(a) Block Disintegration
(b) Exfolation
(c) Pressure Release
(d) Frost Weathering
Answer:
(b) Exfolation

Question 3.
Where does the process of oxidation occur the most?
(a) in arid regions
(b) in humid tropics
(c) in cold regions
(d) in temperate regions
Answer:
(b) in humid tropics

Question 4.
In which process does carbonic acid form?
(a) in oxidation
(b) in carbonation
(c) in hydration
(d) in dissolution
Answer:
(b) in carbonation

Question 5.
Collision of the flowing rock particles and fragments among themselves is called:
(a) Attrition
(b) Corrosion
(c) Plucking
(d) Cavitation
Answer:
(a) Attrition

Question 6.
Where does the process of plucking occur?
(a) in deserts
(b) in plains
(c) in glacial regions
(d) in mountains
Answer:
(c) in glacial regions

Question 7.
The one which is not included in gradual flow, is:
(a) Earth Creep
(b) Rock Creep
(c) Soil Creep
(d) Slumping
Answer:
(d) Slumping

Matching Type Questions

Match Column A with Column B in the following:

Question A.

Column A (Type of Weathering) Column B (Area of Weathering)
(i) Block Disintegration (a) in humid tropics
(ii) Frost Weathering (b) in lime regions
(iii) Oxidation (c) in extreme cold regions
(iv) Carbonation (d) in mining areas
(v) Weathering by humans (e) in desert regions

Answers:

  1. (e)
  2. (c)
  3. (a)
  4. (b)
  5. (d)

Question B.

Column A  (Process) Column B (Relation)
(i) Talus Creep (a) Extreme rapid flow
(ii) Sheet Flow (b) Chemical weathering
(iii)Debris Fall (c) Rapid flow
(iv) Dissolution (d) By glaciers
(v) Plucking (e) Gradual flow

Answers:

  1. (e)
  2. (c)
  3. (a)
  4. (b)
  5. (d)

RBSE Class 11 Physical Geography Chapter 9 Very Short Answer Type Questions

Question 1.
What are Endogenetic Forces?
Answer:
Forces which originate in the internal part of earth are called Endogenetic Forces. New terrains are formed on the earth’s surface by these forces.

Question 2.
What are Exogenetic Forces?
Answer:
Forces originating or acting on the surface of earth are called Exogenetic Forces. The process of leveling is done mainly by these forces.

Question 3.
Of what is denudation a collective process?
Answer:
The denudation is a collective process of weathering, erosion and mass translocation.

Question 4.
Name the controlling factors of weathering.
Answer:
The factors that affect weathering include the structure and organization of rocks, slope of land, diversity of climate and vegetation.

Question 5.
What is the meaning of Mechanical Weathering?
Answer:
The disintegration of rocks by the action of physical factors like insolation, frost, water and wind pressure is called Physical Weathering.

Question 6.
What is the meaning of Disintegration?
Answer:
The action of breaking down of rocks into the fragments by the fragmentation of organized rocks is called Disintegration.

Question 7.
Why does Frost Weathering occur?
Answer:
Frost Weathering occurs in extreme cold regions, when the rocks break down due to the freezing and melting of water in the cracks of rocks on regular basis.

Question 8.
What does Pressure Release mean?
Answer:
When the removal of upper rocks reduces the pressure on the lower rocks, then cracks start appearing in the rocks. This process is called Pressure Release.

Question 9.
What is Chemical Weathering?
Answer:
When the rocks break, dissolve, decay and change into new compounds by the action of water and gas by chemical process, it is called Chemical Weathering.

Question 10.
What does Oxidation mean?
Answer:
When the atmospheric oxygen dissolves in water to change rock minerals into oxides, then this process is called Oxidation.

Question 11.
What does Carbonation mean?
Answer:
The atmospheric carbon dioxide gas (CO2) dissolves in water to form carbonic acid. On coming in contact with it, the lime – rocks dissolve rapidly. This process is called Carbonation.

Question 12.
What is Hydration?
Answer:
The absorption of water in rock minerals is called Hydration.

Question 13.
What does Biological Weathering mean?
Answer:
The fragmentation of rocks by the biotic community (vegetation, living – organisms and humans) found on the earth’s crust is called Biological weathering.

Question 14.
What are the factors responsible for erosion?
Answer:
The factors responsible for erosion include glacier, underground water, sea waves, wind and flowing water.

Question 15.
What does Corrasion mean?
Answer:
A type of erosive action in which pebbles, stones, boulders, rock-particles, etc. along with any factor of erosion, erode the land-surface by rubbing or scraping, is called Corrasion.

Question 16.
What does Attrition mean?
Answer:
The rock fragments or rock particles when carried by the air, water and glacier, smash together and break into smaller particles. This process is called Attrition.

Question 17.
What does Hydraulic Action mean?
Answer:
The process of the erosion of rocks by the heavy pressure of river water or whirlpool is called Hydraulic Action.

Question 18.
What is Deflation?
Answer:
When the sand, dust or other unorganized rocks – flour are carried away by the winds from one place to another, this process is called Deflation.

Question 19.
What is Plunge Pool?
Answer:
Plunge Pool:
Deep trough built in a base rock by the erosion caused by pebbles and stones present in the whirlpool, which fall down from a height with the waterfalls.

Question 20.
What is called Mass Translocation?
Answer:
The process of the movement and transfer of rock debris in massive amount along a slope duetto gravitational force, is called Mass Translocation.

Question 21.
Into how many types is Mass Translocation classified?
Answer:
Mass Translocation is classified into three parts: mass translocation with gradual flow, with rapid flow and with extreme rapid flow.

Question 22.
Where does the process of gradual-flow take place the most?
Answer:
The process of gradual-flow mostly takes place in sub – polar cold regions.

Question 23.
In which forms is the extreme rapid flow seen?
Answer:
The extreme rapid flow is seen mainly in the forms of Landslide, Rock Slide, Rock Fall, Landslip, Debris Fall and Slumping.

Question 24.
Which are known as the Trikoot of Davis?
Answer:
Structure, process and time (stage) are known as the Trikoot of Davis.

Question 25.
What did Penck consider the Cycle of Erosion as the sum of?
Answer:
Penck considered the Cycle of Erosion as the sum of the stages of development of landscapes, their uplift rate and inter-relations of their degradation.

RBSE Class 11 Physical Geography Chapter 9 Short Answer Type Questions (SA-I)

Question 1.
Describe the processes included in Denudation.
Answer:
In Denudation, the following processes are included:

1. Weathering:
It is a spatial process, in which the rocks break down by dis-integration and decomposition at their own place. This process is called Weathering.

2. Erosion:
It is a dynamic process in which rocks keep on dis-integrating by being rubbed, eroded, trans – located or transported through glaciers, underground water, waves, wind and rivers. This process is called Erosion.

3. Mass Translocation:
Movement and transfer of rock debris in massive amount along the slope by gravitational force is called Mass Translocation.

Question 2.
What are the differences between block disintegration and frost weathering?
Answer:
The following differences are found between black disintegration and frost weathering:

Block Disintegration Frost Weathering
1. This process of weathering takes place in desert regions. 1. This process of weathering takes place in polar and sub-polar regions.
2. This process takes place due to high temperature. 2. This process takes place as a result of the condition of low temperature.
3. In this type of weathering, the rocks disintegrate due to high range of temperature. 3. In this type of weathering, the rocks disintegrate due to changing of water into solid and liquid form.

Question 3.
What are the differences between Oxidation and Carbonation?
Answer:
The following differences are found between Oxidation and Carbonation:

Oxidation Carbonation
1. In this process, the atmospheric oxygen dissolves in water. 1. In this process, the atmospheric carbon dioxide dissolves in water.
2. In this process, oxygen changes the rock minerals into oxides. 2. In this process, carbon dioxide dissolves in water to form carbonic acid.
3. Due to oxidation, the process of rusting in iron takes place. 3. Due to carbonation, the process of fragmentation of lime takes place.
4. This process takes place mostly in humid tropics. 4. This process takes place mostly in lime regions.

Question 4.
Explain the types of Biological Weathering.
Answer:
On the basis of their causative factors, Biological Weathering is classified as follows:

1. Weathering by Vegetation:
Roots of different types of trees by entering into the rocks make their particles loose, due to which dis – integration in the rocks takes place. This is called weathering by vegetation.

2. Weathering by Living Organisms:
This weathering takes place when the rocks are disorganized by various living – organisms like earthworm, termite, rats and animals.

3. Weathering by Humans:
When the rocks are disorganized by various human activities like mining, agriculture, construction, atomic explosion, it is included in the weathering by humans.

Question 5.
Explain the importance of weathering in brief.
Answer:
The importance of weathering is explained as follows:

  1. In weathering, the rocks are broken up into small fragments, this facilitates soil formation.
  2. Weathering is responsible for erosion and mass movement.
  3. Depending on the depth of weathered materials, the growth of vegetation and bio-diversity are determined.
  4. Weathering is helpful in massive erosion and degradation of high reliefs.
  5. Topographies are the result of weathering and erosion.
  6. By the weathering and deposition of rocks, the concentration of valuable minerals like iron, manganese, aluminium, copper, etc takes place, by which their exploitation, processing and refinement become easy.
  7. Weathering is an important process of soil formation.

Question 6.
How is weathering responsible for bio-diversity on the earth?
Answer:
Bio – diversity represents the presence of various biological species found in a region on the earth’s surface. Weathering has a significant impact on bio – diversity. Bio – diversity is completely affected by vegetation. Rocks and minerals are trans – located by weathering and new surfaces are formed. This helps in the penetration of moisture and air in the surface by chemical process. Through this, the humus, carbonic and acidic materials enter into the soil, by which bio – diversity is affected. In this way, it can be said that weathering is responsible for bio – diversity on the earth’s surface.

Question 7.
What are the differences between Erosion and Mass Translocation?
Answer:
The following differences are found between Erosion and Mass Translocation:

Erosion Mass Translocation
1. In erosion, the flowing water, glacier, wind, underground water, waves and currents are the major contributors. 1. The process of mass translocation takes place mainly due to gravitational force.
2. Erosion is a continuous process. 2. Mass translocation is not a continuous process.
3. The scope of erosion is wide. 3. The scope of mass translocation is limited in comparison to erosion.
4. Degradation takes place by erosion. 4. It is not necessary that degradation will take place by mass translocation.

Question 8.
Explain the stages of the Cycle of Erosion as described by Davis.
Answer:
Davis classified the Cycle of Erosion into three main stages:

  1. Youth Stage
  2. Mature Stage
  3. Old Stage

1. Youth Stage:
In this stage, the river flows along an uneven surface and there is intensive bottom erosion, the gradients are steep and the erosion is rapid. The rapid deepening of the channel leads to the formation of V – shaped valleys.

2. Mature Stage:
In this stage, river widens the valley by lateral erosion. The bottom is eroded by the lateral erosion in this stage, by which the V – shaped valleys usually converts into a U – shape.

3. Old Stage:
In this stage, unevenness of the earth’s crust reduces and the entire area transforms into a pedeplain. In this stage, the slope is almost non-existent and the water spreads around.

Question 9.
Write the three drawbacks of the Cycle of Erosion concept of Penck.
Answer:
The theory related to the Cycle of Erosion of Penck has the following drawbacks:

  1. Penck presented the theory of the Cycle of Erosion in German language. When it was translated from German to other languages (mainly in English), the meaning became distorted.
  2. The theory becomes non-interesting due to being obscure.
  3. Scholars do not agree on the concept of continuous movement in crust and the parallel withdrawal of slopes as presented by Penck.

RBSE Class 11 Physical Geography Chapter 9 Short Answer Type Questions (SA-II)

Question 1.
What are the components that control weathering.
Answer:
Weathering is a spatial process in which rocks disintegrate at their own place, but during this process many factors or components affect it. In these controlling factors, the structure and composition of rocks, slope of land, diversity of climate and vegetation are mainly included. The short description of all these factors/components is as follows:

1. Structure and Composition of Rocks:
Chemical weathering is more in the rocks with higher porosity and soluble minerals. Mechanical weathering is more in the rocks with vertical base layers, and chemical weathering is more in the rocks with horizontal layers.

2. Slope of Land:
Weathering is less in the land with gentle and low slope relatively to the land with steep slope.

3. Diversity in Climate:
Chemical weathering is more in humid tropics, while mechanical weathering is more in arid and tropical regions.

4. Influence of Vegetation:
Vegetation partially is a factor of weathering and partially is a constraint too. The amount of weathering is also more in vegetation less tropical regions due to more insolation.

Question 2.
Are physical and chemical weathering independent of each other? If not, then why? Explain with example.
Answer:
Weathering is a process of breaking down of rocks, in which due to disintegration and decomposition, rocks become weak and start getting separated. Under weathering, the weakening of rocks takes place in two ways:

  1. By mechanical fragmentation.
  2. By chemical decomposition.

The process of weakening down of rocks by physical factors like temperature, rainfall, air-pressure, etc. is called disintegration. This type of weathering is called Physical Weathering. The process of weakening down of rocks by chemical processes like oxidation, carbonation, hydration, etc. is called Chemical decomposition. This type of weathering is also called chemical Weathering.

The processes of physical and chemical weathering are different from each other, but both types of weathering are not independent of each other. The effects of factors that participate in physical and chemical weathering cannot be separated from each other. For example, temperature plays an important role in physical weathering, due to which the rocks undergo expansion and contraction and they become weak.

Weathering is completely influenced by chemical composition of rocks. On the basis of chemical composition, the ability to absorb heat in rocks is also affected. Similarly, water will not react with any rock, unless it receives heat by temperature or pressure. The processes of chemical weathering are not the same in all the heating boards. Chemical weathering is more active in tropical climatic regions where the temperature is high throughout the year.

Thus, it is clear that the physical and chemical weathering are not independent of each other and they are controlled by atmospheric conditions.

Question 3.
What is mass movement? Explain its various forms with the help of a diagram.
Answer:
All those movements are included in the mass movement, under which massive debris of rocks are displaced in accordance to the slope by gravitational force. In mass movement, there are creeps, flows, slides and falls of debris obtained from rocks. Weathering is not necessary in mass movement because in this, the fall of original rocks also start in accordance to the slope along with the weathered materials.

Mass movement occurs also due to gravitational force. There is no direct connection to the factors of erosion in this. Materials present their resistance on the slope. But due to gravitational force, they are generally not successful. Weak materials, thin bedded rocks, faults, steep slope, lack of adequate vegetation, etc. are the important helping factors of mass movement.
RBSE Solutions for Class 11 Physical Geography Chapter 9 Denudation img-4

Question 4.
What are the major factors of the activation of mass movement?
Answer:
Important factors of the Activation of Mass Movement:
The important factors providing activation to mass movement are as follows:

  1. By natural and artificial agents, the removal of the base, which supports the materials at the top.
  2. Increase in the height and intensity of slope.
  3. Materials to be of heavy weight.
  4. Overload resulting due to the lubrication of saturated materials of slope by excessive rainfall.
  5. Removal of materials from the surface of original slope.
  6. Earthquake
  7. Vibrations caused by explosives or machines.
  8. Excess of natural leakage.
  9. Excessive exploitation of water from river, lake and other water reservoirs, and gradual flow of water from down the river banks and slopes.
  10. Indiscriminate exploitation of vegetation.

Question 5.
Into which stages did Penck classify the process of erosion? Explain.
Answer:
German scholar, Penck, classified his concept related to erosion into the following stages:

1. First Stage:
According to Penck, in this stage, both the processes of upliftment and erosion move together. But the upliftment is more than the erosion.

2. Second stage:
In this stage, uplift and erosion remain equally active. As a result, the valleys start to become wide and deep.

3. Third Stage:
In this stage, due to competitive rate, the dorsal difference of the upper and lower curves of the uplift and erosion processes remains the same.

4. Fourth Stage:
In this stage, the rate of uplift gets inactive and weak, and the erosion is effective at the same rate. As a result, the valleys are deep and move down.

5. Fifth Stage:
In this stage, along with the uplift, the rate of erosion also gets inactive and weak. The dorsal difference of both the curves decreases. The concept of Penck related to erosion is being presented by the following diagram.

Question 6.
“Our Earth is the playground of geomorphic processes of two opposing groups.” Explain.
Answer:
Changes take place in the configuration of the earth’s surface due to the physical pressure and chemical processes by endogenetic and exogenetic forces.

Two types of forces work in the formation of the earth’s surface:

  1. Endogenetic Forces
  2. Exogenetic Forces

Endogenetic forces originate from deep down the surface and work at a very slow speed. By this, mountains are formed and unevenness is created on the earth’s surface. Undesirable incidents also occur by some endogenetic forces like volcanic eruption, earthquake, etc. Exogenetic forces originate above the surface and work right opposite to the endogenetic forces, thereby trying to reduce the relief.

In this process, somewhere they do the work of erosion and somewhere accumulation. In this way, the unevenness of the earth’s surface reduces by exogenetic forces. Thus, these forces are also called ‘Level Establishing Force’. The opposite work of both the forces remains as long as the opposing action of endogenetic and exogenetic forces continue. In this way, it is said that the Earth is the playground of geomorphic processes of two opposing groups.

RBSE Class 11 Physical Geography Chapter 9 Essay Type Questions

Question 1.
Define erosion and describe the modes of factors responsible for it.
Or
Explain the modes of erosion.
Answer:
The word ‘Erosion’ is made from the Latin word ‘Erodere’, the meaning of which is to rub or to nibble. Erosion is a dynamic process. In this process, the rocks keep on disintegrating by being rubbed, eroded, translocated or transported through glaciers, underground water, sea waves, wind and rivers. By this process, a high terrain eventually transforms into a low terrain.

The following modes are held responsible for the process of erosion:

  1. Corrasion
  2. Attrition
  3. Hydraulic Action
  4. Corrosion
  5. Deflation
  6. Cavitation
  7. Plucking

1. Corrasion:
When the erosion factors carry away rocks-debris and rocks – flour in their flow, then these materials cause the friction of surface rocks, which is called Corrasion.

2. Attrition:
The rock fragments or particles when carried by the air, water and glacier smash together and break into smaller particles, this is called Attrition.

3. Hydraulic Action:
The process of erosion of rocks by the heavy pressure of river water or whirlpool is called Hydraulic Action.

4. Corrosion:
By the chemical action of water, the minerals of rocks dissolve in water and flow away, which is called Corrosion.

5. Deflation:
The sand, dust and or other unorganized rocks – flour are carried away by the winds from one place to another place, which is called deflation.

6. Cavitation:
Waves rising from whirlpool in river form many types of holes at the bottom of the river, which is called Cavitation. Pot holes and plunge pools are examples of such holes.

7. Plucking:
When glaciers uproot the rocks coming in their way and transport them along, then this action is called Plucking. The eroded materials usually flow in the three forms:

(a) Dissolving:
Various materials dissolve in water and flow along with it.

(b) Suspension:
Materials float or hang along with erosive factors (water or wind) and flow along.

(c) Sliding:
Large masses of rocks drag or slide and flow on the surface of the river, which is known as sliding.
RBSE Solutions for Class 11 Physical Geography Chapter 9 Denudation img-5

Question 2.
What is the meaning of Mass Translocation? Explain its types.
Answer:
Movement and transfer of rock debris in massive amount along a slope due to gravitational force is called Mass Translocation. The gravitational force is responsible for the rolling down of unorganized rock debris. Creeping down the slopes, the rock particles accumulate in the foothills. This accumulation or pile of rocks – flour is called Talus.

A cone – shaped accumulation of loose rocks is called Talus Cone. On the basis of the amount and speed of creeping or rolling down of unorganized loose materials, mass translocation is classified into the following three groups:

1. Mass translocation with gradual flow:
Due to less amount of moisture, the rocks – flour creeps slowly. The action of gradual flow occurs mostly in sub-polar cold regions. Earth Creep, Rock Creep, Talus Creep and Soil Creep are included under the action of gradual flow.

2. Mass translocation with rapid flow:
Due to the adequacy of water, the rocks – flour gets saturated and creeps rapidly. Earth flow, Mud flow and Sheet flow are included under rapid flow. Mud flow can be seen creeping on the walls of the river valleys.

3. Mass translocation with extreme rapid flow:
Moisture of water is not necessary for extreme rapid flow. Large rock fragments start falling suddenly by gravitational force. Landslide, Rock Slide, Rock Fall, Landslip and Debris Fall are included in extreme rapid flow.

