KKMishra's Tutorials: H C Verma solutions, Laws of Thermodynamics, OBJECTIVE-I, Chapter-26, Concepts of Physics, Part-II

Laws of Thermodynamics

OBJECTIVE-I

1. The first law of thermodynamics is a statement of

(a) Conservation of heat

(b) Conservation of work

(c) Conservation of momentum

(d) Conservation of energy.

Answer: (d)

EXPLANATION: The statement of the first law of thermodynamics is,

ΔQ = ΔU + ΔW.

i.e. if ΔQ amount of heat is given to a gas in a certain process and an amount of ΔW work is done by it then it's internal energy must increase by an amount of ΔQ -ΔW. So it is just a restatement of the law of conservation of energy.

2. If heat is supplied to an ideal gas in an isothermal process,

(a) the internal energy of the gas will increase

(b) the gas will do positive work

(c) the gas will do negative work

(d) the said process is not possible.

Answer: (b)

EXPLANATION: Since the process is isothermal there is no change in the temperature. Also, the internal energy is a function of temperature, there is no change in internal energy. Means ΔU =0. From the first law of thermodynamics, ΔQ =ΔU+ΔW

→ΔQ = ΔW.

So the gas will do positive work.

3. Figure (26-Q1) shows two processes A and B on a system. Let ΔQ₁ and ΔQ₂ be the heat given to the system in processes A and B respectively. Then

(a) ΔQ₁ > ΔQ₂

(b) ΔQ₁= ΔQ₂

(c) ΔQ₁ < ΔQ₂

(d) ΔQ₁ ≤ ΔQ₂.

The figure for Q-3

Answer: (a)

EXPLANATION: Since the initial and final pressure and volume are the same for both the processes the initial and final temperatures of the system will also be the same. Thus the change in the internal energy (ΔU) of the system will also be the same for both the processes.

From the first law of thermodynamics, ΔQ =ΔU+ΔW. Since U is the same for both the processes hence ΔQ will be greater for greater ΔW. From the p-V graph, the area under the curve is equal to the work done. Since the area under the process-A is more than the area under process B, ΔW₁ > ΔW₂. Thus ΔQ₁ > ΔQ₂.

4. Refer to the figure (26-Q1). Let ΔU₁ and ΔU₂ be the changes in internal energy of the system in the processes A and B. Then

(a) ΔU₁ > ΔU₂

(b) ΔU₁ = ΔU₂

(c) ΔU₁ < ΔU₂

(d) ΔU₁ ≠ ΔU₂.

Answer: (b)

EXPLANATION: Since the initial and final pressure and volume are the same for both the processes the initial and final temperatures of the system will also be the same. Thus the change in the internal energy (ΔU) of the system will also be the same for both the processes.

5. Consider the process on a system shown in figure (26-Q2). During the process, the work done by the system

(a) continuously increases

(b) continuously decreases

(c) first increases then decreases

(d) first decreases then increases.

The figure for Q-5

Answer: (a)

EXPLANATION: The work done by the system is equal to the area below the curve in the p-V graph. In the figure (36-Q2), during the process area under the curve continuously increases from the starting point the work done by the system continuously increases.

6. Consider the following two statements.

(A) If heat is added to a system, its temperature must increase.

(B) If positive work is done by a system in a thermodynamic process, its volume must increase.

(a) Both A and B are correct.

(b) A is correct but B is wrong.

(c) B is correct but A is wrong.

(d) Both A and B are wrong.

Answer: (c)

EXPLANATION: If heat is added to the system its temperature depends upon the work done by the system. If the work done by the system ΔW < ΔQ then the temperature of the system will increase but if ΔW = ΔQ then it will not increase. Hence the statement A is not true.

The positive work done by the system is the area under the p-V curve. i.e. it is equal to ∫p*dV. If dV=0, then work done by the system is zero. For a positive work, dV must have positive value i.e. the volume must increase. Statement B is true.

7. An ideal gas goes from the state i to the state f as shown in figure (26-Q3). The work done by the gas during the process

(a) is positive

(b) is negative

(c) is zero

(d) cannot be obtained from this information.

