Tuesday, August 25, 2026

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

 Laws of Thermodynamics


QUESTIONS FOR SHORT ANSWER


   1. Should the internal energy of a system necessarily increase if heat is added to it?   


Answer:  The increase in the internal energy (ΔU) of a system is equal to the amount of heat given (ΔQ) minus the amount of work (ΔW) done by it.

ΔU = ΔQ - ΔW

If ΔW = ΔQ, the change in internal energy is zero. So the internal energy does not necessarily increase if heat is given to the system.    




   2. Should the internal energy of a system necessarily increase if its temperature is increased?   


Answer:  The internal energy of a system increases if the total kinetic energy of all the molecules increases, and this thing increases with the increase in temperature. Hence, the internal energy of a system necessarily increases if its temperature is increased.    




   3. A cylinder containing gas is lifted from the first floor to the second floor. What is the amount of work done on the gas? What is the work done by the gas? Is the internal energy of the gas increased? Is the temperature of the gas increased?    


Answer:  Since there is no change in the volume of the gas, no work is done on the gas, nor any work is done by the gas.

No, the internal energy of the gas is not increased because the systematic movement of a gas sample has no effect on temperature.  



   4. A force is applied on a block of mass M. The block is displaced through a distance d in the direction of the force. What is the work done by the force on the block? Does the internal energy change because of this work?    


Answer:  The work done by the force F here = F*d. 

No, the internal energy of the block does not change. This work done is reflected either as a change in kinetic energy or potential energy.  




   5. The outer surface of a cylinder containing gas is rubbed vigorously by a polishing machine. The cylinder and its gas become warm. Is the energy transferred to the gas heat or work?   


Answer:  Since the volume of the gas does not change, the energy transferred is not due to work. The energy to the gas is transferred by the cylinder, which is in the form of heat that is generated due to friction.     




   6. When we rub our hands, they become warm. Have we applied heat to the hands?   


Answer:  No. It is the mechanical work done by us to overcome the force of friction between the hands that converts into heat energy.   




   7. A closed bottle contains some liquid. The bottle is shaken vigorously for 5 minutes. It is found that the temperature of the liquid is increased. Is heat transferred to the liquid? Is work done on the liquid? Neglect expansion on heating.   


Answer:  No, heat is not transferred to the liquid.

Work done is equal to the force times displacement. Assuming that the bottle is kept at the initial position after shaking, the displacement is zero. So no work is done on the liquid. But in the shaking of the liquid, work is done against the viscous forces; also, it increases the total kinetic energy of the molecules. So the internal energy of the liquid and temperature increase.     




   8. The final volume of the system is equal to the initial volume in a certain process. Is the work done by the system necessarily zero? Is it necessarily nonzero?    


Answer:  If the final volume of the system is equal to the initial volume, then the process is either isochoric or cyclic. The cyclic process may be either reversible or irreversible. In the isochoric process, the work done is zero. In the reversible process, the work done by the system and the work done on the system are compensated, and the final work done is zero. In the irreversible process, the work done by the system is not equal to the work done on the system. Hence, the final work done is not zero.

Thus the answer to both the questions is "no".    



   9. Can work be done by a system without changing its volume?   


Answer:  If the system is not changing its volume, the process is called isochoric. In this process, work can not be done because ΔW = p*ΔV =p*0 =0.   



   10. An ideal gas is pumped into a rigid container having diathermic walls so that the temperature remains constant. In a certain time interval, the pressure in the container is doubled. Is the internal energy of the contents of the container also doubled in the interval?   


Answer:  n = pV/RT; here, V, R, and T are constant, so when the pressure is 2p, the number of moles = 2n. 

Since the mass or number of moles of the gas is doubled, the internal energy associated with the original mass is also doubled in the contents of the cylinder.     



   11. When a tire bursts, the air coming out is cooler than the surrounding air. Explain.   


Answer:  The tire burst is a sudden process in which heat transfer between the system and the surroundings cannot take place. So it is an adiabatic process. The pressure of the air inside the tire is more than that of the surroundings. Due to the tire burst, the volume of the air increases, and some work is done by the air, for which it uses its internal energy. So the internal energy of the system decreases. It results in a decrease in the temperature of the expanded air and makes it feel cooler than the surroundings. 