Question 3.
Compare the ideas of Davis and Penck.
Answer:
The comparison of the ideas of Davis and Penck is as follows:
RBSE Solutions for Class 11 Physical Geography Chapter 9 Denudation img-6

RBSE Solutions for Class 11 Geography

RBSE Solutions for Class 11 Physics Chapter 4 Laws of Motion

May 25, 2022 by Safia Leave a Comment

RBSE Solutions for Class 11 Physics Chapter 4 Laws of Motion 1

Rajasthan Board RBSE Class 11 Physics Chapter 4 Laws of Motion

RBSE Class 11 Physics Chapter 4 Textbook Exercises with Solutions

RBSE Class 11 Physics Chapter 4 Very Short Answer Type Questions

An inclined plane’s normal force formula is never equal to the object’s weight.

Question 1.
If the net force on an object is zero, then, what will be its acceleration?
Answer:
Zero because F = ma ⇒ a = \(\frac{F}{m}\),
when, F = 0 ⇒ a = \(\frac{0}{m}\) = 0 .

Question 2.
Write the formula for momentum of a body.
Answer:
\(\vec{p}=m \vec{v}\) where m = mass of body and \(\vec{v}\) = velocity of the body.

Question 3.
Write the Newton’s second law of motion in terms of a formula.
Answer:
\( \vec{F}=\frac{d \vec{p}}{d t}=\frac{d}{d t}[m \vec{v}]=\frac{m d v}{d t}+\frac{v d m}{d t}\)

Question 4.
What are the directions of action and reaction in Newton’s third law of motion?
Answer:
Opposite to each other but act on different bodies.

Question 5.
Give an example of a system with variable mass.
Answer:
Rocket propulsion.

Question 6.
Of two contact surfaces, whose value is more : Static or kinetic friction?
Answer:
Static because fs > fk i. e., static friction is more than dynamic friction.

Question 7.
Whose value is more μs or μk?
Answer:
μs > μk because fs > fk.

Question 8.
Which force acts on a uniform circular motion?
Answer:
Centripetal.

Question 9.
How does a vehicle obtain a centripetal force on a levelled circular path?
Answer:
By frictional force.

Question 10.
How does a centripetal force other than frictional force is obtained on a banked circular path?
Answer:
By the horizontal component of the normal force i.e., by N sinθ.

RBSE Class 11 Physics Chapter 4 Short Answer Type Questions

Question 1.
Explain why passengers are thrown forward from their seats when a speeding bus stops suddenly?
Answer:
The lower part of the body of passengers (which is in contact with the bus) comes to rest, but because of inertia, the upper part of the body tends to keep on moving. As a result of it the rider falls forward.

Question 2.
Why is Newton’s first law of motion called law of inertia?
Answer:
According to Newton’s first law of motion, in absence of external force, there is no change in state of a body and inertia is the property of a body due to which it opposes the change in its state. That is the reason. Newton’s first law of motion is also known as the law of inertia.

Question 3.
Explain why a cricketer moves his hand backwards while holding a catch?
Answer:
The impulse received by the hand, while taking a catch is equal to the product of force applied and time taken to complete the catch. By moving his hand backward, the cricketer increases time interval so that he does is not hurt due to the applied lower impulsive force on the hand.

Question 4.
Define force.
Answer:
Force is a physical quantity which tries to change the state of body. It is a vector quantity. It’s S.I. unit is Newton (N).

Question 5.
When the acceleration of a particle is measured in an inertial system as zero. Can we say that no force acts on the particle? Explain.
Answer:
In inertial frame of reference, Newton’s first and second laws are valid. The motion is defined on relative basis. Generally, we define the motion assuming the earth to. be stationary and the Earth is supposed to be the inertial frame of reference. Therefore, the gravitational force acts on a body on the , Earth even when its acceleration is zero.

Question 6.
According to Newton’s third law of motion, in a game of tug, each team pulls the opposing team with equal force. Then, why a team wins and the other one loses?
Answer:
In the game of tug, till both the teams pull the rope with the same force, the net force on the system remains zero. As the force applied by one team increases than the other one, the whole system starts moving in the direction of net force. Thus one team wins and the other one loses.

Question 7.
A book is placed on a table. The weight of the book and the normal force by table on the book is equal in magnitude and direction. Can it be an example of Newton’s third law of motion? Explain.
Answer:
Yes; because according to the third law of motion, the magnitude of both the forces should be – equal and directions opposite and both forces should act on two different objects. Here both above conditions are fullfilled.

Question 8.
Define impulse acting on a body.
Answer:
Impulse is total effect of force on motion of a body. It’s value is equal to the product of applied force and time interval for which force is applied.
If a force is applied for a small time interval ‘dt then impulse of force. \(\vec{I}=\vec{F} \cdot d t\)

Question 9.
What is an impulsive force?
Answer:
Those forces of large magnitude which act for a very short time interval (e.g, time of contact), are known as impulsive force.

Question 10.
Write the impulse-momentum theorem?
Answer:
According to this theorem, impulse of a force is equal to change in momentum due to that force.
i.e., I = \(\Delta \vec{p}=\vec{p}_{2}-\vec{p}_{1}\)

Question 11.
Write the law of conservation of momentum.
Answer:
According to this law, net linear momentum of a system remains constant in absence of external force.
According to Newton’s second law of motion
F = \(\frac{d p}{d t}\)
If external force is zero i.e., F = 0
then \(\frac{d p}{d t}\) = 0
⇒ p = constant

Question 12.
What is an isolated system?
Answer:
Such a system which have no relation with external environment is called an isolated system.

Question 13.
Why a gun recoils backward when a bullet is fired?
Answer:
Bullet and gun form a system together. Before shooting the bullet by the gun, net momentum of this system is zero. The momentum of the gun is equal to the momentum of the bullet and is in opposite direction to conserve the momentum, when the gun is fired.

Question 14.
How many types of friction are there?
Answer:
There are two types of friction :
(i) Static friction : This type of friction acts till the body does not move, therefore this type of friction is known as static friction. The maximum static friction is known as “Limiting friction”.

(ii) Dynamic or kinetic friction : When the body starts moving,then the acting friction is known as dynamic or kinetic friction. Its value is some less than the limiting friction.

Question 15.
Define centripetal acceleration.
Answer:
When a particle moves on a circular path then an acceleration acts on the particle towards the centre of the circular path, this is known as centripetal acceleration.

Question 16.
Why is a road banked on a circular turn?
Answer:
Since circular motion of a particle is possible only when a centripetal force acts on it. Without centripetal force, the path cannot be circular. While passing through a circular turn of the road, its path becomes circular, therefore to obtain the required centripetal force, the vehicle bends towards the centre of the road. By this action, if the road is horizontal, then the outer wheels of the vehicle will not remain in contact with the road and it may overturn. To award this unhappening the road is banked towards the centre of the path.

Thus, on banked road, all the wheels of the vehicle remain in contact with the road and required centripetal force is obtained by a horizontal component of normal reaction of the road and the vehicle passes safely through the turn of the road.

Question 17.
Define inertial frame of reference.
Answer:
Inertial frame of reference : The frame of reference in which Newton’s first and second laws are valid is known as inertial frame of reference.

Question 18.
Define non-inertial frame of reference.
Answer:
The frame of reference in which Newton’s first and second laws are not valid, are known as non-inertial frame of reference. .

Question 19.
Is Earth an inertial frame of reference?
Answer:
Earth is revolving round the Sun. Therefore, it possesses centripetal acceleration. Therefore, it should be treated as non-inertial frame of reference. However, value of centripetal acceleration of the Earth,
RBSE Solutions for Class 11 Physics Chapter 4 Laws of Motion 1
Which is negligible small compared to acceleration due to gravity (g= 9.8m/s2). Hence for terrestrial experiments, the Earth can be considered as an inertial frame.

RBSE Class 11 Physics Chapter 4 Long Answer Type Questions

Question 1.
What is Newton’s second law of motion? Define it. Obtain Newton’s first law from it.
Answer:
Momentum and Newton’s Second Law of Motion
Linear momentum : Momentum of body is the physical quantity of motion possessed by the body and mathematically, It is defined as the product of mass and velocity of the body.

As the linear momentum or simply momentum is equal to a scalar times a vector (velocity), it is therefore a vector quantity and is denoted by \(\vec{p}\). The momentum of a body of mass m moving with velocity \(\vec{v}\) is given by the relation :
\(\vec{p}=m \vec{v}\)
Dimensional formula = [M1L1T-1]
Unit = Kg m/s
Suppose that a ball of mass m1 and a car of mass m2 (m2 > m1) are moving with the same velocity v. If P1 and p2 are momentum of ball and car respectively then :
\(\frac{p_{1}}{p_{2}}=\frac{m_{1} v}{m_{2} v}\) or \(\frac{p_{1}}{p_{2}}=\frac{m_{1}}{m_{2}}\)

As m2 > m1: It follows that p2 > p1. If a ball and a car are travelling with the same velocity, the momentum of the car will be greater than that of the ball. Similarly, we can show that if two objects of same masses are thrown at different velocities, the one moving with the greater velocity possesses greater momentum. Finally, if two objects of masses m1 and m2 are moving with velocities v1 and v2 possesses equal momentum.
m1v1 = m2 v2
\(\frac{v_{1}}{v_{2}}=\frac{m_{2}}{m_{1}}\)
In case m2 > m1 then v2 < v1 i. e, two bodies of different masses possess same momentum, the lighter body possesses greater velocity.
The concept of momentum was introduced by Newton in order to measure the quantitative effect of force.
Momentum of body in term of kinetic energy
Ek = \(\frac{1}{2}\) mv2 ……….. (1)
but p = mv
∴ v = p/m
put in equation (1)
Ek = \(\frac{1}{2} m \frac{p^{2}}{m^{2}}=\frac{p^{2}}{2 m}\)

Question 2.
Explain Newton’s third law of motion with the help of two examples.
Answer:
Newton’s Third Law of Motion
If an object ‘A’ exerts force on object ‘B’, then object B must exert a force of equal magnitude and opposite direction back on object ‘A’.

This law represents a certain symmetry in nature : forces always occur in pairs, and one body cannot exert a force on another without experiencing a force itself. We sometimes refer to this law loosely as action-reaction, where the force exerted is the action and the force experienced as a consequence is the reaction.

According to Newton’s third law, “To every action, there is always an equal and opposite reaction”.

It must be remembered that action and reaction always act on different objects. The third law of motion indicates that when one object exerts a force on another object, the second object instantaneously exerts a force back on the object. These two forces are always equal in magnitude, but opposite in direction.

These forces act on different objects and so they do not cancel each other. Thus Newton’s third law of motion describes the relationship between the forces of interaction between two objects.

For example, when we placed a wooden block on the ground, this block exerts a force equal to its weight, W = mg acting downwards to the ground. This is the action force. The ground exerts an equal and opposite force N = mg on the block in upward direction. This is the reaction force.
RBSE Solutions for Class 11 Physics Chapter 4 Laws of Motion 2

Illustrations of Newton’s Third Law
Some of the examples of Newton’s third law of motion are given below :
1. A gun recoils when a bullet is fired form it: When a bullet is fired from a gun, the gun exerts a force on the bullet in the forward direction. This is the action force. The bullet also exerts an equal force on the gun in the backward direction. This is the reaction force. Due to the large mass of the gun it moves only a
RBSE Solutions for Class 11 Physics Chapter 4 Laws of Motion 3
little backward by giving a jerk at the shoulder of the gun man. The backward movement of the gun is called the recoil of the gun.
RBSE Solutions for Class 11 Physics Chapter 4 Laws of Motion 4

2. Walking: In order to walk, we press the ground in backward direction with our feet (action). In turns, the ground gives an equal and opposite reaction R, (figure 4.4 (a). The reaction R can be resolved into two components, one along the horizontal and other along the vertical. The component H = R cosθ along the horizontal, help us to move forward, while the vertical component, V = R sinθ opposes our weight, [figure 4.4 (b)]
RBSE Solutions for Class 11 Physics Chapter 4 Laws of Motion 5

Question 3.
Write impulse-momentum theorem and prove it. How will you find out the impulse from a graph?
Answer:
Impulse-Momentum Theorem .
A force which acts on a body for short interval of time is called impulsive force or impulse.

For Example : Hitting, jumping, diving, catching etc. are all examples of impulsive forces or impulse.

An impulsive force does not remain constant, but changes first from zero to maximum and then from maximum to zero. Thus it is not possible to measure easily the value of impulsive force because it changes with time. In such cases, we measure the total effect of the force, called impulse hence, impulse is defined as the product of the average force and the time interval for which the force acts. If \(\vec{F}\) is the value of force during impact at any time and \(\vec{p}\) is the momentum of the body at that time, then according to Newton’s second law of motion,
\(\vec{F}=\frac{d \vec{p}}{d t}\)
or \(\vec{F} d t=d \vec{p}\)
Suppose that the impact lasts for a small time t and during this time, the momentum of the body changes from \(\vec{p}_{1}\) to \(\vec{p}_{2}\) then integrating the above equation between the proper limits, we have :
RBSE Solutions for Class 11 Physics Chapter 4 Laws of Motion 6
It may be noted that \(\vec{F}\) has not been taken out of the integration sign for the reason that \(\vec{F}\) varies with time and does not remain constant during impact. The integral \(\int_{0}^{t} \vec{F}\) dt is measure of the impulse, when the force of impact acts on the body and from equation (1) we find that it is equal to total change in momentum of the body. Since impulse is equal to a scalar (time) times a vector (force) or equal to the change in momentum (vector), it is a vector quantity and it is denoted by \(\vec{I}\).
Therefore, \(\vec{I}=\int_{0}^{t} \vec{F} d t=\vec{p}_{2}-\vec{p}_{1}\)
However if \(\vec{F}_{a v}\) is the average force (constant) during the impact, then
RBSE Solutions for Class 11 Physics Chapter 4 Laws of Motion 7
i. e., the change in momentum of an object equals to the impulse applied to it. This statement is called
impulse-momentum theorem.
I = Δp
Dimensional Formula and Unit :
I = FΔt = [M1L1T-2] [T1]= [M1L1T-1]
So, the dimensional formula of impulse is same as that of momentum.
The SI unit of impulse are (N-s) and kg m/s.
In C.G.S. system unit of impulse are dyne-sec and g cm/s.

Force Time Curve : In the real world, forces are often not constant. Forces may build up from zero over time and also may vary depending on many factors.

Finding out the overall effect of all these forces directly would be quite difficult. As we calculate impulse, we multiply force by time. This is equivalent to finding out the area under a force time curve. For variable force the shape of the force-time curve would be complicated but for a constant force we will get a simpler rectangle. In any case, the overall net impulse only matters to understand the motion of an object following an impulse.

In the figure, the graph of change in impulsive force with the time is shown :

The force-time curve and the area between the time axis can be divided in the form of many slabs. Suppose the value of force F is considered as constant along the change in time dt, then area of slab is given by F. dt.
RBSE Solutions for Class 11 Physics Chapter 4 Laws of Motion 8
Total effect of the force for time t1 to t2
= \(\int_{0}^{t} \vec{F}\) dt = sum of area of all slabs
= graph of force-time and area covered between the time axis
∵ By the impulse – momentum theorem t
Impulse I = \(\int_{0}^{t} F d t\) = p2 – p1 = change in momentum. Thus force-time graph and the area covered with the time axis is equal to the total change in the momentum of the body.

Question 4.
Write the law of conservation of momentum for a,system of N particles. Obtain it from Newton’s second law of motion. Explain the law of conservation of momentum with the help of an example.
Answer:
Principle of Conservation of Linear Momentum and its Applications
If the net external force acting on a system of bodies is zero, then the momentum of the system remains constant. This is the basic principle of conservation of linear momentum.
According to Newton’s second law
RBSE Solutions for Class 11 Physics Chapter 4 Laws of Motion 9
If net force (or the vector sum of all forces) on system of particle is equal to zero, the vector sum of linear momentum of all particles remains conserved.
RBSE Solutions for Class 11 Physics Chapter 4 Laws of Motion 10
Consider a system of two bodies on which no external force acts. The bodies can mutually interact with each other. Due to the mutual interaction of the bodies the momentum of the individual bodies may increase or decrease according to the situation, but the momentum of the system will always be conserved, as long as there is no external net force acting on it.

Thus, if \(\vec{p}_{1}\) and \(\vec{p}_{2}\) are momentum of the two bodies at any instant, then in absence of external force
\(\vec{p}_{1}+\vec{p}_{2}\) = constant ……………. (3)
If due to mutual interaction, the momentum of two bodies becomes \(\overrightarrow{p_{1}^{\prime}}\) and \(\overrightarrow{p_{2}^{\prime}}\) respectively, then according to principle of conservation of momentum
\(\overrightarrow{p_{1}}+\overrightarrow{p_{2}}=\overrightarrow{p_{1}^{\prime}}+\overrightarrow{p_{2}^{\prime}}\)
or \(m_{1} \overrightarrow{u_{1}}+m_{2} \overrightarrow{u_{2}}=m_{1} \overrightarrow{v_{1}}+m_{2} \overrightarrow{v_{2}}\) ……………. (4)
Where \(\overrightarrow{u_{1}}\) and \(\overrightarrow{u_{2}}\) are initial velocities of the two bodies of masses m1 and m2 and \(\overrightarrow{v_{1}}\) and \(\overrightarrow{v_{2}}\) are their final velocities.
Therefore, the principle of conservation of linear momentum may also be stated as follows :

For an isolated system (a system on which no external force acts), the initial momentum of the systerA is equal to the final momentum of the system.

Practical Applications of the Principle of Conservation of Momentum
1. Recoiling a Gun : Lets consider the gun and bullet in its barrel as an isolated system. In the beginning when bullet is not fired both the gun and the bullet are at rest. So the momentum before firing is zero
or \(\overrightarrow{p_{c}}\) = 0
Now when the bullet is fired, it moves , in the forward direction and gun recoils back in the opposite direction.

Let mb be the mass and vb be the velocity of the bullet and mg and vg be the mass and velocity of the gun after firing.

Total momentum of the system after the firing would be
\(\overrightarrow{p_{f}}=m_{b} \overrightarrow{v_{b}}+m_{g} \overrightarrow{v_{g}}\)
Since, no external forces are acting on the system, we can apply the law of conservation of linear momentum therefore,
Total momentum of gun and bullet before firing
= Total momentum of gun and bullet after firing
\(0=m_{b} \overrightarrow{v_{b}}+m_{g} \vec{v}_{g}\)
or \(\overrightarrow{v_{g}}=-\frac{m_{b} \vec{v}_{b}}{m_{g}}\)
The negative sign shows that \(\vec{v}_{g}\) and \(\vec{v}_{b}\) are in opposite directions i.e., as the bullet moves forward, then the gun will move in backward direction. The backward motion of the gun is called recoil of the gun.

2. While firing a bullet, the gun must be held tight to the shoulder : This would save hurting the shoulder of the man who fires the gun as the recoil velocity of the gun. If the gun is held tight to the shoulder, then the gun and the body of the man recoil as one system. As the total mass is quite large, the recoil velocity will be very small and the shoulder of the man will not get hurt.

3. Rockets works on the principle of conservation of momentum : The rocket’s fuel burns and pushes the exhaust gases downwards, due to this the rocket gets pushed upwards. Motorboats also work on the same principle, it pushes the water backwards and gets pushed forward in reaction to conserve momentum.

Second Law of Motion is the real law of Motion :
(A) First law is contained in the second law :
According to Newton’s second law of motion, the force acting on a body is given by :
RBSE Solutions for Class 11 Physics Chapter 4 Laws of Motion 11
Thus there is no force is applied on the body then the body at rest will remains at rest and a body in uniform motion will continue to move uniformly along the same straight path. Hence first law of motion is contained in the second law.

Question 5.
Explain the motion of a rocket and obtain the important formula for its velocity?
Answer:
Motion of a Rocket: As the fuel in the rocket is burnt and the exhaust gas is expelled out from the rear of the rocket in the downward direction. The force exerted by the exhaust gas on the rocket is equal and opposite to the force exerted by the rocket to expel it. This force exerted by the exhaust gas on the rocket provides an upthrust to the rocket. The more gas is ejected from the rocket, the mass of the rocket decreases.
RBSE Solutions for Class 11 Physics Chapter 4 Laws of Motion 12
To analyze this process let us consider a rocket being fired in upward direction and we neglect the resistance offered by the air to the motion of the rocket and variation in the value of the acceleration due to the gravity with height.