The figure for Q-7

Answer: (c)

EXPLANATION: The p-T graph is a straight line, so the pressure is directly proportional to the temperature.

Since for a closed system, p =nRT/V,

Here p ∝ T, so nR/V = constant. Here n and R are constants, so V is also constant. Since V does not change the work done is zero because work done =∫p.dV.

8. Consider two processes on a system as shown in figure (26-Q4).

The figure for Q - 7

The volumes in the initial states are the same in the two processes and the volumes in the final states are also the same. Let ΔW₁ and ΔW₂ be the work done by the system in the processes A and B respectively.

(a) ΔW₁ > ΔW₂

(b) ΔW₁ = ΔW₂

(c) ΔW₁ < ΔW₂

(d) Nothing can be said about the relation between ΔW₁ and ΔW₂.

Answer: (c)

EXPLANATION: Since the initial and final volumes in both the processes are the same, the change in volume ΔV is the same. The work done by the system in the process A, ΔW₁ =p₁ΔV and in process B, ΔW₂ = p₂ΔV. From the graph, p₁ < p₂, so ΔW₁ < ΔW₂.

9. A gas is contained in a metallic cylinder fitted with a piston. The piston is suddenly moved in to compress the gas and is maintained at this position. As time passes the pressure of the gas in the cylinder

(a) increases

(b) decreases

(c) remains constant

(d) increases or decreases depending on the nature of the gas.

Answer: (b)

EXPLANATION: Since the gas is suddenly compressed, the process is adiabatic. Both the pressure and temperature increase. Since the cylinder is metallic the heat of the gas gets out slowly through the walls. The volume is now constant so no work done. Hence due to the loss of heat energy the temperature slowly decreases. For a closed system,

pV/T =constant. If V is also constant then, p ∝ T. Since T decreases, p also decreases.