   12. When we heat an object, it expands. Is the work done by the object in this process? Is heat given to the object equal to the increase in its internal energy?   


Answer:  ΔW =p*ΔV 

Not only gases, but it is also true for the expansion of solids and liquids.

In the given problem, p is constant, but due to expansion, ΔV has some positive value. Hence some work is done by the object. 

Now the change in the internal energy,

ΔU = ΔQ - ΔW

Since ΔW is not zero, ΔU < ΔQ. So the heat given to the object is not equal to the increase in the internal energy.      




   13. When we stir a liquid vigorously, it becomes warm. Is it a reversible process?   


Answer:  When a liquid is vigorously stirred, it becomes warm because work is done against the viscosity. But the process is dissipative. A dissipative process can not be reversible. Another point is that if mechanical work is done on the system to increase the internal energy, the reverse process is not automatic, but we need a heat engine to do it; also, only a part of the heat energy can be used to do the mechanical work. So it is not a reversible process.

  



   14. What should be the condition for the efficiency of a Carnot engine to be equal to 1?   


Answer:  The efficiency of a Carnot engine, η = 1 - Q₂/Q₁, where Q₁ is the heat applied to the engine, and Q₂ is the heat given to the sink. It is clear that η can be equal to 1 only when Q₂ is zero, but it is not possible according to the second law of thermodynamics, which is also called the Kelvin-Planck statement.   




   15. When an object cools down, heat is withdrawn from it. Does the entropy of the object decrease in the process? If yes, is it a violation of the second law of thermodynamics stated in terms of an increase in entropy?   


Answer:  Yes, when the heat is withdrawn from an object, its entropy decreases.

 The second law of thermodynamics may be stated as "It is not possible to have a process in which the entropy of an isolated system is decreased". Here, the cooling object is not an isolated system. The heat given out by the object is added to the surroundings, and its entropy increases. The universe, along with the object, is taken as an isolated system. And its entropy is not decreasing. Hence, the second law of thermodynamics is not violated.   



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 its 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 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, 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 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, the 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, 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 the work done by the system is zero. For positive work, dV must have a 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 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) 

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 is 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.  



OBJECTIVE-II


   1. The pressure p and volume V of an ideal gas both increase in a process.

(a) Such a process is not possible.

(b) The work done by the system is positive.

(c) The temperature of the system must increase.

(d) Heat supplied to the gas is equal to the change in internal energy.   


Answer:  (b), (c). 

EXPLANATION: Such a process is possible. The work done by the system is pressure x change in volume.

W = ∫p*dV

 Here the volume is increasing; hence the work is done by the system. Option (b) is true.

  For an ideal gas, pV = nRT

→T = pV/nR

Since p and V both increase, T will also increase. Option (c) is true.

      Since the work done by the system is not zero, statement (d) is not true. 




    2. In a process on a system, the initial pressure and volume are equal to the final pressure and volume.

(a) The initial temperature must be equal to the final temperature.

(b) The initial internal energy must be equal to the final internal energy.

(c) The net heat given to the system in the process must be zero. 

(d) The work done by the system in the process must be zero.   


Answer:  (a), (b). 

EXPLANATION: For an ideal gas,

T =pV/nR

Since the value of pV is the same for the initial and final points, T will also be the same. Hence, option (a) is true.

   At a constant volume, the internal energy is proportional to the temperature. Since the initial and final temperatures are the same, internal energy is also the same. Option (b) is true.

          It may be possible that net heat is given to the system and an equal amount of work is done by the system. Hence, the options (c) and (d) are not true.




    3. A system can be taken from the initial state p₁, V₁ to the final state p₂, V₂ by two different methods. Let ΔQ and ΔW represent the heat given to the system and the work done by the system. Which of the following must be the same in both methods?

(a) ΔQ

(b) ΔW

(c) ΔQ+ΔW

(d) ΔQ-ΔW.   


Answer:  (d) 

EXPLANATION: The heat given ΔQ to the system and the work done ΔW by the system depend upon the path followed by the process; hence the options (a), (b) and (c) are not true.