Figure 4.9. (a) shows a rocket of mass ‘m’ at a time ‘t’ after its take off moving with velocity v. Thus at time ‘t’ momentum of the rocket is equal to ‘Mv’
Thus \(\vec{p}_{c}=M \vec{v}\) ………………(1)
Now after a short interval of time dt, gas of total mass dm is ejected from the rocket. If vg represents the downwards speed of the gas relative to the rocket then velocity of the gas relative to the earth is
\(\overrightarrow{v_{g e}}=\vec{v}-\overrightarrow{v_{g}}\)
At time (t + dt), the rocket and unburned fuel gases mass (M – dm) and it moves with the speed of (\(\overrightarrow{V}+d \vec{V}\)). Thus, momentum of the rocket is :
=(M – dm)(\(\vec{v}+d \vec{v}\))
Total momentum of the system at time (t + dt) is
\(\vec{p}_{f}=d m\left(\vec{v}-\vec{v}_{g}\right)+(M-d m)(\vec{v}+d \vec{v})\) ………………. (2)
Here, the ejected gas and rocket constitutes a system at time (t + dt).
External force on the rocket is weight (-mg) of the rocket (the upward direction is taken as positive)
Now Impulse = Change is momentum
RBSE Solutions for Class 11 Physics Chapter 4 Laws of Motion 13
In equation (iii) \(\frac{d \vec{V}}{d t}\) represent the acceleration of dt the rocket, so M\(\frac{d v}{d t}\) = resultant force on the rocket. dt
Therefore,
Resultant force on rocket = Upthrust on the rocket – Weight of the rocket
Here the upthrust on the rocket is proportional to both the relative velocity (\(\vec{v}_{g}\)) of the ejected gas and the mass of the gas ejected per unit time \(\left(\frac{d m}{d t}\right)\)
Again from eq. (3)
\(\frac{d \vec{v}}{d t}=\frac{\vec{v}_{g}}{m} \frac{d m}{d t}-g\) ………………. (4)
As the rocket goes higher and higher, value of the acceleration due to gravity ‘g’ decreases continuously. The values of (vg) and {dm! dt). Practically remains constant while fuel is being consumed but remaining mass decreases continuously. This result increases in acceleration continues until all the fuel is burnt up.

Velocity of the rocket at any time ‘t’
Now we will find out the relation between the velocity at any time ‘t’ and remaining mass. Again from equation (4) we have
\(d \vec{v}=\vec{v}_{g}\left(\frac{d m}{M}\right)-g d t\) ………………. (5)
Initially at time t = 0 if the mass and velocity of the rocket are m0 and u0 respectively. After time ‘t’ if m and v are the mass and velocity of the rocket. On integrating equation (5) within these limits
\(\int_{v_{0}}^{v} d v=-\int_{M_{0}}^{M} v_{g} \frac{d m}{m}-\int_{0}^{t} g d t\) ……………….. (6)
Note : Here dm is a quantity of mass ejected in time ‘dt’ so change in mass of the rocket in time dt is -dm that’s why we have changed the sign of dm in equation (6).
On evaluating this integral we get
RBSE Solutions for Class 11 Physics Chapter 4 Laws of Motion 14
Equation (7) gives the change in velocity of the rocket in terms of exhaust speed and ratio of initial and final masses at any time ‘t’.
At time t = 0 the velocity of the rocket (initial velocity)
RBSE Solutions for Class 11 Physics Chapter 4 Laws of Motion 15
Note : The speed acquired by the rocket when the whole of the fuel is burnt out is called burn-out speed of the rocket.

Question 6.
How many types of friction are there? Write their laws.
Answer:
Explanation of Newton’s Second Law
According to Newton’s second law of motion, the rate of change of linear momentum of a body is directly proportional to the applied external force on the body, and this change takes place always in the direction of the applied force.
Let, m = mass of a body
v = velocity of the body
The linear momentum of the body
\(\vec{p}=m \vec{v}\) ………… (1)
let \(\vec{F}\) = External force applied on the body in the direction of motion of the body. .
\(\Delta \vec{p}\) = a small change in linear momentum of the body in a small time Δt.
Rate of change of linear momentum of the body = \(\frac{\Delta \vec{p}}{\Delta t}\)
According to Newton’s second law
RBSE Solutions for Class 11 Physics Chapter 4 Laws of Motion 16
Where k is proportionality constant. Taking the limit Δt → 0, the term \(\frac{\Delta \vec{p}}{\Delta t}\) becomes the
derivative or differential coefficient of \(\vec{p}\) w.r.t. time t.
It is denoted by \(\frac{d \vec{p}}{d t}\)
\(\vec{F}=k \frac{d \vec{p}}{d t}\)
Where k = 1 in all the system
RBSE Solutions for Class 11 Physics Chapter 4 Laws of Motion 17
As accelerationtion is a vector quantity and mass is scalar, therefore force \(\vec{F}\) being the product of m and \(\vec{a}\) is a vector. The direction of \(\vec{F}\) is the same as the direction of \(\vec{a}\). Equation (4) represents the equation of motion of the body. We can rewrite equation (6) in scalar form as :
F = ma …………. (7)
Thus, magnitude of force can be calculated by multiplying mass of the body and the acceleration produced in it. Hence, second law of motion gives a measure of force.

Dimension and Unit of Force :
As F = ma
∴ F= [M1] [L1T-2 ] = [M1L1T-2 ]
This is the dimensional formula of force.
Unit of force
The unit of force is Newton in (M.K.S.) system and dyne in C.G.S. system and Poundal in (F.P.S.) system.
Definition of 1 Newton
F = ma
When m = 1kg and a = 1 m/s-2
then F = 1 Newton
The force which produces the acceleration of 1m/s-2 in the 1 kg body, is equal to the 1 Newton.
1 Newton = 1 kg m/ s2
In C.G.S. System F = ma
If m = 1 g and a = 1 cm/s2
then F = 1 dyne
Thus 1 dyne is that force which produces 1 cm/s2 acceleration in the mass of 1g body.

Question 7.
Explain how will you find out the direction of static friction?
Answer:
Static Frictional Force : Static friction is the friction that exists between a stationary object and the surface on which it resting. Static friction keeps an object at rest. It must is overcome to start moving the object.

Imagine trying to push an object across the floor. You push on it with a small force, but it does not move, this is because it is not accelerating. However, according to Newton’s second law, the object must move with an acceleration.
a = \(\frac{F}{m}\)
Now, as the body remains at rest, it implies that an opposing force equal to the applied force must have come into play resulting in zero net force on the object. This force is called static friction. It is denoted by Fs.

Thus, static friction is the opposing force that comes into play when one body tends to move over the surface of another, but the actual motion has not started.

  • The static friction depends upon the nature of surfaces of the two bodies in contact.
  • The static friction does not exist by itself. When there is no applied force, there is no static friction. It comes into play only when the applied force tends to move the body.
  • The static friction is a self-adjusting force.

Limiting Friction : If the applied force is increased, the force of static friction also increases. If the applied force exceeds a certain (maximum) value, the body starts moving. This maximum value of static friction up to which the body does not move, is called limiting friction.

1. The magnitude of limiting- friction between any two bodies in contact is directly proportional to the normal reaction between them.
(fs)max ∝ N
(fs) max ∝ μs N ………….. (1)
Where the constant of proportionality p s is called the coefficient of static friction. Its value depends upon the nature of surfaces of the two bodies in contact that means whether dry or wet; rough or smooth; polished or non polished. For example, when two polished metal surfaces are in contact, ps = 0.2, when these surface are lubricated, μs = 0.1
The value of μs lies between 0 and 1 or 0 < μs < 1
∵ N = Mg
(Fs)max = μs mg ……………… (2)

2. Direction of the force of limiting friction is always opposite to the direction in which one body is at the verge of moving over the other.

3. Coefficient of Static Friction :
μs is called coefficient of static friction and is defined as the ratio of force of limiting friction and normal reaction
From(1), μs = \(\frac{\left(f_{s}\right)_{\max }}{N}\)
Dimension: [M0L0T0]
Unit: It has no unit.
Value of μs does not depend upon apparent area of contact.

Question 8.
Define the coefficients of static and kinetic friction. How will you find out their value?
Answer:
1. Static Frictional Force : Static friction is the friction that exists between a stationary object and the surface on which it resting. Static friction keeps an object at rest. It must is overcome to start moving the object.

Imagine trying to push an object across the floor. You push on it with a small force, but it does not move, this is because it is not accelerating. However, according to Newton’s second law, the object must move with an acceleration.
a = \(\frac{F}{m}\)
Now, as the body remains at rest, it implies that an opposing force equal to the applied force must have come into play resulting in zero net force on the object. This force is called static friction. It is denoted by Fs.

Thus, static friction is the opposing force that comes into play when one body tends to move over the surface of another, but the actual motion has not started.

  • The static friction depends upon the nature of surfaces of the two bodies in contact.
  • The static friction does not exist by itself. When there is no applied force, there is no static friction. It comes into play only when the applied force tends to move the body.
  • he static friction is a self-adjusting force.

Limiting Friction : If the applied force is increased, the force of static friction also increases. If the applied force exceeds a certain (maximum) value, the body starts moving. This maximum value of static friction up to which the body does not move, is called limiting friction.

1. The magnitude of limiting- friction between any two bodies in contact is directly proportional to the normal reaction between them.
(fs)max ∝ N
(fs) max ∝ μs N ………….. (1)
Where the constant of proportionality p s is called the coefficient of static friction. Its value depends upon the nature of surfaces of the two bodies in contact that means whether dry or wet; rough or smooth; polished or non polished. For example, when two polished metal surfaces are in contact, ps = 0.2, when these surface are lubricated, μs = 0.1
The value of μs lies between 0 and 1 or 0 < μs < 1
∵ N = Mg
(Fs)max = μs mg ……………… (2)

2. Direction of the force of limiting friction is always opposite to the direction in which one body is at the verge of moving over the other.

3. Coefficient of Static Friction :
μs is called coefficient of static friction and is defined as the ratio of force of limiting friction and normal reaction
From(1), μs = \(\frac{\left(f_{s}\right)_{\max }}{N}\)
Dimension: [M0L0T0]
Unit: It has no unit.
Value of μs does not depend upon apparent area of contact.

2. Kinetic Friction : We know that when the applied force on a body is small, it may not move but as the applied force becomes greater than the force of limiting friction, the body is set into motion. The force of friction acting between the two surfaces in contact which are moving relatively, so as to oppose their motion, is known as kinetic frictional force.
1. Kinetic friction is directly proportional to the normal reaction i. e.,
fk ∝ N
or fk ∝ μkN ………….. (1)
Where μk, is called the coefficient of kinetic friction.
∵ N – mg
∴ fk = μkmg …………… (2)
2. Value of μk depends upon the nature of surface in contact.
3. Kinetic friction is always lesser than the limiting friction
fk < (fs)max
∴ μk < μs
i. e., Coefficient of kinetic friction is always less than coefficient of static friction. Thus we require more force to start a motion than to maintain it against friction. This is because once the motion starts actually; inertia of rest has been overcomed. Also when motion has actually started, irregularities of one surface have little time to get locked again into the irregularities of the other surface.
4. Kinetic friction does not depend upon the velocity of the body.

Question 9.
Explain the circular motion of an object in horizontal plane and determine the formula for its time period.
Answer:
Motion in a Horizontal Plane
Figure 4.19 shows a mass m tied to an end of a string of length L. The mass m moves in a horizontal plane with a constant speed. As the mass moves in the circle, the string sweeps a cone of an angle θ with the surface where θ is the angle made by the string with the normal. The forces that act on the mass m at any instant are shown in figure 4.19. If T is the tension in the string, then the components of T will be T cosθ and T sinθ. There is no vertical acceleration on m. So, the component T cos θ balances the weight W of m. This way
RBSE Solutions for Class 11 Physics Chapter 4 Laws of Motion 18
RBSE Solutions for Class 11 Physics Chapter 4 Laws of Motion 19
Here, the value of θ cannot be 90°, because then x will be zero or v = ∞.
The maximum value of τ will be
τmax = \(2 \pi \sqrt{\frac{L}{g}}\)
It is possible for very small angle (θ ≈ 0°) so that cosθ = cos 0° = 1.

Equation (5) represents the formula for the time period of a simple pendulum (we will study in chapter 8). With this similarty, the above device is called conical pendulum.

Question 10.
Explain the circular motion of an object in vertical plane. Find out the formula for the tension produced at the highest and lowest points in the string.
Answer:
Motion in a Vertical Plane Consider a body of mass m tied at the end of a string and whirled in a vertical circle of radius r. Let iq & v2 be velocities of the body and T1and T2 be tensions in the string at the lowest point A and the highest point B respectively. The velocities of the body at points A and B will be directed along tangents to the circular path at these points while tensions in the string will always act towards the fixed point O as shown in figure 4.20. At the lowest point A, a part of tension T, balances the weight of the body and the remaining part provides the necessary centripetal force. Therefore.
T1 – mg = \(\frac{m v_{1}^{2}}{r}\) ……….. (1)
At the highest point, the tension in the string and the weight of the body together provide the necessary centripetal force.
RBSE Solutions for Class 11 Physics Chapter 4 Laws of Motion 20
Let us now find the out minimum velocity, the body should possess at the lowest point, so that the string does not slack when it is at the highest point. The body is then said to just loop the vertical circle. It is obvious that the velocity at the lowest point will be minimum. When the velocity at the highest point is also minimum.

Minimum velocity at the highest point :
From equation (2), it follows that the velocity at the highest point will be minimum when the tension in the string at the highest point is zero.
T2 = 0 . ………… (3)
In that case, the whole of the centripetal force will be provided by the weight of body. Therefore in such a case, the equation (2) becomes :
0 + mg = \(\frac{m v_{2}^{2}}{r}\)
or v2 = \(\sqrt{g r}\) …………… (4)
This is the mininum velocity the body should possess at the top, so that it can just loop the vertical circle without the slackening of the string. In case the velocity of the body at point B is less than \(\sqrt{g r}\). The string will slack and the body will not loop the circle. Therefore, a body will just loop the vertical circle if it possesses velocity equal to \(\sqrt{g r}\) at the top.

Minimum velocity at the lowest point : According to the principle of conservation of energy.
K.E. of the body at point A = (P.E. + K.E) of the body at point B
\(\frac{1}{2} m v_{1}^{2}=m g(2 r)+\frac{1}{2} m v_{2}^{2}\)
or \(v_{1}^{2}=4 g r+v_{2}^{2}\)
As said earlier, when the velocity at the highest point is minimum. The velocity at the lowest point will also be minimum.
Setting v2 = \(\sqrt{g r}\), we have
\(v_{1}^{2}\) = 4gr + gr
or v1 = \(\sqrt{5 g r}\) ……….. (5)
’The equation (5) gives the magnitude of the velocity at the lowest point with which body can safely go round the vertical circle of radius r or can loop the circle of radius r. Let us find out the tension in the string at the lowest point in such a case.
In equation (1),
setting v1 = \(\sqrt{5 g r}\), we have
T1 – mg = \(\frac{m \times(\sqrt{5 g r})^{2}}{r}\)
or T1 – mg = 5mg
or T1 = 6mg

Question 11.
Explain the motion of a vehicle on a circular path and give the formula for maximum velocity.
Answer:
Motion of Vehicle on a Plane on a  Circular Path
A circular turn on a level road : Consider a car of weight ‘mg’ going around on a circular turn of radius ‘r’ with velocity nona level road as shown in the figure 4.21. While rounding the curve, the wheels of the vehicle have a tedency to leave the curved path and regain the straight line path.
RBSE Solutions for Class 11 Physics Chapter 4 Laws of Motion 21
Force of friction between the wheel and the road opposes this tendency of the wheel. Therefore, this frictional force acts towards the centre of the circular path and provides the necessary centripetal force.
Three forces are acting on the car Figure 4.21.
RBSE Solutions for Class 11 Physics Chapter 4 Laws of Motion 22
(i) The weight of the car, mg, acting vertically downwards,
(ii) Normal reaction N of the road on the car, acting vertically upwards,
(iii) Frictional force F, along the surface of the road, towards the centre of the turn,
As there is no acceleration in the vertical direction,
N – mg = 0
or N = mg ……….. (i)
Since, for safe driving of a car, on the circular path, the centripetal force must be equal to or less than friction force.
RBSE Solutions for Class 11 Physics Chapter 4 Laws of Motion 41
Here, μ is coefficient of static friction between the tyres and the road.
Hence, the maximum velocity with which a vehicle can go round a level curve; without slidding is
v = \(\sqrt{\mu r g}\)
The value of v depends upon :
(i) Radius r of the curve.
(ii) Coefficient of friction (μ) between the tyre and the road.
Clearly v is independent of the mass of the car

Question 12.
Why is a road banked on a circular turn? Obtain the maximum velocity of a vehicle on such a turn. If we assume that the friction on road is negligible, then obtain the banking angle.
Answer:
Banking of roads : The value of maximum velocity for a vehicle to take a circular turn (without skidding) on a level road is quite low. This limiting value of the velocity decreases further due to decrease in the value of the coefficient of friction p on a slippery road and for a vehicle, whose tyres have worn out. Therefore especially in hilly areas, where the vehicle has to move constantly along the curved tracks, the maximum speed at which it can be run, will be very low. If any attempt is made to run it at greater speed, the vehicle is likely to skid and go off the track. In order that the vehicle can go round the curved tracks at reasonable speed without skidding, the sufficient centripetal force is managed for it by raising the outer edge of the track a little above the inner one. It is called the banking of the circular tracks.

Consider a vehicle of weight ‘mg’ moving round a curved path of radius r with speed v on a road banked through angle θ. If OA is banked road and OX is a horizontal line, then ∠AOX = θ is called angle of banking, (figure 4.22).
RBSE Solutions for Class 11 Physics Chapter 4 Laws of Motion 23
RBSE Solutions for Class 11 Physics Chapter 4 Laws of Motion 24
The vehicle is under the action of following forces :
(i) Weight ‘mg’ of the vehicle acting vertically downwards.
(ii) Normal reaction W of the ground to the vehicle acting along normal to the banked road OA in the upward direction.
(iii) Force of friction F’ between the banked road and the tyres, acting along OA
‘N’ can be resolved into two rectangular components :
(a) N cosθ, along vertically upward direction.
(b) N sinθ, along the horizontal, towards the centra of the curved round.
F’ can also be resolved into two rectangular components :
(a) Fsinθ, along vertically downward direction
(b) Fcosθ, along the horizontal, towards the centre of curved road.
As there is no acceleration along the vertical direction, the net force along this direction must be zero.
Therefore, N cosθ = mg + Fsinθ ………… (1)
The horizontal component N sinθ and Fcosθ will provide the necessary centripetal force to the vehicle,
Thus N sinθ + F cosθ = \(\frac{m v^{2}}{r}\) …………….. (2)
But F ≤ μsN, where μs is coefficient of static friction between the banked road and tyres. To obtain vmax, we Put F = μsN in equation (1) and (2)
∴ Ncosθ = mg + μsNsinθ …………. (3)
N sinθ + μsN cosθ = \(\frac{m v^{2}}{r}\) ……………….. (4)
From equation (3)
N (cosθ – μs sinθ) = mg
⇒ N = \(\frac{m g}{\left(\cos \theta-\mu_{s} \sin \theta\right)}\) …………… (5)
From equation (4) N (sinθ – μs cosθ) = \(\frac{m v^{2}}{r}\)
RBSE Solutions for Class 11 Physics Chapter 4 Laws of Motion 25
Equation (6) represents the maximum velocity of vehicle on a banked road.

Discussion :
(i) If μs = 0 i. e., if banked road is perfectly smooth, then from equation (6)
vmax = (rgtanθ)1/2 ………… (7)
This is the speed at which a banked road can be rounded even when there is no friction. Driving at this speed on banked road will cause no wear and tear of the tyres.
From equation (7) \(v_{\max }^{2}\) = rgtanθ
or tan = \(\frac{v_{\max }^{2}}{r g}\) …………….. (8)
(ii) If h is the height AB of outer edge of the road above the inner edge and l = OA is breadth of the road then from the figure 4.23.
RBSE Solutions for Class 11 Physics Chapter 4 Laws of Motion 26
Roads are usually banked for the average speed of vehicles passing over them. However, if the speed of a vehicle is some what less or more than this, the self adjusting static friction will operate between the tyres and the road, and the vehicle will not skid.