------------------------------------------------

Click here for all links → kktutor.blogspot.com

For beginners - Vectors & its addition -YouTube

===<<<O>>>===

My Channel on YouTube → SimplePhysics with KK

Links to the Chapters

ALL LINKS

CHAPTER- 26-Laws of Thermodynamics

Questions for Short Answer

CHAPTER- 25-CALORIMETRY

Questions for Short Answer

OBJECTIVE-I

OBJECTIVE-II

EXERCISES - Q1 to Q10

EXERCISES - Q-11 to Q-18

CHAPTER- 24-Kinetic Theory of Gases

Questions for Short Answer

OBJECTIVE - I

OBJECTIVE - II

EXERCISES- Q1 to 10

EXERCISES- Q11 to 20

EXERCISES- Q21 to 30

EXERCISES- Q31 to Q40

EXERCISES- Q41 to Q50

EXERCISES- Q51 to Q62

CHAPTER- 23 - Heat and Temperature

Questions for Short Answer

OBJECTIVE - I

OBJECTIVE - II

EXERCISES - Q-1 TO Q-10

EXERCISES - Q-11 TO Q-20

EXERCISES - Q-21 TO Q-34

CHAPTER- 22 - Photometry

Questions for Short Answer

OBJECTIVE - I

OBJECTIVE - II

EXERCISES

CHAPTER- 21 - Speed of Light

Questions for Short Answers

OBJECTIVE-I

OBJECTIVE-II

EXERCISES

CHAPTER- 20 - Dispersion and Spectra

Questions for Short Answers

OBJECTIVE-I

OBJECTIVE-II

EXERCISES

CHAPTER- 19 - Optical Instruments

Questions for Short Answers

OBJECTIVE-I

OBJECTIVE-II

EXERCISES - Q 1 to Q 12

EXERCISES - Q 13 to Q 24

CHAPTER- 18 - Geometrical Optics

Questions for Short Answers

OBJECTIVE-I

OBJECTIVE-II

EXERCISES - Q 1 to Q10

EXERCISES- Q11 to Q20

EXERCISES- Q21 to Q30

EXERCISES- Q31 to Q40

EXERCISES- Q41 to Q50

EXERCISES- Q51 to Q60

EXERCISES- Q61 to Q70

EXERCISES- Q71 TO 79

CHAPTER- 17 - Light Waves

Questions for Short Answers

OBJECTIVE - I

OBJECTIVE - II

EXERCISES - Q1-Q10

EXERCISES - Q11-Q20

EXERCISES - Q21-Q30

EXERCISES - Q31-Q41

CHAPTER- 16 - Sound Waves

Questions for Short Answers

OBJECTIVE-I

OBJECTIVE-II

EXERCISES - Q-1 TO Q-10

EXERCISES - Q-11 TO Q-20

EXERCISES -Q-21 TO Q-30

EXERCISES -Q-31 TO Q-40

EXERCISES -Q-41 TO Q-50

EXERCISES -Q-51 TO Q-60

EXERCISES -Q-61 TO Q-70

EXERCISES - Q-71 TO Q-80

EXERCISES - Q-81 TO Q-89

CHAPTER- 15 - Wave Motion and Waves on a String

Questions for Short Answers

OBJECTIVE-I

OBJECTIVE-II

EXERCISES - Q-1 TO Q-10

EXERCISES - Q-11 TO Q-20

EXERCISES - Q-21 TO Q-30

EXERCISES - Q-31 TO Q-40

EXERCISES - Q-41 TO Q-50

EXERCISES - Q-51 TO Q-57

CHAPTER- 14 - Fluid Mechanics

Questions for Short Answers

OBJECTIVE-I

OBJECTIVE-II

EXERCISES- Q-1 TO Q-10

EXERCISES- Q-11 TO Q-20

EXERCISES- Q-21 TO Q-32

CHAPTER- 13 - Fluid Mechanics

Questions for Short Answers

OBJECTIVE-I

OBJECTIVE-II

EXERCISES Q-1 TO Q-10

EXERCISES- Q11 TO Q20

EXERCISES Q-21 TO Q30

EXERCISES Q-31 TO Q35

CHAPTER- 12 - Simple Harmonic Motion

Questions for Short Answers

OBJECTIVE-I

OBJECTIVE-II

EXERCISES- Q1 TO Q10

EXERCISES- Q11 TO Q20

EXERCISES- Q21 TO Q30

EXERCISES- Q31 TO Q40

EXERCISES- Q41 TO Q50

EXERCISES- Q51 TO Q58 (2-Extra Questions)

CHAPTER- 11 - Gravitation

Questions for Short Answers

OBJECTIVE - I

OBJECTIVE - II

EXERCISES - Q_1 to Q_10

EXERCISES -Q-11 TO Q-20

EXERCISES -Q 21 TO 30

EXERCISES -Q 31 TO 39

CHAPTER- 10 - Rotational Mechanics

Questions for Short Answers

OBJECTIVE - I

OBJECTIVE - II

EXERCISES - Q-1 TO Q-15

EXERCISES - Q-16 TO Q-30

EXERCISES Q-31 TO Q-45

EXERCISES Q-46 TO Q-60

EXERCISES Q-61 TO Q-75

EXERCISES Q-76 TO Q-86

CHAPTER- 9 - Center of Mass, Linear Momentum, Collision

Questions for Short Answers

Objective - I

Objective - II

EXERCISES Q-1 TO Q-10

EXERCISES Q-11TO Q-20

EXERCISES Q-21 TO Q-30

EXERCISES Q-31 TO Q-42

EXERCISES Q-43 TO Q-54

CHAPTER- 8 - Work and Energy

Click here for → Question for Short Answers

Click here for → OBJECTIVE-I

Click here for → OBJECTIVE-II

Click here for → Exercises (1-10)

Click here for → Exercises (11-20)

Click here for → Exercises (21-30)

Click here for → Exercises (31-42)

Click here for → Exercise(43-54)

CHAPTER- 7 - Circular Motion

Click here for → Questions for Short Answer

Click here for → OBJECTIVE-I

Click here for → OBJECTIVE-II

Click here for → EXERCISES (1-10)

Click here for → EXERCISES (11-20)

Click here for → EXERCISES (21-30)

CHAPTER- 6 - Friction

Click here for → Questions for Short Answer

Click here for → OBJECTIVE-I

Click here for → Friction - OBJECTIVE-II

Click here for → EXERCISES (1-10)

Click here for → Exercises (11-20)

Click here for → EXERCISES (21-31)

For more practice on problems on friction solve these- "New Questions on Friction".