     The internal energy of a system is a state function. Hence, in both processes, the initial and the final internal energies will be the same, and hence the change in the internal energy  ΔU will be the same. From the first law of thermodynamics,
ΔQ = ΔU + ΔW
→ΔU = ΔQ - ΔW will be the same in both processes. Option (d) is true.

  




    4. Refer to the figure (26-Q5). Let ΔU₁ and ΔU₂ be the change in internal energy in processes A and B, respectively, ΔQ be the net heat given to the system in process A+B, and ΔW be the net work done by the system in the process A+B. 
The figure for Q-4

(a) ΔU₁+ΔU₂ = 0.

(b) ΔU₁-ΔU₂ = 0.

(c) ΔQ-ΔW = 0.

(d) ΔQ+ΔW = 0.   


Answer:  (a), (c). 

EXPLANATION: The process A+B is a cyclic process. Hence, the system comes to the initial state A and the change in the internal energy ΔU = 0. Thus 

ΔU =ΔU₁+ΔU₂ =0.

Option (a) is true. 

And option (b) is not true.

From the first law of thermodynamics, ΔU =ΔQ-ΔW. Since ΔU = 0 in the process A+B, hence ΔQ-ΔW = 0. Option (c) is true.  




    5. The internal energy of an ideal gas decreases by the same amount as the work done by the system.

(a) The process must be adiabatic.

(b) The process must be isothermal.

(c) The process must be isobaric.

(d) The temperature must decrease.  


Answer:  (a), (d). 

EXPLANATION: From the first law of thermodynamics,

ΔQ =ΔU +ΔW

Given that, ΔW = -ΔU, hence,

ΔQ = 0. So the heat exchanged with the system is zero, and the process is adiabatic, not isothermal. Option (a) is correct. 

  The process can not be isobaric because in this process, pressure is constant and the work is done due to the heat energy given. Since (a) is true, there will not be heat transfer.

    Since the internal energy decreases, the temperature in this adiabatic process must decrease. Option (d) is true.  


EXERCISES


   1. A thermally insulated, closed copper vessel contains water at 15°C. When the vessel is shaken vigorously for 15 minutes, the temperature rises to 17°C. The mass of the vessel is 100 g, and that of the water is 200 g. The specific heat capacities of copper and water are 420 J/kg-K and 4200 J/kg-K, respectively. Neglect any thermal expansion. (a) How much heat is transferred to the liquid-vessel system? (b) How much work has been done on this system? (c) How much is the increase in the internal energy of the system?      


Answer:  (a) Since the system is insulated, no heat is transferred to the water-vessel system. The increase in temperature is due to the energy transfer by work done on the water-vessel system; the heat transferred to the system is zero.


     (b) The work done on the system is equal to the increase in the heat energy of the system. 

Increase in heat energy of water 

=(0.20 kg)*(4200 J/kg-K)*(17° - 15°)

=1680 J

Increase in heat energy of vessel

=(0.10 kg)*(420 J/kg-K)*(17° - 15°)

=84 J

Total = 1680+84 J =1764 J. 


       (c) ΔU = ΔQ - ΔW, Since ΔQ = 0, 

ΔU = ΔW = 1764 J. 



 

   2. Figure (26-E1) shows a paddlewheel coupled to a mass of 12 kg through fixed frictionless pulleys. The paddle is immersed in a liquid of heat capacity 4200 J/K kept in an adiabatic container. Consider a time interval in which the 12 kg block falls slowly through 70 cm. (a) How much heat is given to the liquid? (b) How much work is done on the liquid? (c) Calculate the rise in the temperature of the liquid, neglecting the heat capacity of the container and the paddle. 
The figure for Q-2
 


Answer: (a) The container is adiabatic, so heat transfer is zero.


     (b) The work done on the liquid =  Decrease in the potential energy of the block

=mgh

=(12 kg)*(10 m/s²)*(0.70 m)

=84 J. 


          (c) The rise in temperature will depend on the mass of the liquid, which is not mentioned here. Assuming the mass of the liquid = 1 kg. Specific heat capacity of the liquid, s = 4200 J/K. If the rise in temperature =ΔT, then

Q = ms*Δt

→84 = 1*4200*Δt

→Δt = 84/4200

→Δt = 0.02°C.  