On the same basis, curved railway tracks are also banked. The level of outer rail is raised a little above the level of inner rail, while lying a curved railway track.

Question 13.
Explain the motion of a body on an inclined plane.
Answer:
Motion on an Inclined Plane
Consider an inclined plane as shown in the figure (4.24) that makes an angle 0 with the horizontal plane OA Assume an object of mass m moving on this plane. Initially, let us suppose that the object does not move due to friction. Various forces act on the object.
(i) Weight of the object, W = mg
(ii) Normal reaction, N
(iii) Frictional force, fs
In the vertical direction, there is no acceleration. Balancing the components of weight, W = mg,
RBSE Solutions for Class 11 Physics Chapter 4 Laws of Motion 27
mg sinθ = fs …………. (1)
and, mg cosθ = N …………. (2)
Now, if we move the plane OB upwards so that 0 increases, then at a position, the object will start moving on the inclined plane OR At this postion
(fs)max = μsN
Where, μs is the coefficient of static friction between the object and the plane. In this position, let θ = θs (maximum value of the angle) and (fs)max are placed in the above expressions,
mg sinθs = μsN
mg cosθs = N
From above equations,
tanθs = μs ……….. (3)
From equation (3), we can determine the coefficient of static friction between the inclined plane and the surface of the object.
When the value of θ is slightly more than θs, then a small net force acts on the object and object starts to slide. We further increase the value of θ so that the object moves downward with accelerated motion.

Then, we keep on decreasing the value of θ so that the object moves with a constant velocity at θ = θk,. At this position, kinetic frictional force acts on the object.
Substituting fk in place of fs, μk in place of μs and θk in place of θs in equation (1 & 2), we get the following equations :
mg sinθk = fk ………….. (4)
mg cosθk = N ………….. (5)
Again, putting fk = μkN in above expressions,
tanθk = μk ……………. (6)
From equation (6), we can determine the coefficient of kinetic friction between the inclined plane and the surface of the object.
Knowing the distances of the horizontal plane OA and inclined plane OR we can find tanθs and tanθk using the relation, tanθ = \(\frac{A B}{O A}=\frac{\sqrt{O B^{2}-O A^{2}}}{O A}\) this way, we can find out the coefficient of friction.
Note that BA is perpendicular to OA.

Question 14.
Distinguish between the inertial and non-inertial frames of reference. Is Earth an inertial frame of reference? Explain.
Answer:
Inertial and Non-inertial Frames of Reference ;
Frame of Reference
Motion of a body is always described with reference to some well defined coordinate system. This coordinate system is referred to as ‘frame of reference’.

In three dimensional space a frame of reference consists of three mutually perpendicular lines called ‘axis’ or ‘frame of reference’ meeting at a single point or origin. The coordinates of the origin are O (0, 0, 0) and that of any other point ‘P in space are P(x, y, z). The line joining the points O and P is called the position vector of the point P with respect to ‘O’.
RBSE Solutions for Class 11 Physics Chapter 4 Laws of Motion 28
Inertial Frame of Reference
A frame of reference that remains at rest or moves with a constant velocity with respect to the other frame of reference is called ‘Inertial Frame of Reference’. An inertial frame of reference is actually an unaccelerated frame of reference. Newton’s law of motion are valid in all inertial frames of reference. In this frame of reference, a body is not acted upon by external forces. All inertial frames of reference are equivalent for the measurement of physical phenomena.
Examples :
1. Our Earth.
2. A space shuttle moving with a constant velocity relative to the earth.
3. A rocket moving with a constant velocity relative, to the Earth.

Non-inertial Frame of Reference
A frame of reference is said to be a non-inertial frame of reference when a body, not acted upon by an external force, is accelerated. In non-inertial frame of reference, Newton’s law of motion are not valid unless we introduce a force, called pseudo force.
For example : A freely falling elevator may be taken as a non-inertial frame.

Is Earth an Inertial-Frame of Reference?
The Earth rotates around its axis and also revolves around the Sun. In both these motions, centripetal acceleration is present. Therefore, strictly speaking the Earth or any frame of reference fixed on Earth can not be taken as an inertial frame. However, as we are dealing with speed ≈× 108m/s, (speed of light) and speed of Earth is only about 3 × 104 m/s, therefore when small time intervals are involved, effect of rotation and revolution of Earth can be ignored. Further more, this speed of the Earth can be assumed to be constant. Hence, the Earth or any other frame of reference set up on the Earth can be taken as approximately inertial frame of reference.

On the contrary, a frame of reference which is accelerated is non-inertial frame.
Other examples of inertial frames of reference are :

  • A frame of reference remaining fixed w.r.t. Stars.
  • A spaceship moving in outer space, without spinning and with its engine cut off.

RBSE Class 11 Physics Chapter 4 Numerical Questions

Question 1.
Determine the maximum acceleration of the train in which a box lying on its floor will remain stationary. It is given that the coefficient of static friction between the box and the floor of the train is 0.13. (g = 9.8m/s2)
Solution:
Given ; μs = 0.13; g = 9.8m/s2; amax =?
The frictional force acting between the floor of the train and the surface of the box, will oppose the slipping of the box on the floor of the train.
∴ Limiting friction force
fs = mamax
and fs = μsN = μsmg
RBSE Solutions for Class 11 Physics Chapter 4 Laws of Motion 29
∴ m.amax = μsmg
or amax = μsg = 0.13 × 9.8
or amax = 1.274 ms-2
or amax = L27 ms-2

Question 2.
A cyclist speeding at 18 km/h on a level road takes a sharp circular turn of radius 3 m without reducing the speed. The coefficient of static friction between the tyres and the road is 0. 1. Will the cyclist slip while taking the turn?
Solution:
Given; v = 18km/h = 18 × \(\frac{5}{18}\) = 5 m/s,
r = 3m
μs = 0.1; g = 9.8 m/s2
The limiting friction available to cyclist,
fs = μsN = μsmg
Required centripetal force,
RBSE Solutions for Class 11 Physics Chapter 4 Laws of Motion 30
Therefore, cyclist will slip on the turn and fall down.

Question 3.
A bomb at rest explodes into three fragments. Two fragments fly off at right angles to each other. These two fragments: the one with mass 2 kg moves with a speed of 12 m/s, and the other one with mass 1 kg moves with 8 m/s. The speed of the third fragment is 20 m/s, then calculate its mass.
Solution:
Given; m1 = 2 kg; v1 = 12 m/s; m2 = 1kg;
v2 = 8 m/s; m3 = ?; v3 = 20 m/s
RBSE Solutions for Class 11 Physics Chapter 4 Laws of Motion 31
The momentum of all the three fragments,
P1 = m1v1 = 2 × 12 = 24 kg m/s
p2 = m2v2 = 1 × 8 = 8 kg m/s
p3 = m3v3 = m3 × 20
or p3 = 20 m3 kg m/s
RBSE Solutions for Class 11 Physics Chapter 4 Laws of Motion 32
According, to the conversation of linear momentum,
RBSE Solutions for Class 11 Physics Chapter 4 Laws of Motion 33

Question 4.
Two masses 8 kg and 12 kg are connected at the two ends of a light inextensible string that goes over a frictionless pulley. Find out the acceleration of the masses and the tension in the string when the masses are released.
Solution:
Given; m1 = 12 kg; m2 = 8kg
RBSE Solutions for Class 11 Physics Chapter 4 Laws of Motion 34
When the masses are released, mass being heavier will move downwards and m2 will move upwards with the same acceleration.
Now equation of motion for mass m1,
m1g – T = m1a ……………. (1)
and that for m2,
T – m2g = m2a …………….. (2)
On adding equation (1) & (2), we get
RBSE Solutions for Class 11 Physics Chapter 4 Laws of Motion 35
or a = 1.96 m/s 2
≈ 2m/s2
From equation (2),
T2 = m2g + m2a
= m2 (g + a)
= 8 (10 + 2)
= 8 × 12
= 96N

Question 5.
A batsman deflects a ball by an angle of 45° without changing its initial speed which is equal to 54 km/h. What is the impulse imparted to the ball when the mass of the ball is 0.15 kg?
Solution:
Given; v = 54 km/h
or v = 54 × \(\frac{5}{18}\) = 15 m/s
RBSE Solutions for Class 11 Physics Chapter 4 Laws of Motion 36
As is clear from the adjoining diagram, that there is no change in momentum in Y-direction,
Therefore, Δpy = 0
∴ Iy = 0
Change in momentum in X-direction
Δpx = p2 – p1 = -mv cos 22.5° – mv cos 22.5°
or Δpx = -2 mv cos 22.5°
∴ Impulse, Ix = Δpx
(From – Impulse-momentum theorem).
or Ix = -2mv cos 22.5°
= -2 × 0.15 × 15 × 0.9239
= -4.15755
= – 4.2 N.s.
Total impulse,
I = Ix + Iy
= -4.2 + 0 = – 4.2 N.s.
or I = – 4.2 N.s.
Here, the negative sign indicates that this impulse is acting in -X direction.
∴ Magnitude of impulse I = 4.2 N.s.

Question 6.
A constant retarding force of 50 N is applied to a body of mass 20 kg moving initially with a speed of 15 m/s. How long does the body takes to stop?
Solution:
Given; u = 15m/s; v = 0; t = ?
F = 50N, m = 20kg
∴ Retardation,
a = \(\frac{F}{m}=\frac{50}{m}=\frac{50}{20}\) = 2.5 ms-2
Now applying the equation, v = u + at
0 = 15 – 2.5t ⇒ 2.5t = 15
∴ t = \(\frac{15}{2.5}\) = 6 sec
or t = 6 sec

Question 7.
A shell of mass 0.020 kg is fired by a gun of mass 100 kg. If the muzzle speed of the shell is 80 m/s, then what is the recoil speed of the gun?
Solution:
Given; Mass of the gun M = 100kg; mass of shell m = 0.02 kg; speed of the shell v = 80 m/ s; speed of the gun v = ?
Applying the law of conservation of momentum,
Total momentum after fire = Total momentum before fire
RBSE Solutions for Class 11 Physics Chapter 4 Laws of Motion 37

Question 8.
The motion of a particle of mass 0.1 kg is described by y = \(0.3 t+\frac{9.8}{2} t^{2}\). Find out the force acting on the particle.
Solution:
Given; m = 0.1kg; y = \(0.3 t+\frac{9.8}{2} t^{2}\); F = ?
∴ Velocity v = \(\frac{d y}{d t}=\frac{d}{d t}\left(0.3 t+\frac{9.8}{2} \cdot t^{2}\right)\)
or v = 0.3 + \(\frac{9.8}{2}\) × 2t
or v = 0.3 + 9.8t
Now, the acceleration,
a = \(\frac{d v}{d t}=\frac{d}{d t}(0.3+9.8 t)\) = 0 + 9.8
or a = 9.8 m/ s2
Therefore, the force acting on the particle,
F = ma = 0.1 × 9.8
or F = 0.98 N

Question 9.
A stone of mass 0.25 kg tied to the end of a string is whirled round in a circle of radius 1.5 m with a speed of 40 rev/min in a horizontal plane. What is the tension in the string? What is the maximum speed with which the stone can be whirled around if the string can withstand a maximum tension of 200 N?
Solution:
Given; m = 0.25kg; r= 1.5 m; Tmax = 200N;
RBSE Solutions for Class 11 Physics Chapter 4 Laws of Motion 38
RBSE Solutions for Class 11 Physics Chapter 4 Laws of Motion 39

Question 10.
A body of mass 5 kg is acted upon by two perpendicular forces 8N and 6N. Give the magnitude and direction of acceleration of the body.
Solution:
Given; m = 5 kg; F1 = 8N; F2 = 6N;
a = ?; θ = 90°
RBSE Solutions for Class 11 Physics Chapter 4 Laws of Motion 40

RBSE Solutions for Class 11 Physics

RBSE Solutions for Class 11 Physics Chapter 13 Thermodynamics

May 25, 2022 by Fazal Leave a Comment

RBSE Solutions for Class 11 Physics Chapter 12 Thermal Properties

Rajasthan Board RBSE Class 11 Physics Chapter 13 Thermodynamics

RBSE Class 11 Physics Chapter 13 Textbook Exercises With Solutions

RBSE Class 11 Physics Chapter 13 Very Short Answer Type Questions

Question 1.
If two systems A and B are in thermal equilibrium with C, then will A and B also be in thermal equilibrium?
Answer:
Yes; according to Zeroth law of Thermodynamics.

Question 2.
On which law is the first law of thermodynamics based on?
Answer:
Law of conservation of energy.

Specific heat calculator is the amount of heat per unit mass required to raise the temperature by one degree Celsius.

Question 3.
Write the Mayer’s relation.
Answer:
Mayer’s relation ⇒ Cp – Cv = R
where R = universal gas constant
Cp = molar specific heat at constant pressure
Cv = molar specific heat at constant volume

Question 4.
Write the P-V relation for adiabatic expansion of ideal gas.
Solution:
PVγ = constant, where γ = \(\frac { { C }_{ p } }{ { C }_{ v } } \) which is known as atomicity of gas

Question 5.
What are the dimensions of efficiency of heat engine?
Answer:
Efficiency of heat engine,
RBSE Solutions for Class 11 Physics Chapter 13 Thermodynamics
Q1 = heat taken out from source.
∵ W and Q have same unit,
∴ efficiency η is dimensionless.
∴ Dimensional formula for η
= [M0L0T0]

Question 6.
What is the effect on pressure on change of state in a system in an isobaric process?
Answer:
Answer:
In an isobaric process, the pressure remains constant hence the change in pressure ∆P = 0.

Question 7.
Can heat be supplied without increasing the temperature of gas?
Answer:
Yes, when the gas is allowed to do the work on the surroundings.

Question 8.
Which thermodynamical variable is defined by the Zeroth law of Thermodynamics?
Answer:
Temperature.

Question 9.
What is the specific heat of a gas in isothermal and adiabatic process?
Answer:
RBSE Solutions for Class 11 Physics Chapter 13 Thermodynamics
In isothermal process ∆T = 0 ∴s = ∞ (infinity)
and in adiabatic process Q = 0 ∴ s = 0 (zero)

Question 10.
What type of process is a Carnot cyclic?
Answer:
Reversible cyclic process.

Question 11.
On what factors does the efficiency of a Carnot engine depend?
Answer:
Efficiency of a carnot heat engine,
RBSE Solutions for Class 11 Physics Chapter 13 Thermodynamics
∴ It depends on the temperatures of source and sink.

RBSE Class 11 Physics Chapter 13 Short Answer Type Questions

Question 1.
Reversibility is the basis of an ideal engine. Briefly explain the statement.
Answer:
In a reversible process, the system and its surroundings return to the original states if the forward cycle is followed by the reverse cycle. Carnot engine is a theoretical thermodynamical cycle proposed by Leonard Carnot. It gives the estimate of the maximum possible efficiency that a heat engine during the conversion process of heat into work and conversely, working between two reservoirs can possess. Thus we can say that reversibility is the basis of an ideal engine.

Question 2.
Explain the efficiency of heat engine.
Answer:
For a heat eng`ine, the total thermodynamical process is generally irreversible. The ratio of work done to convert heat and the heat absorbed by the source is called efficiency.
RBSE Solutions for Class 11 Physics Chapter 13 Thermodynamics
From the above relation, it is clear that the efficiency of engine depends upon the temperature of source (T1) and sink (T2).
Here T1 > (T1 – T2). Efficiency is always less than one, i. e., less than 100%.
If T1 = T2, then η = 0, which represents that when source and sink are at same temperature. Then, in this situation, heat energy can not be transformed to mechanical energy. It is to be noted here that the value of η will be 100% in two situations : if T1 = ∞ or T2 = 0, and both these situations are not possible. So, efficiency can not be 100%.

Question 3.
What is Thermodynamics? Explain the Zeroth law of Thermodynamics and give its importance.
Answer:
Principle : It is based on the principle of temperature.

To understand this law, consider two systems A and B, separated by an adiabatic wall, each of these are in contact with a third system C by a conducting wall. The states of the system will change until A and B are in equilibrium with C. Now, let the adiabatic wall between A and B be replaced by a conducting wall and C is insulated from A and B by an adiabatic wall. It is found that A and B are in thermal equilibrium with each other. This is the basis of Zeroth law of thermodynamics. This law states that the two systems are in thermal equilibrium with a third system separately, then both are in thermal equilibrium with each other.

The zeroth law apparently suggests that when two systems A and B are in thermal equilibrium, then, there must be a physical quantity that has the same value for both. The thermodynamic variable whose value is equal for the two systems (in thermal equilibrium) is called temperature (T). If A and B are separately in equilibrium with C, then, TA = TC and TB = TC – This means that TA = TB. So, the systems A and B are in thermal equilibrium.
RBSE Solutions for Class 11 Physics Chapter 13 Thermodynamics
RBSE Solutions for Class 11 Physics Chapter 13 Thermodynamics

Question 4.
Explain heat, work and internal energy of a system.
Answer:
Temperature determines the direction of flow of heat when two bodies are in thermal contact. Heat flows from the body at a higher temperature to a lower temperature. The flow stops when the two bodies are in thermal equilibrium. This type of heat transfer is a non-mechanical energy transfer.

When thermal energy is transferred from an object into the atmosphere then the thermal energy of the object is taken to be negative while on giving heat to an object from the atmosphere the thermal energy of the object is taken to be positive. If the temperature of the object remains constant then due to the exchange of heat in the object the state of the object changes.

Question 5.
Explain the work done in an isothermal process.
Answer:
Essential conditions for an isothermal process:
(i) The process of compression or expansion should be infinitely slow. So as to provide a sufficient time for the exchange of heat.
(ii) The walls of the container must be perfectly conducting to allow free exchange of heat between container and surroundings.

Consider a cylinder with µ, moles of an ideal gas at isothermal expansion, then it reaches its final state (P2, V2) from initial state (P1, V1). Meanwhile, in the stage when pressure is P and change in volume is V + dV from V, then pressure (P).
RBSE Solutions for Class 11 Physics Chapter 13 Thermodynamics
RBSE Solutions for Class 11 Physics Chapter 13 Thermodynamics
Since pressure and volume are inversely proportional to each other in an isothermal process, then from equation (13.20 a), from Boyle’s law for ideal gas
RBSE Solutions for Class 11 Physics Chapter 13 Thermodynamics
The above expression shows the work done for p moles of an ideal gas in an isothermal process. State equation for isothermal process,
RBSE Solutions for Class 11 Physics Chapter 13 Thermodynamics
The slope for P.V curve in an isothermal process is negative.

Question 6.
Explain the First Law of Thermo-dynamics.
Answer:
According to figure 13.6, consider a cylinder with n moles of ideal gas made of completely insulated material in which a frictionless insulated piston is placed. Let the pressure of gas be P and the area of cross-section of piston be A If piston is moved upwards by a distance dx, i.e:, the work done by the gas on expansion of the gas,
RBSE Solutions for Class 11 Physics Chapter 13 Thermodynamics
dW = P A dx = P dV …(13.33)
Here, dV = Adx represents the increase in the volume. Now, let the gas expands adiabatically and the initial state (P1 V1 T1) becomes final state (P2, V2, T2). Then, the total work done by the gas
RBSE Solutions for Class 11 Physics Chapter 13 Thermodynamics
The above expression shows that work done by p moles of an ideal gas in adiabatic process. It is clear that if work is done by the gas, then the value of Wadiabatic is positive and it is possible when T1 > T2. When gas is compressed, work is done on the gas, then the value of Wadiabatic is negative and it is possible when T1 < T2, i. e., there is increase in the temperature of gas.