---------------------------------------------------------------------------------

CHAPTER- 5 - Newton's Laws of Motion

Click here for → QUESTIONS FOR SHORT ANSWER

Click here for→ Newton's laws of motion - Objective - I

Click here for → Newton's Laws of Motion - Objective -II

Click here for → Newton's Laws of Motion-Exercises(Q. No. 1 to 12)

Click here for→Newton's Laws of Motion,Exercises(Q.No. 13 to 27)

Click here for→Newton's Laws of Motion,Exercises(Q.No. 28 to 42)

-------------------------------------------------------------------------------

CHAPTER- 4 - The Forces

The Forces-

"Questions for short Answers"

Click here for "The Forces" - OBJECTIVE-I

Click here for "The Forces" - OBJECTIVE-II

Click here for "The Forces" - Exercises

--------------------------------------------------------------------------------------------------------------

Sunday, February 2, 2020

H C Verma solutions, Laws of Thermodynamics, OBJECTIVE-I, Chapter-26, Concepts of Physics, Part-II

1. The first law of thermodynamics is a statement of

(a) Conservation of heat

(b) Conservation of work

(c) Conservation of momentum

(d) Conservation of energy.

Answer: (d)

EXPLANATION: The statement of the first law of thermodynamics is,

ΔQ = ΔU + ΔW.

i.e. if ΔQ amount of heat is given to a gas in a certain process and an amount of ΔW work is done by it then it's internal energy must increase by an amount of ΔQ -ΔW. So it is just a restatement of the law of conservation of energy.

2. If heat is supplied to an ideal gas in an isothermal process,

(a) the internal energy of the gas will increase

(b) the gas will do positive work

(c) the gas will do negative work

(d) the said process is not possible.

Answer: (b)

EXPLANATION: Since the process is isothermal there is no change in the temperature. Also, the internal energy is a function of temperature, there is no change in internal energy. Means ΔU =0. From the first law of thermodynamics, ΔQ =ΔU+ΔW

→ΔQ = ΔW.

So the gas will do positive work.

3. Figure (26-Q1) shows two processes A and B on a system. Let ΔQ₁ and ΔQ₂ be the heat given to the system in processes A and B respectively. Then

(a) ΔQ₁ > ΔQ₂

(b) ΔQ₁= ΔQ₂

(c) ΔQ₁ < ΔQ₂

(d) ΔQ₁ ≤ ΔQ₂. The figure for Q-3

Answer: (a)

EXPLANATION: Since the initial and final pressure and volume are the same for both the processes the initial and final temperatures of the system will also be the same. Thus the change in the internal energy (ΔU) of the system will also be the same for both the processes.

4. Refer to the figure (26-Q1). Let ΔU₁ and ΔU₂ be the changes in internal energy of the system in the processes A and B. Then

(a) ΔU₁ > ΔU₂

(b) ΔU₁ = ΔU₂

(c) ΔU₁ < ΔU₂

(d) ΔU₁ ≠ ΔU₂.

Answer: (b)

EXPLANATION: Since the initial and final pressure and volume are the same for both the processes the initial and final temperatures of the system will also be the same. Thus the change in the internal energy (ΔU) of the system will also be the same for both the processes.

5. Consider the process on a system shown in figure (26-Q2). During the process, the work done by the system

(a) continuously increases

(b) continuously decreases

(c) first increases then decreases

(d) first decreases then increases.

Answer: (a)

EXPLANATION: The work done by the system is equal to the area below the curve in the p-V graph. In the figure (36-Q2), during the process area under the curve continuously increases from the starting point the work done by the system continuously increases.

6. Consider the following two statements.

(A) If heat is added to a system, its temperature must increase.