 


  3. A 100 kg block is started with a speed of 2.0 m/s on a long, rough belt kept fixed in a horizontal position. The coefficient of kinetic friction between the block and the belt is 0.20. (a) Calculate the change in the internal energy of the block-belt system as the block comes to a stop on the belt. (b) Consider the situation from a frame of reference moving at 2.0 m/s along the initial velocity of the block. As seen from the frame, the block is gently put on a moving belt, and in due time the block starts moving with the belt at 2.0 m/s. Calculate the increase in the kinetic energy of the block as it stops slipping past the belt. (c) Find the work done in this frame by the external force holding the belt.   


Answer:  (a) m = 100 kg. u = 2.0 m/s. µ = 0.2, v = 0.

Change in internal energy ΔU =ΔQ - ΔW
Here ΔQ = 0, So
ΔU = -ΔW =-{½mv² - ½mu²}
     = -½m{v²-u²}
     = -½*100*{0²-2²}
     = 200 J.


(b) The initial speed of the block with respect to the moving frame, u = 0,
The final speed of the block, v = 2 m/s.
The increase in the kinetic energy of the block = ½mv² -½mu²
=½*100*2² - 0
=200 J.

(c) Let the force of friction between the block and belt = F. Given, µ = 0.2,
So, F = µ*R =µ*mg
→F = 0.2*100*10 N =200 N.
Acceleration, a = F/m =200/100 m/s² 
=2 m/s².
      Let the distance traveled by the block = s. From v²-u² =2as,
2² -0² =2*2*s
→s = 4/4 =1 m.
So the work done by the external force in moving the block by 1 m = Force*distance= (200 N)*(1 m) =200 J.
But the external force not only moves the block by 1 m but also gives it kinetic energy. The kinetic energy gained by the block is also due to the work done by the external force. This is equal to ½mv² =½*100*2² =200 J.
       So the total work done by the external force =200 J +200 J =400 J.    

    


 

  4. Calculate the change in internal energy of a gas kept in a rigid container when 100 J of heat is supplied to it.   


Answer:  Since the container is rigid, there will be no change in volume. So, ΔV = 0. And the work done by the system, ΔW =p*ΔV =0.

Given that ΔQ = 100 J. From the first law of thermodynamics,
ΔQ = ΔU + ΔW

→100 J = ΔU +0

→ΔU = 100 J.

So the change in the internal energy of the gas =100 J. 



 

  5. The pressure of a gas changes linearly with volume from 10 kPa, 200 cc to 50 kPa, 50 cc. (a) Calculate the work done by the gas. (b) If no heat is supplied and extracted from the gas, what is the change in the internal energy of the gas?    


Answer:  (a) Let us draw the p-V curve, 
Diagram for Q-5

The work done by the gas is the area under the p-V curve. It is shaded. Since the volume decreases, the work done by the gas is negative and equal to,

= -½(50 + 10)*(200-50) J
= -30*150 J
= -4500 J
= -4.5 kJ. 

(b) The change in internal energy,
ΔU =ΔQ - ΔW
Here ΔQ = 0 and ΔW = -4.5 kJ,
so, ΔU = 4.5 kJ. 

  6. An ideal gas is taken from an initial state i to a final state f in such a way that the ratio of the pressure to the absolute temperature remains constant. What will be the work done by the gas? 


Answer:  Since pV = nRT

→V = nR(T/p), 

Since n, R, and p/T and hence T/p are constant, the volume also remains constant from process i to f. So ΔV =0.
The work done by the gas ΔW =p*ΔV =0 (zero). 


 

  7. Figure (26-E2) shows three paths through which a gas can be taken from state A to state B. Calculate the work done by the gas in each of the three paths. 
The figure for Q-7

Answer:  The work done by the gas can be calculated in a p-V diagram by the area bounded between the p-V curve and the V axis between the two ordinates of volumes at the initial and final points.