Question 7.
Derive the Mayer’s relation.
Answer:
Consider one mole of an ideal gas. Let dQ be the amount of heat is given to the system to raise the temperature by dT, and change in internal energy be dV. Then, from first law of Thermodynamics,
dQ = dU + PdV …(13.14)
If heat is supplied to one mole at constant volume, i.e.,
V = constant,
then, dV = 0
From equation (13.14),
dQ = dU …(13,15)
Molar specific heat (CV ) at constant volume,
RBSE Solutions for Class 11 Physics Chapter 13 Thermodynamics
From equation (13.15),
dQ = dU = Cv dT …(13.16)
If one mole of gas is supplied heat at constant pressure, i.e., from equation (13.13) of molar specific heat Cp at constant pressure,
RBSE Solutions for Class 11 Physics Chapter 13 Thermodynamics
The above expression is called Mayer’s relation where R = 8.31 J.mol-1 K-1. This relation is for one mole of gas.

Question 8.
Differentiate between reversible and irreversible process.
Answer:
Reversible Process :

  1. The process is carried out infinitesimally slowly
  2. At any stage, the equilibrium is not disturbed
  3. It takes infinite time for completion.
  4. Work obtained in this process is maximum.

Irreversible Process :

  1. It is carried out rapidly
  2. Equilibrium may exist only after the completion of the process.
  3. It takes a finite time for completion.
  4. Work obtained in this process is not maximum

Question 9.
Explain the work done by gas in an adiabatic expansion.
Answer:
Adiabatic process is that process in which temperature (T), pressure (P) and volume (V) can be changed, but there is no transfer of heat between a thermodynamical system and its surroundings. In this process, energy is transferred only as work. This process provides a rigorous conceptual basis to explain the first law of Thermodynamics.

A process in which transfer of heat is not involved in the system, so that Q = 0, is called an adiabatic process. For example, the compression of a gas within an engine’s cylinder is assumed to take place so fast that on the time scale of the process of compression, little energy of the system can be transferred out as heat.

Essential conditions for an adiabatic process:

  • The process of expansion or compression should be sudden, so that heat does not get time to get exchanged with the surroundings.
  • The walls of the container must be perfectly insulated so that there cannot be any exchange of heat between the gas and surroundings.

Question 10.
Derive the formula for efficiency of Carnot engine.
Answer:
In 1824 scientist Sadi carnot imagined a theoretical engine by which the heat taken from high temperature object when changed completely into useful work then the efficiency of the engine will be more. Carnot tried to find out if the maximum efficiency of an engine could be 100%. An ideal reversible heat engine working between two temperatures is called carnot’s engine.

Carnot Cycle : In carnot’s engine, starting from a definite state of working material passing through different reversible processes again gaining the initial position is called carnot cycle. Carnot engine has following parts :
(i) Working substance
(ii) Source
(iii) Insulating Stand
(iv) Sink
RBSE Solutions for Class 11 Physics Chapter 13 Thermodynamics
(i) Working substance: The substance by which work is done is called working substance. In carnot engine ideal gas is taken as working substance in a cylinder with a friction less piston P.

(ii) Source: This is an object of high temperature whose temperature T1K is constant. Is heat capacity should be infinite so that if the takes heat is no change in its temperature.

(iii) Insulating Stand: This is a completely non conducting object on which the working substance is kept and passed through adiabatic process.

(iv) Sink: This is a low temperature object with constant temperature T2K. It shodld also have infinite heat capacity so that on giving heat to its temperature does not change.

Question 11.
Derive the formula for the efficiency of carnot heat engine and explain it.
Answer:
For a heat engine, the total thermodynamical process is generally irreversible. The ratio of work done to convert heat and the heat absorbed by the source is called efficiency.
RBSE Solutions for Class 11 Physics Chapter 13 Thermodynamics
From the above relation, it is clear that the efficiency of engine depends upon the temperature of source (T1) and sink (T2).
Here T1 > (T1 – T2). Efficiency is always less than one, i. e., less than 100%.

If T1 = T2, then η = 0, which represents that when source and sink are at same temperature. Then, in this situation, heat energy can not be transformed to mechanical energy. It is to be noted here that the value of q will be 100% in two situations : if T1 = ∞ or T2 = 0, and both these situations are not possible. So, efficiency can not be 100%.

RBSE Class 11 Physics Chapter 13 Long Answer Type Questions

Question 1.
Explain briefly the zeroth, first and second law of Thermodynamics.
Answer:
Principle : It is based on the principle of temperature.

To understand this law, consider two systems A and B, separated by an adiabatic wall, each of these are in contact with a third system C by a conducting wall. The states of the system will change until A and B are in equilibrium with C. Now, let the adiabatic wall between A and B be replaced by a conducting wall and C is insulated from A and B by an adiabatic wall. It is found that A and B are in thermal equilibrium with each other. This is the basis of Zeroth law of thermodynamics. This law states that the two systems are in thermal equilibrium with a third system separately, then both are in thermal equilibrium with each other.

The zeroth law apparently suggests that when two systems A and B are in thermal equilibrium, then, there must be a physical quantity that has the same value for both. The thermodynamic variable whose value is equal for the two systems (in thermal equilibrium) is called temperature (T). If A and B are separately in equilibrium with C, then, TA = TC and TB = TC – This means that TA = TB. So, the systems A and B are in thermal equilibrium.
RBSE Solutions for Class 11 Physics Chapter 13 Thermodynamics
RBSE Solutions for Class 11 Physics Chapter 13 Thermodynamics
The first law of thermodynamics states the equivalence of heat and work. This law is based on the principle of conservation of energy.

The internal energy U of a system can change through two modes of energy transfer : heat and work.

Let ∆Q = Heat supplied to the system by the surroundings,
∆W = Work done by the system on the surroundings.
∆U = Change in internal energy of the system.
From the general principle of conservation of energy, i.e..
∆Q = ∆U + ∆W …(13.9)
The energy ∆Q, supplied to the system goes in partly to increase the internal energy of the system (∆U) and work (∆W). Equation (13.9) is called the first law of Thermodynamics. This is the general law of conservation of energy applied to any system in which the energy transfer from or to the surroudnings.
We can write equation (13.9) as
∆Q – ∆W = ∆U …(13.9a)

Now, the system go from an initial state to final state in a number of ways.

To change the state of a gas from (P1,V1) to (p2, V2), we first change the volume from V1 to V2, keeping the pressure constant. This means that we first go to state (P1 ,V2) Changing the pressure from P1 and P2, keeping volume constant to the state (P2,V2). We can also first keep the volume constant and then keep the pressure constant.

U is a state variable. ∆U depends only on the initial and final states and not on the path taken by the gas to .move from one to the other. However ∆Q and ∆W depends on the path in moving from initial to final states.

From equation (13.9 a), (∆Q – ∆W) is path independent for a system with ∆U = 0,
∆Q = ∆W
This implies that heat supplied to the system is completely used up by the system in doing work on the environment. Work done by the system against a constant pressure P is
∆W = P∆V
RBSE Solutions for Class 11 Physics Chapter 13 Thermodynamics
Here, ∆V is the small change in volume of gas.
Sis, equation (13.9 a) gives
∆Q = ∆U + P∆V …(13.9 b)

The second law of Thermodynamics states that the total entropy of an isolated system increases over time, or it remains same in ideal cases (where the system undergoes a reversible process or when it is in a steady state). The first law of Thermodynamics tells the basic definition of internal energy and states the law of conservation of energy. The second law of Thermodynamics deals with the direction of natural process. It states that a natural process is not reversible. For e.g., heat flows from hot to cold region, and never the reverse, unless some external work is done on the system.

In Terms of Entropy
In a reversible process, a very small increase in the entropy (dS) of a system is defined to result from a small heat transfer to a closed system divided by the common temperature (T) of the system and the surroundings that supply of the heat :
dS = \(\frac { dQ }{ T } \) (closed system)

Question 2.
Explain the various processes of thermodynamics and work done in these processes.
Answer:
Every equilibrium state of a thermodynamical system is completely described by specific values of some macroscopic variables. If in a thermodynamic system, variables like temperature (T), pressure (P), volume (V), etc. changes with time, then the process is called thermodynamical process. There are various types of thermodynamical processes :
(i) Isothermal process
(ii) Isochoric process
(iii) Isobaric process
(iv) Adiabatic process
(v) Cyclic process

(i) Isothermal process :
An isothermal process is a change of a system, in which the temperature remains constant, ∆T = O.This typically occurs when a system is in contact with an outside thermal reservoir and the change takes place slowly to allow the system to continually adjust to the reservoir’s temperature through heat exchange.

Isothermal process takes place in any type of system that regulates the temperature. In the thermodynamic analysis of chemical reactions, it is first analyzed what happens under isothermal conditions. In an isothermal process, the internal energy of an ideal gas is constant. This is due to the fact that there are no intermoleculaf forces in an ideal gas. In the isothermal compression’of a gas, there is work done on the system to decrease the volume and increase the pressure. Doing work on the gas, can increase the internal energy and increase the temperature. For constant temperature, energy must leave the system. For an ideal gas, the amount of energy entering the environment is equal to the amount of work done on the gas, as the internal energy does not vary.

(ii) Isochoric process :
If in a physical change of thermodynamic system, there is change in pressure (P) and temperature but volume remains constant, then, it is called an isochoric process. So, no work is done by the gas in this process. No work is done on the gas because V is constant. Then, dV = 0

(iii) Isobaric process :
If temperature (T) and volume (V) changes in a physical change in a thermodynamic system and pressure (P) is constant, the process is called isobaric i.e, when the state of the substance changes, then the pressure (P) is constant. For example, freezing of water, formation of steam, etc.

(iv) Adiabatic process :
Adiabatic process is that process in which changed, but there is no transfer of heat between a thermodynamical system and its surroundings. In this process, energy is transferred only as work. This process provides a rigorous conceptual basis to explain the first law of Thermodynamics.

A process in which transfer of heat is not involved in the system, so that Q = 0, is called an adiabatic process. For example, the compression of a gas within an engine’s cylinder is assumed to take place so fast that on the time scale of the process of compression, little energy of the system can be transferred out as heat.

Essential conditions for an adiabatic process :

  • The process of expansion or compression should be sudden, so that heat does not get time to get exchanged with the surroundings.
  • The walls of the container must be perfectly insulated so that there cannot be any exchange of heat between the gas and surroundings.

(v) Cyclic process :
The process of thermodynamics in which system undergoes various changes and comes back to the initial state, is called a cyclic process. In this process, dU = 0. Since, internal energy is a function of state. Now, according to the first law of Thermodynamics:
dQ = dU + dW
dQ = 0 + dW
dQ = dW
This means that the total absorbed heat in the system is equal to the work done by the system.

Question 3.
Differentiate between the isothermal and adiabatic processes and calculate the work done in these process.
Answer:
An isothermal process is a change of a system, in which the temperature remains constant, ∆T = O.This typically occurs when a system is in contact with an outside thermal reservoir and the change takes place slowly to allow the system to continually adjust to the reservoir’s temperature through heat exchange.

Isothermal process takes place in any type of system that regulates the temperature. In the thermodynamic analysis of chemical reactions, it is first analyzed what happens under isothermal conditions. In an isothermal process, the internal energy of an ideal gas is constant. This is due to the fact that there are no intermoleculaf forces in an ideal gas. In the isothermal compression’of a gas, there is work done on the system to decrease the volume and increase the pressure. Doing work on the gas, can increase the internal energy and increase the temperature. For constant temperature, energy must leave the system. For an ideal gas, the amount of energy entering the environment is equal to the amount of work done on the gas, as the internal energy does not vary.

Adiabatic process is that process in which changed, but there is no transfer of heat between a thermodynamical system and its surroundings. In this process, energy is transferred only as work. This process provides a rigorous conceptual basis to explain the first law of Thermodynamics.

A process in which transfer of heat is not involved in the system, so that Q = 0, is called an adiabatic process. For example, the compression of a gas within an engine’s cylinder is assumed to take place so fast that on the time scale of the process of compression, little energy of the system can be transferred out as heat.

Essential conditions for an adiabatic process :

  • The process of expansion or compression should be sudden, so that heat does not get time to get exchanged with the surroundings.
  • The walls of the container must be perfectly insulated so that there cannot be any exchange of heat between the gas and surroundings.

Question 4.
Write the working principle of Carnot’s reversible engine and plot the P-V curve for work in each process, and obtain the formula for efficiency.
Answer:
In 1824 scientist Sadi carnot imagined a theoretical engine by which the heat taken from high temperature object when changed completely into useful work then the efficiency of the engine will be more. Carnot tried to find out if the maximum efficiency of an engine could be 100%. An ideal reversible heat engine working between two temperatures is called carnot’s engine.

Carnot Cycle : In carnot’s engine, starting from a definite state of working material passing through different reversible processes again gaining the initial position is called carnot cycle. Carnot engine has following parts :
(i) Working substance
(ii) Source
(iii) Insulating Stand
(iv) Sink
RBSE Solutions for Class 11 Physics Chapter 13 Thermodynamics
(i) Working substance : The substance by which
work is done is called working substance. In carnot engine ideal gas is taken as working substance in a cylinder with a friction less piston P.

(ii) Source : This is an object of high temperature
whose temperature T1K is constant. Is heat capacity should be infinite so that if the takes heat is no change? in its temperature.

(iii) Stand : This is a completely non conducting object on which the working substance is kept and passed through adiabatic process.

(iv) Sink : This is a low temperature object with constant temperature T2K. It should also have infinite heat capacity so that on giving heat to its temperature does not change.

Question 5.
Write the statements of Kelvin-Planck and Clausius of second law of Thermodynamics and show the equivalence of the two statements.
Answer:
The second law of Thermodynamics states that the total entropy of an isolated system increases over time, or it remains same in ideal cases (where the system undergoes a reversible process or when it is in a steady state). The first law of Thermodynamics tells the basic definition of internal energy and states the law of conservation of energy. The second law of Thermodynamics deals with the direction of natural process. It states that a natural process is not reversible. For e.g., heat flows from hot to cold region, and never the reverse, unless some external work is done on the system.

In Terms of Entropy
In a reversible process, a very small increase in the entropy (dS) of a system is defined to result from a small heat transfer to a closed system divided by the common temperature (T) of the system and the surroundings that supply of the heat :
dS = \(\frac { dQ }{ T } \) (closed system)

Various Statements of the Planck’s Law :

1. Statement of Kelvin : According to this statement “It is not possible to make an engine in which the working substance can take heat from the source and change it completely into .work and remain unchanged itself in one complete cycle.” It is concluded from this statement that it is not possible to get work continuously from heat source. Actually the sink in necessary and it is necessary to release. Some heat to the sink. This statement is based on the principle of carnot heat engine.

2. Statement of Clausius : According to this statement, “It is an impossible process that in cyclic process the working substance transfers heat from low temperature object to high temperature object directly in the absence of external work.”

It is concluded from this statement that transfer of heat from a low temperature object to a high temperature object on its own is not possible. For this it is important that external work be done on the working substance. This statement is based on the principle of refrigerator.

Equivalence of Clausius and Kelvin Statements :

Let there be an engine violating the Kelvin statement : i.e., one that drains heat and converts it completely into work in a cyclic manner without any other result. Now, pairing it with a reverse Carnot engine.
RBSE Solutions for Class 11 Physics Chapter 13 Thermodynamics
The newly created engine (consists of two engines) transfers heat ∆Q = Q(\(\frac { 1 }{ \eta } \) – 1) from the colder region (sink) to the hotter region (source), which violates the statement of Clausius. Thus, violation of Kelvin statement means violation of Clausius statement i.e., the Clausius statement implies the Kelvin statement. Similarly, we can prove that the Kelvin Statement implies the Clausius statement and hence both the statements are equivalent.
RBSE Solutions for Class 11 Physics Chapter 13 Thermodynamics

Question 6.
Write and obtain the Carnot’s theorem.
Answer:
Carnot’s theorem states :
(a) All heat engines between two heat reservoirs are less efficient than a Carnot heat engine operating between the same reservoirs. The efficiency of Carnot (reversible) engine is maximum.
(b) The efficiency of and each every Carnot heat engine working between a pair of heat reservoirs are equal, regardless of the working substance employed.

Proof:
Figure 13.11 shows two reservoirs of different temperatures. The engine with less efficiency is being driven as a heat pump and hence it is a reversible engine. If the less efficient engine is not reversible, then the device could be built, but the expressions for work and heat flow would not be valid.
RBSE Solutions for Class 11 Physics Chapter 13 Thermodynamics
If one of the engines is irreversible, it must be the engine (I), placed so that it reverse drives the less efficient but reversible engine (R). But if the irreversible engine is more efficient than the reversible engine, then the second law of Thermodynamics is violated.

Efficiency of irreversible engine (ηI)
RBSE Solutions for Class 11 Physics Chapter 13 Thermodynamics
If the reversible engine takes the heat Q’1 from the source at temperature T1 and does work W and provides heat (Q’1 – W) to the sink. Then, the efficiency of reversible process (ηR),
RBSE Solutions for Class 11 Physics Chapter 13 Thermodynamics
According to figure 13.11, both the engines are connected so that the irreversible engine does work in the same direction and the reversible engine performs work in the opposite direction, then the reversible engine does the work as the refrigerator (working on irreversible engine). The temperature T2 takes the heat (Q’1 – W) from the sink and work W is done on it. Here at temperature T1 it gives Q’1 heat to the source. Work done (W) on the reversible engine is fulfilled by the irreversible engine; and irreversible engine and refrigerators works as an automated device.

Since, the source provides Q1 heat to the irreversible engine and takes Q’1 heat from the reversible engine. Then,

Heat taken by the source = Q’1 – Q1

Similarly, the sink takes (Q1 – W) heat from the irreversible engine and gives (Q’1 – W) heat to the reversible engine. Then, loss in heat by the sink,

= (Q’1 – W) – (Q1 – W) = Q’1 – Q1
Since, Q’1 > Q1
∴ It is a positive quantity.

This means that the automated device, in every cycle, is transferring (Q’1 – Q1) heat from sink at T2 (low temperature) to the source at T1 (high temperature), which is not possible as second law of thermodynamics, i.e., our assumption ηI > ηR is not possible. Therefore, we can say that of all the engines working between similar temperature, reversible engines have maximum efficiency.

For the second statement of Carnot’s theorem, we replace the irreversible engine with another reversible engine (R’), i.e., two reversible engines are working between similar temperatures T1 (High) and T2 (low). Now, if we consider the efficiency of R engine (ηR) to be greater than that of engine R’ (ηR’)-

Then, the result again violates the second law of Thermodynamics. This implies the transfer of heat from low temperature to high temperature without heat loss. Therefore ηR > ηR, then the result is same and violation is obtained, i. e., it must be ηR = ηR’. So, we can say that all the reversible engines working at similar temperatures will have similar efficiency.

Refrigerator : A refrigerator is a carnet’s heat engine working in the reverse direction.
RBSE Solutions for Class 11 Physics Chapter 13 Thermodynamics
In a refrigerator or heat pump, the working substance absorbs an amount of heat O12 from the sink at temperature (T2). An amount of work W is done on it by the external agency and rejects a larger quantity of heat to the source at temp T1.