(B) If positive work is done by a system in a thermodynamic process, its volume must increase.

(a) Both A and B are correct.

(b) A is correct but B is wrong.

(c) B is correct but A is wrong.

(d) Both A and B are wrong.

Answer: (c)

EXPLANATION: If heat is added to the system its temperature depends upon the work done by the system. If the work done by the system ΔW < ΔQ then the temperature of the system will increase but if ΔW = ΔQ then it will not increase. Hence the statement A is not true.

The positive work done by the system is the area under the p-V curve. i.e. it is equal to ∫p*dV. If dV=0, then work done by the system is zero. For a positive work, dV must have positive value i.e. the volume must increase. Statement B is true.

7. An ideal gas goes from the state i to the state f as shown in figure (26-Q3). The work done by the gas during the process

(a) is positive

(b) is negative

(c) is zero

(d) cannot be obtained from this information.

Answer: (c)

EXPLANATION: The p-T graph is a straight line, so the pressure is directly proportional to the temperature.

Since for a closed system, p =nRT/V,

Here p ∝ T, so nR/V = constant. Here n and R are constants, so V is also constant. Since V does not change the work done is zero because work done =∫p.dV.

8. Consider two processes on a system as shown in figure (26-Q4).

The volumes in the initial states are the same in the two processes and the volumes in the final states are also the same. Let ΔW₁ and ΔW₂ be the work done by the system in the processes A and B respectively.

(a) ΔW₁ > ΔW₂

(b) ΔW₁ = ΔW₂

(c) ΔW₁ < ΔW₂

(d) Nothing can be said about the relation between ΔW₁ and ΔW₂.

Answer: (c)

EXPLANATION: Since the initial and final volumes in both the processes are the same, the change in volume ΔV is the same. The work done by the system in the process A, ΔW₁ =p₁*ΔV and in process B, ΔW₂ = p₂*ΔV. From the graph, p₁ < p₂, so ΔW₁ < ΔW₂.

9. A gas is contained in a metallic cylinder fitted with a piston. The piston is suddenly moved in to compress the gas and is maintained at this position. As time passes the pressure of the gas in the cylinder

(a) increases

(b) decreases

(c) remains constant

(d) increases or decreases depending on the nature of the gas.

Answer: (b)

pV/T =constant. If V is also constant then, p ∝ T. Since T decreases, p also decreases.

------------------------------------------------

Click here for all links → kktutor.blogspot.com

For beginners - Vectors & its addition -YouTube

===<<<O>>>===

My Channel on YouTube → SimplePhysics with KK

(d) ΔQ₁ ≤ ΔQ₂.

The figure for Q-3

EXPLANATION: Since the initial and final volumes in both the processes are the same, the change in volume ΔV is the same. The work done by the system in the process A, ΔW₁ =p₁ΔV and in process B, ΔW₂ = p₂ΔV. From the graph, p₁ < p₂, so ΔW₁ < ΔW₂.

Click here for → Question for Short Answers

Click here for → OBJECTIVE-I

Click here for → OBJECTIVE-II

Click here for → Exercises (1-10)

Click here for → Exercises (21-30)

Click here for → Exercises (31-42)

Click here for → Exercise(43-54)

Click here for → Questions for Short Answer

Click here for → OBJECTIVE-I

Click here for → OBJECTIVE-II

Click here for → EXERCISES (1-10)

Click here for → EXERCISES (11-20)

Click here for → EXERCISES (21-30)

CHAPTER- 6 - Friction

Click here for → OBJECTIVE-I

Click here for → Friction - OBJECTIVE-II

Click here for → EXERCISES (1-10)

Click here for → Exercises (11-20)

Click here for → EXERCISES (21-31)

For more practice on problems on friction solve these- "New Questions on Friction".

Click here for→ Newton's laws of motion - Objective - I

Click here for → Newton's Laws of Motion - Objective -II

Click here for → Newton's Laws of Motion-Exercises(Q. No. 1 to 12)

Click here for→Newton's Laws of Motion,Exercises(Q.No. 28 to 42)

-------------------------------------------------------------------------------