Coefficient of performance : It may be defined as the ratio of the amount of heat removed (Q2) percycle to the mechanical work (W) required to be done on it :

RBSE Class 11 Physics Chapter 13 Numerical Questions

Question 1.
The efficiency of an ideal engine is 75% and it emits heat 2 × 103 W to sink at 283 K, then calculate :
(i) Temperature of source.
(ii) Work done per minute by the engine.
(iii) Absorbed heat from the source in a cycle.
Solution:
RBSE Solutions for Class 11 Physics Chapter 13 Thermodynamics

Question 2.
Calculate the efficiency of a Carnot engine working between freezing point and boiling point.
Solution:
Given, T1 = 100°C = 100 + 273 = 373K;
T2 = 0°C = 273K
RBSE Solutions for Class 11 Physics Chapter 13 Thermodynamics

Question 3.
The efficiency of a Carnot engine is 40%. If the temperature of the source is 193.6°C, then determine the temperature of the sink.
Solution:
RBSE Solutions for Class 11 Physics Chapter 13 Thermodynamics

Question 4.
A Carnot refrigerator is operated between 260 K and 400 K. It takes 600 cal heat from sink at low temperature, then calculate the heat given to source at high temperature and the work in every cycle.
Solution:
RBSE Solutions for Class 11 Physics Chapter 13 Thermodynamics
RBSE Solutions for Class 11 Physics Chapter 13 Thermodynamics

Question 5.
The efficiency of a Carnot engine at temperature 100 K and T K; and 180 K and 900 K is same, then calculate T.
Solution:
RBSE Solutions for Class 11 Physics Chapter 13 Thermodynamics

Question 6.
Two Carnot engines A and B are working in series. First engine A takes heat at 900 K and gives heat to the sink at T K. Another engine B, takes the heat from A and emits it to sink at 400 K. Calculate the temperature T in following conditions :
(i) when same work is done by both the engines.
(ii) when efficiency of both the engines is equal.
Solution:
(i) When same work is done by both the engines,
Let’s assume that engine A takes heat Q1 from the source and gives heat Q2 to the sink, then
Carnot cycle
RBSE Solutions for Class 11 Physics Chapter 13 Thermodynamics
RBSE Solutions for Class 11 Physics Chapter 13 Thermodynamics
Now, lets assume engine Stakes heat Q2 from its source (at temperature T) and gives away Q3 to its sink (at temperature T = 400K). Then again from Carnot’s cycle.
RBSE Solutions for Class 11 Physics Chapter 13 Thermodynamics
RBSE Solutions for Class 11 Physics Chapter 13 Thermodynamics

Question 7.
A gas (γ = 1.5) is compressed in an adiabatic process, then its volume changes from 1600 cm3 to 400 cm3. If the initial pressure is 150 kPa, then calculate the final pressure, and the work done on the gas.
Solution:
Given, γ = 1.5;
V1 = 1600 cm3 = 1600 × 10-6 m3 = 1.6 × 10-3 m3;
V2 = 400 cm3 = 400 × 10-6 m3 = 0.4 × 10-3 m3;
P1 = 150kPa
or P1 = 150 × 103 Pa = 1.5 × 105 Pa; P2 = ?
∵ In an adiabatic process
RBSE Solutions for Class 11 Physics Chapter 13 Thermodynamics

Question 8.
If a gas (γ = 1.5) is compressed at pressure 27 times the initial pressure, then calculate the change in temperature if the initial temperature is 27°C.
Solution:
Given, γ = 1.5; P1 = P; P2 = 27P;
T1 = 27 + 273 = 300K; T2 = ?
∴ Relation between pressure and temperature in an adiabatic process
RBSE Solutions for Class 11 Physics Chapter 13 Thermodynamics
RBSE Solutions for Class 11 Physics Chapter 13 Thermodynamics

Question 9.
A refrigerator transfers 200 J average heat per second from temperature -10°C to 27°C. Taking the cycle to be reversible, calculate the average power when there is no loss of heat.
Solution:
RBSE Solutions for Class 11 Physics Chapter 13 Thermodynamics

Question 10.
P-V curve of a cyclic process is shown below, then calculate the work done in the cyclic process.
Solution:
The given cyclic process is anticlockwise, hence work done will be positive.
RBSE Solutions for Class 11 Physics Chapter 13 Thermodynamics
RBSE Solutions for Class 11 Physics Chapter 13 Thermodynamics

Question 11.
A Carnot engine operates between 373K and 283 K. Then, calculate the efficiency and deduce that when would its efficiency be 100%.
Solution:

RBSE Solutions for Class 11 Physics Chapter 13 Thermodynamics

RBSE Solutions for Class 11 Physics

RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium

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Rajasthan Board RBSE Class 11 Chemistry Chapter 7 Equilibrium

RBSE Class 11 Chemistry Chapter 7 Text Book Questions

RBSE Class 11 Chemistry Chapter 7 Multiple Choice Questions

Question 1.
For the reaction A + 2B ➝ C, the equilibrium constant will be :
RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium 1
Answer:
RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium 2

RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium

Question 2.
For the reaction, A (soild) + 2B (Gas) ➝ 3C (solid) + 2D (gas) :
(a) Kp = Kc(RT)°
(b) Kp = Kc(R2T2)
(c) Kp = Kc(RT)
(d) Kp = Kc(R-2T-2)
Answer:
(a) Kp = Kc(RT)°

Question 3.
For the reaction N2 + 3H2 ➝ 2NH3 + xkJ, The conditions to make more ammonia are :
(a) high temperature and high pressure
(b) low temperature and high pressure
(c) low temperature and low pressure
(d) high temperature and low pressure
Answer:
(b) low temperature and high pressure

RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium

Question 4.
The solubility product of two electrolytes AB and AB2 is 1 × 10-10 . The molar conductivity of AB is …… AB2.
(a) equal to
(b) more than
(c) less than
(d) no relation with
Answer:
(c) less than

Question 5.
The pH of a solution on addition of 50 ml. water slowly to a solution
(a) 1
(b) 5
(c) 7
(d) 10
Answer:
(c) 7

RBSE Class 11 Chemistry Chapter 7 Very Short Answer Type Questions

Question 6.
Write the expression for the equlibrium constant, KC for each of the following reaction
(i) 2NOCl(g) ➝ 2NO(g) + Cl2(g)
(ii) 2CU(NO3)2 ➝ 2CuO(s) + 4NO2(g) + O2(g)
(iii) CH3COOC2H5aq + H2O(l) ➝ CH3COOH(aq) + C2H5OH(aq)
(iv) Fe3+ (aq) + 30H– (aq) ➝ Fe(OH)3(s)
(v) I2(s) + 5F2 ➝ 21F5
Answer:
RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium 3
RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium 4

RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium

Question 7.
What is chemical equilibrium ?
Answer:
A dynamic state of a chemical reaction at which the rate of the forward reaction is equal to the rate of backward reaction is called chemical equilibrium.

Question 8.
Write the examples of the reaction in which
(a) Product increases on increasing the pressure
(b) Product increases on increasing the temperature.
Answer:
RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium 5

Question 9.
What will be the effect on equilibrium when ∆n is negative and pressure is decreased ?
Answer:
When ∆n is negative and pressure is decreased then the equilibrium will be shifted to left and the yield of product will decrease.

RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium

Question 10.
What are the conditions for more yield of ammonia by Haber’s proress?
Answer:
Haber’s Process
RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium 6
Low temperature favours the exothermic reection because it releases energy. So the forward reaction is forward and the yield of ammonia will increase.

High pressure favours the reaction that decreares the number of gas molecules so the forward reaction is favoured. The equilibrium will shift to the right and the yield of ammonia will increase. .

Question 11.
Which states of a substance are at equilibrium at its melting point?
Answer:
RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium 7

Question 12.
Which states of a substance are at equilibrium at its boiling point?
Answer:
RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium 8

RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium

Question 13.
If the degree of dissociation of PCl3 and Cl2 is x, them how many moles of PCl5 is present at equilibrium?
Answer:
RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium 9
Moles
No of moles at (1 – x) x x equilibrium

Question 14.
Write the relation for Kww, kα and Kh for KCN salt.
Answer:
RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium 10

Question 15.
Write the conjugate acid of NH2.
Answer:
NH3

RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium

Question 16.
Write the conjugate base of HCO3
Answer:
\(\mathrm{CO}_{3}^{2-}\)

Question 17.
Calculate the pH of 0.001 N HCl.
Answer:
pH = – log10 [H+] = – log10 [10-3]
= 3log10 = 3 × 1 = 3

Question 18.
What will be the effect of presence of HCl on dissociation of H2S?
Answer:
The dissociation of H2S is suppressed by the presence of HCl due to common ion effect.

Question 19.
What are reversible and irreversible reactions?
Answer:
Reversible reaction : A reaction which takes place in both forward and backward direction is called reversible reaction.

Irreversible reaction: A reaction in which entire amount of reactant is changed into product and no reaction from product side occurs in called irreversible reaction.

RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium

Question 20.
For the following equilibrium, Kc = 63 × 1014 at 1000K
RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium 11
Both the forward and backward reactions in the equilibrium are elementary bimolecular reactions. What is Kc, for the backward reaction?
Answer:
NO(g) + O3 (g) ➝ NO2 (g) + O2 (g)
Kc = 6.3 × 1014 for forward reaction
For backward reaction,
RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium 12

RBSE Class 11 Chemistry Chapter 7 Short Answer Type Questions

Question 21.
Which of the following reactions will get affected by increasing the pressure ? Also, mention whether change will cause the reaction to go into forward or backward direction.
(i) COCl2 ➝ CO(g) + Cl2(g)
(ii) CH4 (g) + 2S2 (g) ➝ CS2 (g) + 2H2S(g)
(iii) CO2(g) + C(s) ➝ 2CO(g)
(iv) 2H2 (g) + CO(g) ➝ CH3OH(g)
(v) CaCO3 (s) ➝ CaO(s) + CO2 (g)
(vi) 4NH3 (g) + SO2 (g) ➝ 4NO(g) + 6H2O (g)
Answer:
(i) COCl2 ➝ CO(g) + Cl2(g)
If pressure increases, the reaction will go into backward direction.
(ii) CH4 (g) + 2S2 (g) ➝ CS2 (g) + 2H2S(g)
If pressure increases, the reaction will go into backward direction.
(iii) CO2(g) + C(s) ➝ 2CO(g)
If the pressure increases, the reaction will go into backward direction.
(iv) 2H2 (g) + CO(g) ➝ CH3OH(g)
If pressure increases, the reaction will go in to forward direction.
(v) CaCO3 (s) ➝ CaO(s) + CO2 (g)
If pressure increases, the reaction will go in to backward direction.
(vi) 4NH3 (g) + SO2 (g) ➝ 4NO(g) + 6H2O (g)
If pressure increases, the reaction will go into backward direction.

RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium

Question 22.
Describe the factors which affect chemical equilibrium?
Answer:
There are following factors which affect chemical equilibrium :
(1) Concentration: If at equilibrium, the concentration of reactants is increased, then rate of forward reaction increases and more amount of product is obtained. If there is increase in concentration of products, then rate of backward reaction increases, more amount of reactant is obtained.

(2) Pressure : The effect of pressure on equilibrium takes place when the reactants are in gaseous state and in those reactions in which number of molecules of reactants and products are different.
RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium 13
In this reaction, the forward reaction is accompanied by a decrease in the total number of moles of reactants. If the pressure of the system is increased, then the equilibrium will shift in that direction where pressure decreases i. e., decrease in number of moles taken i. e., in the favour of formation of ammonia.

If pressure decreases, the equilibrium will shift in the direction where pressure increases i.e., increase in number of moles occurs i. e., in backward direction. So, a decrease in pressure will fovour dissociation of NH3 to form N2 and H2.

When there is no change in number of molecules of reactants and products in a reaction, then there is no effect of pressure on it.

(3) Temperature:

(4) Catalyst : A catalyst increases the rate of the chemical reaction by making available a new low energy pathway for the conversion of reactants to products. It increases the rate of forward and reverse reactions that pass through the same transition state and does not affect equilibrium. Catalyst lowers the activation energy for the forward and backward reactions by exactly the same amount. Catalyst does not affect the equilibrium composition of a reaction mixture. It does not appear in the balanced chemical equation or in the equilibrium constant expression.

RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium

Question 23.
Under what conditions the synthesis of SO3 is more ? Explain.
Answer:
RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium 14
The conditions for more synthesis of SO3 are given as below:
(i) Increase in concentration of SO2 or O2
(ii) Decerase in temperature
(iii) Increase in pressure
(iv) Removal of SO3 from reaction vessel
(i) Increase in concentration of SO2 or O2
If the concentration of SO2 or O2 will increase, then according to Le-Chatelier’s principle, the equilibrium will shift to forward direction and more SO3 will be formed.

(ii) Decrease in temperature A decrease in temperature favours the exothermic reaction so the equilibrium will shift to forward direction and more amount of SO3 will be formed.

(iii) Increase in Pressure : An increase in pressure will favour the reactions that decrease the number of gaseous molecules. The equilibrium will shift to the right and the yield of SO3 will be increased.

Question 24.
The equilibrium for the synthesis of ammonia is
RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium 15
What is the effect of pressure, temperature and concentration on this equilibrium ?
Answer:
Effect of pressure :
High pressure favours the formation of ammonia. An increase in pressure will favour the reaction that decreases the number of gas molecules. The equilibrium will shift to the right and the yield of ammoina will be increased.

Effect of temperature:
Low temperature favours the formation of ammonia, a decrease in temperature favours the exothermic reaction because it releases energy, it will favour the forward reaction and the yield of ammonia will be increased.

Effect of concentration
High concentration of reactants favours the formation of ammonia, According to Le-Chatelier’s principle, if the concentration of N2 or H2 is increased then equilibrium will shift to forward direction and the yield of ammonia will be increased.

RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium

Question 25.
Identify acid-base according to Lewis concept
S2-, H+, OH–, BF3, Ni2+ , F–
Answer:
According to Lewis concept, acids are electron pair acceptor and bases are electron pair donor.
Lewis acid : H+, BF3, Ni2+
Lewis basis : S2-, OH–, F–

Question 26.
Write the conjugate acid of the following:
S2-, NH3, H2 \(\mathrm{PO}_{4}^{-}\), CH3NH2
Answer:
RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium 16

Question 27.
Write any two factors which affect ionization.
Answer:

  • Concentration : In a solution, on increasing the concentration of electrolyte, ionization decreases. The extent of ionization of an electrolyte is inversely proportional to the concentration of its solution.
  • Temperature : The degree of ionization increases with increase in temperature.

RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium

Question 28.
In precipitation of saturated solution of common salt, HCl gas is added and not HCl acid. Explain why?
Answer:
In precipitation of saturated solution of a common salt, HCl gas is added and not HCl acid because if we add HCl acid to a saturated solution of common salt, the equilibrium of equation (1) will be shifted to left side by increasing concetration of Cl– (eq). As a result, NaCl will recrystallize in the solution.
NaCl ➝ Na+ + Cl– …(1)
HCl ➝ H+ + Cl– …(2)

Question 29.
Derive the formula for Kh and [H+] concentration for a solution of strong acid and weak base.
Answer:
Consider an example of NH4 Cl. It ionizes in water completely into \(\mathrm{NH}_{4}^{+}\) and Cl– ions. \(\mathrm{NH}_{4}^{+}\) ions react with water to form a weak base (NH4 OH) and H3O+ ions.
RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium 17
∴ H3O+ ion concentration increases and the solution becomes acidic.
Applying law of mass action,
RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium 18
RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium 19
RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium 20

Question 30.
Mg (OH)2 is soluble in NH4Cl but insoluble in NaCl. Explain why?
Answer:
Mg(OH)2 is soluble in NH4 Cl but insoluble in NaCI because the Mg2+ is less reactive than Na+ ion. Ammonium chloride, being a soluble salt, dissociates completely to form ammonium cation and chloride ions.
RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium 21
The \(\mathrm{NH}_{4}^{+}\) ions can act as an acid and neutralize the OH– ions to form ammonia and water.
RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium 22
∴ The \(\mathrm{NH}_{4}^{+}\) cations will neutralize the hydroxide anions present in solution. Due to which, the equilibrium will shift to the right to create more OH– ions. This will determine more ions to be dissolve from the solid.
Mg(OH)2 + 2NH4 Cl ➝ MgCl2 + 2NH4OH
However, magnesium ion can not displace the sodium ion since the later is more reactive than the former.
Mg(OH)2 +NaCl ➝ No reaction

RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium

Question 31.
Experimentally how it is proved that chemical equilibrium is dynamic in nature?
Answer:
RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium 23
In the above equilibrium, on adding radioactive isotope of iodine, there is no change in relative concentrations of hydrogen, iodine and hydrogen iodide. This shows that the reaction is taking place in both the directions at the same rate. The formation of radioactive reaction is taking place in forward direction but the same concentrations shows same rate of reaction in both the directions. Thus, chemical equilibrium is a dynamic equilibrium.
RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium 24

Question 32.
Establish the relation between concentration equilibrium constant KC and pressure equilibrium constant KP for the following homogeneous reaction—
4NH3 (g) + 5O2 (g) ➝ 4NO (g) + 6H2O (g)
Answer:
RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium 25
RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium 26

RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium

Question 33.
Explain law of mass Action by taking the following example,
CH3COOH + C2H2OH ➝ CH3COOC2H5 +H2O
Answer:
Law of Mass Action
According to this law, “the rate of reaction of a substance is proportional to the product of molar concentration of reactants at a constant temperature at any given time”. Molar concentration is called active mass. Active mass is the number of moles dissolved in one litre of solution.
eg. CH3COOH + C2H5OH ➝ CH3COOC2 H5 + H2O
According to law of mass action,
Rate of forward reaction a [CH3COOH] [C2H5OH]
= K1 [CH3COOH] [C2H5OH]
When K2 = rate constant for backward reaction At equilibrium,
Rate of backward reaction a [CH3COOC2H5] [H2O]
= K2 [CH3COOC2H5] [H2O]

Where K2= rate constant for backward reaction At equilibrium,
Rate of forward reaction = Rate of backward reaction K2 [CH3COOH] [C2H6OH] – K2 [CH3COOC2H5] [H2O]
RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium 27

Question 34.
For the system, N2 + O2 ➝ 2NO – 44 kcal, at equilibrium, what will be the effect of the following
(i) Increasing the temperature
(ii) Decreasing the pressure
(iii) Increasing the concentration of NO
(iv) Presence of catalyst
Answer:
For the system,
N2 + O2 ➝ 2NO – 44 kcal
At equilibrium,
(i) Increasing the temperature : An increase in temperature favours the endothermic reaction because it takes up energy so it will favour the forward reaction and the yield of NO will increase.
(ii) Decreasing the pressure : Since the number of molecules of reactants and products are same, so decrease in pressure will not affect the equilibrium.
(iii) Increasing the concentration of NO : On increasing the concentration of NO, it will favour backward reaction and the yield of NO.
(iv) Presence of catalyst : It increase the rate of reaction by lowering the activation energy.

RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium

Question 35.
What will be the effect of temperature and pressure on the following equilibrium?
N2 + 3H2 ➝ 2NH3
Answer:
Effect of temperature
Low temperature favours the forward reaction. A decrease in temperature favours the exothermic reaction because it releases energy. So, it will favour the forward reaction. The yield of ammonia will increase.

Effect of pressure
An increase in pressure will favours the reaction that decreases the number of gaseous molecules. There are fewer molecules of product then reactants so it will favour the forward reaction. The yield of ammonia will increase.

RBSE Class 11 Chemistry Chapter 7 Long Answer Type Questions

Question 36.
(i) Explain equilibrium in physical process and chemical process with examples.
(ii) Prove that degree of dissociation of PC15 is inversely proportional to square root of its pressure.
Answer:
Equilibrium in Physical Process
In a reaction, when there is change in physical state only, it is known as physical process. When equilibrium is established in a physical process, it is called physical equilibrium. The characteristics of a system of equilibrium are better understood if we examine some physical processes. The most familiar examples are phase transformation processes. For example,
Solid ➝ Liquid
Liquid ➝ Gas
Solid ➝ Gas
Solid – Liquid Equilibrium
Conversion of ice into water is an important example of solid- liquid equilibrium.
RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium 28
Ice and water kept in a perfectly insulated thermos flask (no exchange of heat between its contents and the surroundings) at 273K and the atmospheric pressure are in equilibrium state. The melting and freezing point of water is 273K. It is obvious that the ice and water are in equilibrium only at particular temperature and pressure. At equilibrium, the mass of ice and water does not change with time and the temperature remains constant as rate of transfer of molecules from ice to water and of reverse transfer from water into ice. are equal at atmospheric pressure and 273 K. The equilibrium is not static. Molecules from the liquid water collide against ice and adhere to it and some molecules of ice escape into liquid phase.

Freezing Point: For a pure substance, at atmospheric pressure, the temperature at which the solid and liquid phases are at equilibrium is called the normal melting point or normal freezing point of the substance. The system here is in dynamic equlibrium. There is change in melting point of substance on changing the pressure.

Equilibrium in Chemical Processes
The state of chemical equilibrium depends on temperature, pressure, concentration of reactants and products and presence of other substances. At equilibrium, all the above factors are constant. If there is any change in any of the above factors, then it affects the equilibrium also. As a result, rate of forward or backward reaction increases. As the equilibrium is reversible in nature, new equilibrium state is attained in the changing direction. It appears as equilibrium is shifted from one state to another. It also shows dynamic nature of equilibrium.
RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium 29
Example:
RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium 30
In the above equilibrium, on adding radioactive isotope of iodine, there is no change in relative concentrations of hydrogen, iodine and hydrogen iodide. This shows that the reaction is taking place in both the directions at the same rate. The formation of radioactive hydrogen iodide shows that reaction is taking place in forward direction but the same concentrations show same rate of reaction in both the directions. Thus, chemical equilibrium is a dynamic equilibrium.

(ii) Let us consider that one mole of PCl5 is present initially. At equilibrium, let us assume that x mole of PClx dissociates to give x mole of PCl5 and x mole of Cl2. Let the total pressure at equilibrium be P atmosphere. The number of moles of PCl5, PCl3 and Cl2 present at equilibrium can be gives as follows :
RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium 31
Totla number of moles at equilibrium,
= 1 – x + x + x
= 1 + x
We know that partial pressure is the product of mole fraction and the total pressure. Mole fraction is the number of moles of that component divided by the total number of moles in the mixture. Therefore,
RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium 32
Substituting the values of partial pressure in this expression
RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium 33
When x << 1
The value of x2 can be neglected when compared to one.
RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium 34

The Buffer Capacity is a measure of resistant a particular solution is resistant to change in pH when an acid or a base is added to it.

Question 37.
What is buffer solution ? Write any two properties of buffer solution. Derive the formula for calculating the pH for acidic buffer. Write any two examples of simple buffer.
Answer:
Buffer solution : The solution which resists change in pH on dilution or with the addition of small amounts of acid or alkali is called buffer solution.
Properties of buffer solution

  • Buffer solution resists change in pH.
  • It is used to prepare a buffer solution of particular pH.

pH of Acidic buffer
RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium 35
Examples of Simple buffere

  • CH3COONH4
  • (NH4)2CO3

RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium

Question 38.
What is solubility product? Establish the relation between solubility and solubility product for CdS type compounds. In third group analysis, NH4 Cl is added before NH4 OH, Why?
Answer:
Solubility Product: It is defined as the product of the concentration of ions in a saturated solution of an electrolyte at a given temperature. It is repersented by the symbol Ksp.

Relation between solubility and solubility product for CdS type compounds
Let x be the solubility of CdS
RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium 36
In third group analysis, NH4 Cl is added before NH4 OH because \(\mathrm{NH}_{4}^{+}\) ions furnished by NH4Cl lower the ionization of NH4 OH and hence the concentration of OH– ion. At low concentration of OH– ion, III group hydroxides precipitate.

Whereas, when NH4OH is added in the presence of NH4 Cl, the precipitation of II group hydroxides takes place.

RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium

Question 39.
Explain law of mass action taking suitable chemical reaction. Derive the relationship between Kp and Kc for a homogeneous
Answer:
Law of Mass Action:
According to this law,
“The rate of reaction of a substance is proportional to the product of molar concentration of reactants at a constant temperature at any given time.”
Molar concentration is called active mass. Active mass is the number of moles dissolved in one litre of solution. Let a reaction;
A + B ➝ C + D

According to law of mass action ;
Rate α [A] [B]
Rate = k1 [A] [B]
Here, k1 = constant of proportionality or rate constant for forward reaction.
When the concentration of [A] and [B] is unity, k1 shows rate of forward reaction
Rate of backward reaction α [C][[D]
Rate = k2 [C] [D]
Here k2 is rate constant for backward reaction.
At equilibrium,
Rate of forward reaction = Rate of backward reaction
RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium 37
Relationship between Kp and Kc for a homogeneous reaction,
RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium 38
Suppose ‘a’ moles of H2 and ‘b’ moles of I2 be heated in sealed glass bulb having volume V litres in a thermostat till equilibrium is established. If the concentration of HI formed after analysis is 2x, then according to above reaction, 2x moles of HI will be obtained from x moles of each H2 and I2.

The equilibrium concentration per litre of various reactions and products may be shown as follows :
RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium 39
RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium 40

Question 40.
What is an indicator? Explain Ostwald theory of indicator. Explain titration between HNO3 and KOH using suitable indicator with the help of a curve.
Answer:
Indicator : An indicator is a chemical substance that undergoes a colour change at the end point. The end point of an acid-base titration can be determined using acid-base indicators. Acid Base indicators are either weak organic acids or weak organic bases. They change their colour within a certain pH range. Two theories have been proposed to explain the change of colour of acid-base indicators with change in pH.

Ostwald Theory of Indicators:
According to this theory :
(a) The colour change is due to ionisation of the acid-base indicator. The unionised form has different colour than the ionised form.
(b) The ionisation of the indrcation is largely affected in acids and bases as it is either a weak acid or a weak base. In case, the indicator is a weak acid, its ionisation is very much low in acids due to common H+ ions while it is fairly ionised in alkalies. Similarly if the indicator is a weak base, its ionisation is large in acids and low in alkalies due to common OH– ions.

Considering two important indicators, Phenolphthalein (a weak acid) and Methyl orange (a weak base). Ostwald theory can be illustrated as follows:

Phenolphtalein : It can be represented as HPh. It ionises in solution to a small extent as:
RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium 41
The undissociated molecules of phenolphthalein are colourless while Ph– ions are pink in colour. In presence of an acid, the ionisation of HPh is practically negligible as the equilibrium shifts to left hand side due to high concentration of H+ ions. Thus, the solution would remain colourless. On addition of alkali, hydrogen ions are removed by OH– ions in the form of water molecules and the equilibrium shifts to right hand side. Thus, the concentration of Ph– ions increases in solution and they impart pink colour to the solution.

Titration between HNO3 (strong acid) and KOH (Strong bose)
For this titration, base is taken in a burette and acid is taken in a beaker. The pH of strong acid is very less.

Initially when base is added pH changes gradually but at end point, pH changes rapidly from 3 to 10. The curve obtained when volume of base used and pH change are plotted is known as titration curve. The range of strong acid and strong base is very large. Many indicators come in this range.

The diagram shows the pH curve for adding a strong acid to a strong base. Superimposed on it are the pH ranges for methyl orange and phenolphthalein
RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium 42
You can see that indicator does not change colour at the equivalence point.

However, the graph is so steep at that point that there will be virtually no difference in the volume of acid added whichever indicator you choose. However, it would make sense to titrate to the best possible colour with each indicator.

If you use phenolphthalein, you would titrate until it just becomes colourless (at pH 8.3) because that is as close as you can get to the equivalence point.

RBSE Class 11 Chemistry Chapter 7 Numerical Problems

Question 41.
The first ionization constant of H2S is
9.1 × 10-8. Calculate the concentratioin of HS– ions in its 0.1 M solution. How will this concentration be affected if the solution is 0.1 M in HCl also? If the second dissociation constant of H2S is 1.2 × 10-13. Calculate the concentration of S2- under both conditions.
Answer:
To calculate the concentration of HS– ion.
Case I: In case of absence of HCl
RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium 43
RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium 44
RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium 45
RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium 46
RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium 47

RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium

Question 42.
The ionization constant of propanoic acid is 1.32 × 10-5. Calcualte the degree of ionization of the acid in its 0.05 M solution and also its pH. What will be its degree of ionzation if the solution is 0.01 M HCl also?
Answer:
RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium 48
RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium 49
RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium 50

Question 43.
Calculate pH of following mixtures :
(i) 10 mL of 0.2 M Ca(OH)2 + 25 mL of 0.1 M HCl
(ii) 10 mL of 0.01 m H2SO4 + 10 mL of 0.01 m Ca(OH)2
(iii) 10 mL of 0.1 m H2SO4 + 10 mL of 0.1 M KOH
Answer:
(i) 10 mL of 0.2 m Ca(OH)2 + 25 mL of 0.1 m HCl
RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium 51
RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium 52

RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium

Question 44.
The solubility product of Ag2CrO4 and AgBr are 11 × 10-12 and 5.0 × 10-13 respectively. Calculate the ratio of the molarities of their saturated solutions.
Answer:
Let solubility of Ag2 CrO4 = S
RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium 53
RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium 54

Question 45.
The ionization constant of benzoic acid is 6.46 × 10-5 and Ksp for silver benzoate is 2.5 × 10-13. How many times is silver benzoate more soluble in a buffer of pH 3.9 compared to its solubility in pure water?
Answer:
RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium 55
RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium 56
RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium 57
Hence, silver benzoate in 3.32 times more soluble in low pH solution.

RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium

Question 46.
In a reaction, A + 2B ➝ 2C + D, A and B are heated in a flask at 25°C. The inital concentration of B is 1.5 times the concentration of A. At equilibrium, the concentration of A and D is same. Calculate equilibrium constant at this temperature.
Answer:
Let the initial concentration is 1 g mol/ L. So, the initial concentration of B is 1.5 g mol/L. If x mole of D are added, then the concentration in equilibrium can be respresented as :
RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium 58

Question 47.
At 25°C and one atmospheric pressure, 20% of N2O4 is dissociated into NO2. Calculate equilibrium constant Kp for this equilibrium.
Answer:
RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium 59

RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium

Question 48.
At normal temperature and pressure, 5.29mL hydrogen reacts with 0.040 gm iodine at 444°C and gives 6.35 mL hydrogen iodide. Calculate equilibrium constant for the synthesis of HI at this temperature.
Answer:
RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium 60

RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium

Question 49.
The molecular mass of PCl5 is 208.3. At 200°C, the mass of partial dissociated vapours is 62 times that of mass of hydrogen. Calculate degree of dissociation of PCl5.
Answer:
RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium 61

Question 50.
One mole of H2O and one mole of CO are taken in 10L vessel and heated to 725 K. At equilibrium. 40% of water (by mass) reacts with CO according to the equation,
H2O(g) + CO(g) ➝ H2 (g) + CO2 (g)
Calculate the equilibrium constant for the reaction.
Answer:
RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium 62

RBSE Solutions for Class 11 Chemistry

RBSE Solutions for Class 11 Chemistry Chapter 1 Basic Concepts of Chemistry

May 24, 2022 by Prasanna 1 Comment

Rajasthan Board RBSE Class 11 Chemistry Chapter 1 Basic Concepts of Chemistry

RBSE Class 11 Chemistry Chapter 1 Text Book Questions

RBSE Class 11 Chemistry Chapter 1 Multiple Choice Questions

Question 1.
Number of significant figures in 0.0287 is
(a) 5
(b) 2
(c) 3
(d) 4
Answer.
(c) 3

RBSE Solutions for Class 11 Chemistry Chapter 1 Basic Concepts of Chemistry

Question 2.
Molecular mass of glucose is
(a) 342 u
(b) 110 u
(c) 90 u
(d) 180 u
Answer:
(d)

Question 3.
Volume of 2 g methane at STP will be
(a) 2.8 L
(b) 5.6 L
(c) 11.2 L
(d) 22.4
Answer:
(a) 2.8 L

RBSE Solutions for Class 11 Chemistry Chapter 1 Basic Concepts of Chemistry

Question 4.
At STP, the number of molecules in 1 ml of ideal gas will be
(a) 6.023 × 1023
(b) 2.69 × 1019
(c) 2.69 × 1023
(d) 458 × 1026
Answer:
(b)

Question 5.
Which of the following has least weight?
(a) 108 g of silver
(b) 1 mol of sulphur
(c) 1 g atom of nitrogen
(d) 3.011 × 1023atoms of carbon
Answer:
(d)

RBSE Class 11 Chemistry Chapter 1 Very Short Answer Type Questions

Question 6.
What is the unit and symbol for quantity of matter in SI system?
Answer:
In SI system, the unit and symbol for quantity of matter are mole and mol respectively.

Question 7.
What is the difference between 8.0 g and,8.000 g weight?
Answer:
8.0 g has two significant figures which shows precision up to 1 decimal place where as 8.000 g, has 4 significant figures which shows precision up to 3 decimal place.

RBSE Solutions for Class 11 Chemistry Chapter 1 Basic Concepts of Chemistry

Question 8.
Express 3600 g in three significant figures.
Answer:
3.60 × 103

Moles to Grams Calculator is a free online tool that displays the conversion of moles to grams.

Question 9.
How many gram molecules of hydrogen are present in 20 g hydrogen?
Answer:
2 g of hydrogen molecules = 1 mol
1 g of hydrogen molecules = 1/2 mol
20 g of hydrogen molecules=1/2 × 20= 10 mol

Question 10.
How many molecules will be present in 64 g oxygen?
Answer:
Gram molecular weight of oxygen = 32 g
32 g of O2 contains = 6.023 × 1023 molecules
Hence, 64 g of O2 will contain 6.023 × 101023 × 64/32
= 6.023 × 1023 × 2
=12.046 × 1023 molecules

RBSE Solutions for Class 11 Chemistry Chapter 1 Basic Concepts of Chemistry

Question 11.
Calculate equivalent weight of H2SO4 Molecular mass of H2SO4 is 98.
Answer:
Equivalent weight of H2SO4 = Molecular mass / number of replaceable H+
= 98/2 = 49

Question 12.
State Avogadro’s law,
Answer:
Avogadro’s law states that “equal volumes of all gases, at the same temperature and pressure, have the same number of molecules”.

Question 13.
How many atoms of carbon will be present in 0.1 mole of C6H12O6.
Answer:
Number of moles of carbon in 1 mol of C6H12O6 = 6
Number of carbon atoms in 1 mole of C6H12O6
= 6 × 6.023× 1023 atoms
Number of carbon atoms is 0.1 mol of C6H12O6
= 0.1 × 6 × 6.023 × 1023 atoms
= 3.6138 × 1023 atoms

RBSE Solutions for Class 11 Chemistry Chapter 1 Basic Concepts of Chemistry

Question 14.
What are significant figures?
Answer:
Each of the digits of a number that are used to express it to the required degree of accuracy or precision, starting from the first non-zero digit, are significant figures.

Question 15.
What is atomicity of a gas?
Answer:
Atomicity of a gas is the number of atoms of an element present in one molecule of that element. For example, helium is monoatomic and hydrogen is diatomic.

Question 16.
Calculate the number of electrons present in 36 g water.
Answer:
Gram molecular weight of H2O= 18 g
According to molecular weight concept
18 g H2O= 6.023 × 1023 molecules of water
RBSE Solutions for Class 11 Chemistry Chapter 1 Basic Concepts of Chemistry img 1
1 molecule of water contains=10 electrons
12.046 × 1023 molecules will contain
=10 × 12.046 × 1023 electrons
=12.046 × 10 24

RBSE Solutions for Class 11 Chemistry Chapter 1 Basic Concepts of Chemistry

Question 17.
What do you understand by molecular mass?
Answer:
Molecular mass is a number equal to the sum of the atomic masses of the atoms in a molecule. The molecular mass gives the mass of a molecule relative to that of the 12 C atom, which is taken to have a mass of 12. Example: molecular mass of H2O = 18.

Question 18.
What is the mass of water which contains 50% heavy water (D2O)?
Answer:
Water containing half H2O and half D2O will have mass = average of masses of,H2O (18) and D2O (20)
Mass of water = (18 + 20)/ 2 = 38/ 2 = 19

Question 19.
What do you understand by standard temperature and pressure?
Answer:
The standard temperature is 273.15 K (0° Celsius or 32° Fahrenheit) and the standard pressure is 1 atm pressure or 760 mm. This is the freezing point of pure water at sea level atmospheric pressure. At STP, one mole of gas occupies 22.4 L of volume (molar volume).

RBSE Solutions for Class 11 Chemistry Chapter 1 Basic Concepts of Chemistry

Question 20.
At STP, x mL N2 gas reacts Completely with xmL O2 gas and forms gas A. What will be the molecular formula of A if volume is unchanged after reaction?
Answer:
According to Avogadro’s Law, equal volume of gases contains equal number of molecules at STP. Therefore number of oxygen and nitrogen molecules will be equal. Hence the formula will be NO.

RBSE Class 11 Chemistry Chapter 1 Short Answer Type Questions

Question 21.
What is the weight of 1 mole of the following
(i) NaCl
(ii) CaCO3
Answer:
(i) Molecular mass of NaCl : 23 amu + 35.5 amu = 58.5 amu
Molar of mass : NaCl: 58.5 g/mol
No. of Moles = Mass / Molar mass
Mass = 1 × 58.5g
Therefore, mass of 1 mole of NaCl is 58.5 g.

(ii) Molecular mass of CaCO3= (1 × Ga + 1 × C + 3 × O)
= (1 × 40 + 1 ×12 + 3 × 16)
= 40 + 12 + 48 = 100
Molar mass of CaCO3= 100 g
Nor of Moles = Mass / Molar mass
Mass = 1 × 100 = 100 g
Therefore, mass of 1 mole of CaCO3 is 100 g.

RBSE Solutions for Class 11 Chemistry Chapter 1 Basic Concepts of Chemistry

Question 22.
Calculate number of significant figures in each of the following:
(i) 0.00468
(ii) 753
Answer:
Both numbers have 3 significant figures.

Question 23.
What will be the mass of 2 moles of following:
(i) MgSO4
(ii) KCl
Answer:
(i) Mass of 1 mole = molar mass
Mass of 1 mole of MgSO4 = 120 g
Mass of 2 moles of MgSO4 = 120 g x 2 = 240 g
(ii)Mass of 1 mole = molar mass
Mass of 1 mole of KCI = 74.5 g
Mass of 2 mole of KCI = 74.5 x 2 = 149 g

Question 24.
Calculate number of significant figures, in each of the following:
(i) 0.868
(ii) 3.865 × 104
Answer:
(i) 3 significant figures
(ii) 4 significant figures

RBSE Solutions for Class 11 Chemistry Chapter 1 Basic Concepts of Chemistry

Question 25.
What is mole? Explain.
Answer:
1 mole is the amount of substance that contains the same number of entities as there are atoms in exactly 12 g of the 12 C isotope.
∴ Number of atoms in 12 g of 12C-isotope
= 6.023 × 1023 atoms/mol
The term ‘mole’ is given by the “Amedeo Avaogadro”. So it is also known as Avogadro’s number (NA).
∴ 1 mole =NA= 6.023 × 1023 atoms, molecules, ions etc.

Question 26.
What is the Limiting Reagent? Explain with the help of example.
Answer:
A limiting reagent is a reactant which is present in lesser amount in a reaction. When it gets consumed, the reaction will not proceed irrespective of the amount of other reactant present in the reaction. It limits the amount of product formed. In other words, it determines the extent of reaction. Let us consider a chemical reaction which is initiated by passing a spark through a reaction vessel containing 10 moles of H2 and 7 moles of O2.
RBSE Solutions for Class 11 Chemistry Chapter 1 Basic Concepts of Chemistry img 2
The reaction stops only after consumption of 5 moles of O2 as no further amount of H2 is left to react with
unreacted O2. Thus H2 is a limiting reagent in this reaction.

RBSE Solutions for Class 11 Chemistry Chapter 1 Basic Concepts of Chemistry

Question 27.
Give formula for calculation of equivalent weight of the compound.
Answer:
Equivalent weight of acid = molecular weight of acid/basicity
For example, Equivalent weight of H2SO4 = \(\frac{98}{2}\) = 49 g/eq
Equivalent weight of base = molecular weight of base/acidity
Example, Equivalent weight of NaOH = \(\frac{40}{1}\) = 40 g/eq Equivalent weight of salt = molecular weight of salt/charge present on ionic form ‘
Example, Equivalent weight of Na2CO3= \(\frac{106}{2}\)
= 53 g/eq

Normality formula is the gram equivalent of the solute dissolved per liter of the solution.

Question 28.
Define molarity, molality and normality of a solution.
Answer:
Molarity: It is the number of moles, of the solute dissolved per litre of the solution. It is represented by the symbol ‘M’. It is given as:
M = Mass of solute/Volume of solution in litre.
Unit of Molarity is mol/L.
Molality = It is the number of moles of the solute dissolved in 1 kg of the solvent. It is represented by the symbol ‘m’. It is given as :
m = Moles of solute / Weight of solvent in kg
Unit of molality is mol/kg.
Normality: Normality is number of gram equivalents of the solute dissolved per litre of solution. It is denoted by N.
Normality = No. of gram equivalent of solute / Volume of solution
The unit of Normality is gram equivalent per litre,

RBSE Solutions for Class 11 Chemistry Chapter 1 Basic Concepts of Chemistry

Question 29.
Write the formula of the following compounds and calculate molecular mass :
(i) Calcium Carbonate
(ii) Magnesium Phosphate
(iii) Ferric chloride ,
Answer:
(i) Calcium carbonate = CaCO3.
Molecular mass = (1 x Atomic mass of calcium) + (1 x Atomic mass of carbon) + (3 x atomic mass of oxygen)
= (1 × 40+1 × 12 + 3 ×16) .
= 40 + 12 + 48= 100 g/Aol
(ii) Magnesium Phosphate : Mg3(PO4)2
Molecular mass = (3 x Atomic mass of Mg) + 2 x (1 x atomic mass of P + 4 x atomic mass of 0)
= (3 × 24) + 2[1 × 31 + 4 × 16]
= 72+2(31 + 64)
= 72+2(95)
= 72+190
= 262 g/mol
(iii) Ferric Chloride: FeCl3
= 12.046 × 1023 atoms
Molecular mass = (1 × Atomic mass of Fe + 3 × atomic Now, gram atomic mass of Nitrogen = 14 g mass of Cl) . So, 14 g nitrogen contains 6.023 × 023atoms
= (55.8 + 3 × 35.5)
= (55.8 + 106.5)
= 162.35 g/mol

Question 30.
Calculate the number of moles in the following:
(i) 100 g of CaCO3
(ii) 80 g of Oxygen
(iii) 10 g of C12H22O11
Answer:
(i) Gram molar mass of CaCO3is 100 g.
So, 100 g CaCO3 contains 1 mole of it.

(ii) Gram atomic mass of Oxygen is 16 g.
So, 16 g of Oxygen contains 1 mole of it
Therefore, 80 g of Oxygen will contain = 80/ 16 = 5 moles.

(iii) Gram molar mass of C12H22O11 (Sucrose) is 342 g So, 342 g of sucrose contains 1 mole of it Therefore, 10 g will contain 10/342 = 0.029 mole

RBSE Solutions for Class 11 Chemistry Chapter 1 Basic Concepts of Chemistry

Question 31.
Write the formula of the following compounds and calculate the molecular mass:
(i) Ammonium Oxalate
(ii) Sodium Sulphate
(iii) Aluminium Nitrate
Answer:
(i) Ammonium Oxalate-(NH4)2C2O4Molecular mass = (2 x (1 x atomic mass of N + 4 x atomic mass of H) + (2 x atomic mass of C) + (4 × atomic mass of O)
= 2(14+4) +(2 × 12) +(4 × 16)
= 36 + 24+ 64 = 124 g/mol

(ii) Sodium Sulphate :Na2SO4
Molecular mass = (2 x atomic mass of Na + 1 x atomic mass of S + 4 x atomic mass of O)
= (2 × 23 + 1 × 32 + 4 × 16)
= (46 + 32 + 64) = 142 g/mol

(iii) Aluminium Nitrate : Al(N03)3
Molecular mass = (1 x atomic mass of A1 + 3 (1 × atomic mass of N + 3 × atomic mass of O)
= 26.9+3(14+3 × 16)
= 26.9+ 3 × 62
= 26.9+ 186
= 212.9 g/mol

Question 32.
What is the number of atoms present in 32 g of oxygen gas and 14 g of nitrogen gas?
Answer:
Gram atomic mass of oxygen is 16 g
So, 16 g oxygen contains 6.023 × 1023 atoms
Hence 32 g oxygen will contain 6.023 × 1023 × 2 atoms
=12.046 x 1023
Now, gram atomic mass of Nitrogen = 14 g
So, 14 g nitrogen contains 6.023 × 1023 atoms

RBSE Solutions for Class 11 Chemistry Chapter 1 Basic Concepts of Chemistry

Question 33.
Calculate molar mass of the following:
(i) HNO3
(ii) CO2
(iii) C2H6
Answer:
(i) Molar mass of HNO3 = (1 × atomic mass of H +1 × atomic mass of N + 3 atomic mass of O)
= (1 × 1) + (14 x 1) + (16 × 3)
= 1 + 14 + 48= 63 g/mol
(ii) Molar mass of CO2 = (1 x atomic mass of C + 2 × atomic mass of O)
= 1(1 × 12)+ (2 × 16)
= 12 + 32= 44 g/mol.
(iii) Molar mass of C2H6 = (2 x atomic mass of C + 6 × atomic mass of H) = ( 2 × 12) + (6 × 1) = 24 + 6
= 30 g/mol

Question 34.
Which of the following contains maximum number of molecules :
(i) 36 g water
(ii) 28 g carbon monoxide
Answer:
(i) Molar mass of water (H2O) is 18 g/mol 18 g of water contains 1 mole which contains 6.023 × 1023 molecules of it.
Therefore, 36 g water will contain 2 moles i.e, 6.023 × 1023 × 2 molecules
= 12.046 × 1023molecules
(ii) Molar mass of carbon monoxide (CO) is 28 g/mol So, 28 g of CO will contain 1 mole i.e, 6.023 × 1023  molecules
Therefore 36 g of water will contain maximum number of molecules.

RBSE Solutions for Class 11 Chemistry Chapter 1 Basic Concepts of Chemistry

Question 35.
Which of the following contains minimum number of molecules?
(i) 46 g of ethyl alcohol
(ii) 54 g of nitrogen pentaoxide
Answer:
(i) Molar mass of ethyl alcohol (C2H5OH) is 46 g/mol
So, 46 g of ethyl alcohol will have 6.023  × 1023 molecules,
(ii) Molar mass of nitrogen pentaoxide (N2O5) is 108 g/mol
So, 54 g of N2O5 will contain 1/ 2 mole
i. e., 6.023  × 1023  × (1/2) molecules.
= 3.0115  × 1023 molecules
Hence, 54 g of nitrogen pentaoxide will have minimum number of molecules.

RBSE Class 11 Chemistry Chapter 1 Long Answer Type Questions

Question 36.
Calculate the mass percentage of different elements present in sodium sulphate.
Answer:
Molecular mass of sodium sulphate (Na2SO4) is 142 g mol.
The elements present in sodium sulphate are sodium, sulphur and oxygen.
Mass of Sodium : 23  × 2 = 46 g
Mass of Sulphur : 32  × 1 = 32 g
Mass of Oxygen : 16  × 4 = 64 g
Mass percentage of sodium = \(\frac{32}{142}\)  × 100
= 32.38 or 32.4%
Mass percentage of sulphur = \(\frac{32}{142}\) × 100
= 22.57 or 22.6
Mass percentage of oxygen = \(\frac{64}{142}\) × 100 .
= 45.05%

RBSE Solutions for Class 11 Chemistry Chapter 1 Basic Concepts of Chemistry

Question 37.
What is Limiting reagent? Identify limiting reagent in reaction between 3.0 g H2 and 29 g O2.
Answer:
A limiting reagent is a reactant which is present in lesser amount in a reaction. When it gets consumed, the reaction will not proceed irrespective of the amount of other reactant present in the reaction. It limits the amount of product formed. In other words, it determines the extent of reaction. Consider the reaction between hydrogen and oxygen to form water. It can be represented as:
2H2 + O2 = 2H2O
From the above reaction, it is clear that 2 moles of hydrogen react with 1 mole of oxygen.
Molar mass of H2 = 2 g
Molar mass of O2= 32 g
This means that
4 g of H2 reacts with 32 g of O2
3 g of H2 reacts with = (32/4) x 3 g of O2 gas = 24 g
As the given amount of O2 is more than required.
Therefore, O2 is in excess and H2 is limiting reagent
here.

RBSE Solutions for Class 11 Chemistry Chapter 1 Basic Concepts of Chemistry

Question 38.
Two compounds are formed by carbon and oxygen In one of these, carbon content is 42.9% and in other 27.3%. Verify law of multiple proportion.
Answer:
Step 1: Calculate the percentage composition of carbon and oxygen in each of the two oxides :
RBSE Solutions for Class 11 Chemistry Chapter 1 Basic Concepts of Chemistry img 3

Step 2: Calculate the weights of carbon which combine with a fixed weight i.e., one part by weight of oxygen in each of the two oxides. In the first oxide, 57.1 parts by weight of oxygen combine with carbon = 42.9 parts.
1 part by weight of oxygen will combine with carbon
= \(\frac{42.9}{57.1}\)
= 0.751
In the second oxide, 72.7 parts by weight of oxygen combine with carbon = 27.3 parts,
.-. 1 part by weight of oxygen will combine with carbon
=\(\frac{27.3}{72.7}\)
=0.376 parts

Step 3: Compare the weights of carbon which combine with the same weight of oxygen in both the oxides— The ratio of the weights of carbon that combine with the same weight of oxygen (1 part) is 0.751 : 0.376 or 2 : 1 Since this is a simple whole.number ratio, so the above data illustrate the law of multiple proportions.

Question 39.
Write short notes on:
(i) Dalton’s Atomic Theory
(ii) Gay Lussac’s Law of Combining Volumes
(iii) Avogadro’s Hypothesis and its applications.
Answer:
(i) Dalton’s Atomic Theory: Main postulates of
Dalton’s atomic theory are as follow :

  • Matter is composed of very tiny or microscopic particles called “Atom”.
  • Atom is an indivisible particle.
  • Atom can neither be created nor be destroyed.
  • Atoms of an element are identical in size, shape, mass and in other properties.
  • Atoms of different elements are different in their properties.
  • Atoms combine with each other in small whole numbers.
  • All chemical reactions are due to combination or separation of atoms.

(ii) Gay Lussac’s Law of Combining Volumes: This law states that the volume of gases taking part in a chemical reaction show simple whole number ratio to. one another when those volumes are measured at the same temperature and pressure.

For the reactions :
RBSE Solutions for Class 11 Chemistry Chapter 1 Basic Concepts of Chemistry img 4
1 volume of nitrogen combines with 3 volumes of hydrogen to form 2 volumes of ammonia.
RBSE Solutions for Class 11 Chemistry Chapter 1 Basic Concepts of Chemistry img 5
2 volumes of hydrogen combine with 1 volume of oxygen to form 2 volumes of steam.

(iii) Avogadro’s-Hypothesis and its applications : In 1811, Italian physicist and mathematician Anedeo Avogadro published a hypothesis (also termed Avogadro’s law or principle) states that the volume of a gas is directly proportional to the number of molecules of the gas. This is represented by the formula ,
V = aN,
where a is a constant, V is the volume of the gas, and N is the number of gas molecules.
Applications
Avogadro’s law has been useful in substantiating a number of important laws and concepts. It has helped in the following areas:

  • In explaining Gay Lussac’s law of gaseous volumes.
  • In determining the atomicity of gases.
  • In determining the molecular formula of a gas.
  • In establishing the relationship between relative molecular mass and vapour density.

RBSE Solutions for Class 11 Chemistry Chapter 1 Basic Concepts of Chemistry

Question 40.
Briefly explain Law of chemical combinations.
Answer:
There are five laws of chemical Combinations :
1. Law of Conservation of Mass: This law states that matter can neither be created nor be destroyed. In other words, the total mass, that is, the sum of mass of all reactants and the products formed remains constant. This law is also called as the law of indestructibility of matter. Example
RBSE Solutions for Class 11 Chemistry Chapter 1 Basic Concepts of Chemistry img 6

2. Law of Definite Proportions: Joseph Proust, a French chemist stated that the proportion of elements by weight in a given compound will always remain exactly the same. In simple terms, we can say that, irrespective of its source, origin or its quantity, the percent composition of elements by weight in a given compound will always remain the same.
Example: A pure water obtained from which ever source or any country will always be made up of only hydrogen and oxygen elements combined together in the same fixed ratio of 1 : 8 by mass.

3. Law of Multiple Proportions: This law states that if two elements combine to form more than one compound, the masses of these elements in the reaction are in the ratio of small whole numbers. This law was given by Dalton in the year 1803.

For example, Hydrogen and Oxygen combine to form water H20 and hydrogen peroxide H202. Two atoms of hydrogen combine with one atom of oxygen in the case of water, while-two atoms of hydrogen combine with two atoms of oxygen in the case of hydrogen peroxide. The ratio of oxygen atoms combining with a fixed number of hydrogen atoms in these two compounds is 1 : 2.

4. Law of Reciprocal Proportion: This law states that “when two elements combine separately with third element and form different types of molecules, their combining ratio is directly reciprocal if they combine directly.”

5. Gay Lussac’s Law of Gaseous’Volumes: In 1808, Gay Lussac gave this law which states that when gases are produced or combined in a chemical reaction, they do so in simple ratio by volume given that all the gases are at same temperature and pressure.

Example: Combination between nitrogen and hydrogen: One volume of nitrogen always combines With 3 volumes of hydrogen to form two volumes of ammonia. This reaction also indicates a simple ratio of 1: 3 : 2 between the volume of the reactants and products.

RBSE Class 12 Chemistry Chapter 1 Numerical Problems

Question 41.
In an organic compound, 40% carbon, 6.66% hydrogen and rest oxygen is present. Its vapour destiny is 30. Calculate the empirical formula and molecular formula.
Answer:
Percentage of oxygen=100-(40+6.66)=100-46.6=53.4%
RBSE Solutions for Class 11 Chemistry Chapter 1 Basic Concepts of Chemistry img 7
Hence, empirical formula=CH2O
Now, empirical formula mass
= 12 + (2 × 1) + 16 = 12 + 2 +16 = 30″
Molecular mass = 2 x Vapour density = 2 × 30 = 60
n = Molecular mass / Empirical formula mass
= \(\frac{60}{30}\) = 2
Thus, Molecular formula = (Empirical formula) × n = (CH2O) × 2 = C2H4O2

RBSE Solutions for Class 11 Chemistry Chapter 1 Basic Concepts of Chemistry

Question 42.
In an organic compound, the ratio of masses of C, H and N are 9:1:3.5 and molecular mass is 108. What will be the empirical formula and molecular formula of the compound.
Answer:
RBSE Solutions for Class 11 Chemistry Chapter 1 Basic Concepts of Chemistry img 8
Empirical formula =C3H4N
Empirical formula mass = (3 × 12) + (4 × 1)+ 14 = 54
Thus, Molecular formula of the compound = (Empirical formula) x n
= (C3H4N)2=C6H8N2

Question 43.
In chemical analysis, it is found that in 10 g iron chloride, 3.438 g iron and 6.650 g chlorine are present. Calculate the empirical formula of iron chloride. (Fe = 55.8, Cl= 35.5)
Answer:
Percentage of iron = \(\frac{3.438}{10}\) × 100 = 34.38%
Percentage of Chlorine = \(\frac{6.65}{10}\) × 100 = 66.5%
RBSE Solutions for Class 11 Chemistry Chapter 1 Basic Concepts of Chemistry img 9
Thus, the empirical formula of iron chloride is FeCl3

Question 44.
Calculate number of oxygen atoms in 88 g of CO2. Also calculate mass of CO having same number of oxygen atoms.
Answer:
Number of moles of CO2 in 88 g of CO2
= Mass of CO2 / Molar mass of CO2
= 88 g / 44 g/mol = 2 mol CO2
Since one mole of CO2 contains two moles of oxygen atoms, two moles of CO2contain four moles of oxygen atoms. Hence,
1 mole of oxygen atoms contains 6.023 ×1023 oxygen atoms.
4 moles of oxygen atoms contain 6.023 × 1023 × 4 atoms = 2.4092 × 1024 oxygen atoms.
Since 1 mole of oxygen atom is present in 1 mole of CO, 4 moles of oxygen atoms are present in 4 moles of CO.
Mass of 4 moles of CO = number of moles of CO x gram molecular mass of CO
= 4 × (12+ 16)
= 4 × 28
= 112g

RBSE Solutions for Class 11 Chemistry Chapter 1 Basic Concepts of Chemistry

Question 45.
Define atomic mass, molecular mass and equivalent weight. In 500 mL solution, 20.7 g potassium carbonate is dissolved. Calculate its molarity. (Molecular mass of K2CO3=138)
Answer:
Atomic mass: The atomic mass of an element is defined as the number of times an atom of that element is heavier than an atom of carbon-12. It is expressed in a.m.u. or atomic mass unit. In place of a.m.u., symbol u is used which is known as unified mass.
Atomic mass = Mass of an atom × 1 /12 mass of carbon atom

Molecular mass: Molecular mass of any substance is defined as the average relative mass of its molecules as compared to mass of an atom of carbon-12. Molecular mass is expressed as a.m.u. It is given by,
Molecular mass = Mass of one molecule of substance /(1/12) x mass of carbon -12. ,

Equivalent mass: Quantity of a substance in which is numerically equal to its equivalent mass is called its gram equivalent mass.
Calculation of Molarity:
Molar mass of K2CO3= 138 g/mol
Volume of solution = 500 mL or 0.5 L
Mass of  K2CO3= 20.7
Number of moles = 20.7/138 = 0.149
Now,
Molarity = Number of moles/Volume of solution
= \(\frac{0.149}{0.5}\) = 0.299 M
Thus, molarity of solution is 0.299 M.

Question 46.
In industrial preparation of nitric acid, how. many moles of N02 are required to form 7.33 moles of HNO3 if the reaction is.
3NO2(g) + H2O(l) → 2HNO3 (l) + NO(g)
Answer:
According to the above equation, 3 moles of NO2 react with one moles form 2 moles of HNO3.
Therefore, number of moles of NO2 produced by 7.33 moles of HNO3, is given as :
3 moles of NO2-+ 2 moles of HNO3,
x moles of NO2-+ 7.33 moles of HNO3
By Cross multiplication, x = 7.33 × 3/2 = 10.99 moles.

RBSE Solutions for Class 11 Chemistry Chapter 1 Basic Concepts of Chemistry

Question 47.
1.68 g iron contains how many moles of iron? Calculate, the number of atoms in same amount of iron. (At. mass of Fe = 56).
Answer:
Number of moles = mass of substance/mass of one mole
= \(\frac{1.68}{56}\) = 0.03 moles
1 mole of iron contains 6.023 × 1023 atoms
0.03 moles of iron will contain 0.03 x 6.023 × 1023  atoms
= 1.8 ×1022  atoms

Question 48.
Calcium carbonate reacts with aqueous HCl to give CaCl2 and CO2 according to the reaction, CaCO3 (s) + 2HCl (aq) → CaCl2 (aq) + CO2 (g) + H2O (l). What mass of CaCO3 is required to react completely with 25 mL of 0.75 M HCl ? .
Answer:
1000 mL of.0.75 M HCl have 0.75 mol of HCl = 0.75 x 36.5 g = 24.375 g
(Molar mass of HCl is 36.5)
Mass of HCl in 25 mL of 0.75 M
HCl = 24.375/1000 x 25 g = 0.6844 g
According to the given chemical equation,
CaCO3 (s) + 2HCl(ag) →CaCl2 (aq) + CO2(g) + H2O(Z)
2 mol of HCl i.e. 73 g HCl react completely with 1 mol of CaCO3g i.e. 100 g
.’. 0.6844 g HCl reacts completely with CaCO3 = 100/73 x 0.6844 g = 0.938 g

Question 49.
In laboratory preparation of chlorine, manganese dioxide (Mn02) reacts with aqueous hydrochloric acid according to the reaction:
4HCl (aq) + MnO2 (s) → 2H2O(l) + MnCl2 (aq) + Cl2 (g)
How many grams of HCl react with 5.0 g of manganese dioxide?
Answer:
1 mol of MnO2 = (Atomic mass of Mn + 2 x atomic mass of oxygen)
= 55 + 32 g = 87 g
87g of MnO2 react with 4 moles of HCl i.e. 4 x 36.5 g
= 146 g of HCl.
(Molar mass of HCl is 36.5)
5.0 g of MnO2 will react with HCl = 146/87 x 5.0 g
=8.40 g.

RBSE Solutions for Class 11 Chemistry Chapter 1 Basic Concepts of Chemistry

Question 50.
Hydrochloric acid is 38% by mass. If the density of solution is 1.19 gm cm”3, then what will be the molarity of solution?
Answer:
38% by mass = 38 g HCl in 100 g solution
38 g HCl in 100/1.19 mL = 84 mL solution = 0.084 L .
Molarity = \(\frac{38}{36.5}\) × \(\frac{1}{0.084}\) = 12.39 M.
Thus, the molarity of solution is 12.39 M.

RBSE Solutions for Class 11 Chemistry

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