In the process AB,

Work done by the gas

=½(10 kPa+30 kPa)*(25 cc -10 cc)

=20 kPa * 15 cc

=20000 N/m² * 15x10⁻⁶ m³

=0.30 J


In the process ADB

Work done by the gas = Area under AD + Area under DB

= (10 kPa) * (25 cc -10 cc) + 0

=10000 N/m² * 15x10⁻⁶ m³

= 0.15 J.


In the process ACB

The work done by the gas =Area under AC +Area under CB
= 0 + (30 kPa)*(25 cc -10 cc)
= 30000 N/m² * 15x10⁻⁶ m³
=0.45 J.


 

  8. When a system is taken through the process abc shown in Figure (26-E3), 80 J of heat is absorbed by the system, and 30 J of work is done by it. If the system does 10 J of work during the process adc, how much heat flows into it during the process?   
The figure for Q-8


Answer: The difference in internal energies (ΔU) between states at points a and c will be the same whatever the process is adopted.

     Work done in the process abc, ΔW =30 J (Given), and heat given ΔQ =80 J.

Since ΔU =ΔQ -ΔW =80 -30 J =50 J.

In the second process, adc, ΔW = 10 J.

Now ΔU =ΔQ -ΔW

→50 =ΔQ - 10 

→ΔQ = 50+10 =60 J.

Hence, 60 J of heat flows in this process. 




 

  9. 50 cal of heat should be supplied to take a system from the state A to the state B through the path ACB as shown in figure (26-E4). Find the quantity of heat to be supplied to take it from A to B via ADB.   
The figure for Q-9


Answer:  For the process ACB,

ΔQ = 50 cal=50*4.2 J= 210 J, 

ΔW =area between the curve ACB and the V-axis 

=(50 kPa)*(400-200)cc

=(50000 N/m²)*(200x10⁻⁶ m³)

=10 J,

Now ΔU =ΔQ -ΔW =210 -10 =200 J.

The change in the internal energy between the states A and B, ΔU, will be the same in both processes. For the process ADB (from the figure)

ΔW =(155 kPa)*(400-200)cc

     =(155x1000 N/m²)*(200x10⁻⁶ m³)

     =31 J

From ΔU =ΔQ -ΔW

→200 J =ΔQ -31 J

→ΔQ =200+31 =231 J =(231/4.2) cal

        =55 cal.

So 55 cal of heat is needed to take the system from state A to B through the process ADC.   


 



  10. Calculate the heat absorbed by a system in going through the cyclic process shown in figure (26-E5). 
The figure for Q-10
 


Answer:  Since it is a cyclic process, the initial and final states are the same; hence the change in the internal energy ΔU =0. 

Now ΔQ -ΔW =ΔU =0

→ΔQ = ΔW

So the heat absorbed is equal to the work done. In the full cycle, sometimes work is done by the system, and sometimes work is done on the system. So the work done by the system is sometimes positive and sometimes negative. Since the work done by the system is the area under the curve and the V-axis, the net work done by the system is equal to the area of the circle,

=πd²/4

=π(200)²*1000*10⁻⁶/4 J

=40π/4 J
=31.4 J = Heat absorbed.  




   11. A gas is taken through a cyclic process ABCA as shown in Figure (26-E6). If 2.4 cal of heat is given in the process, what is the value of J?
The figure for Q-11
   

Answer:  From the laws of thermodynamics,

ΔQ = ΔU +ΔW, (Here ΔU =0, for a cyclic process).

→ΔQ = ΔW =ΔWₐᵦ+ΔWᵦ₍+ΔW₍ₐ 

{Area under the curve and volume graph}  

→(2.4 cal)*J = 0 + {½(200x10⁻⁶*100x10³)+(200x10⁻⁶*100x10³)} - (200x10⁻⁶*100x10³)

→(2.4 cal)*J =(3/2 -1)*(20) =10 Joules

→J = 10/2.4 = 4.17 J/cal 

 




  12. A substance is taken through a process abc as shown in Figure (26-E7). If the internal energy of the substance increases by 5000 J and heat of 2625 cal is given to the system, calculate the value of J.
(26-E7)The figure for Q-12
   


Answer:  ΔQ =ΔU +ΔW 

Here ΔQ = 2625 cal =2625*J joules

ΔU = 5000 J

ΔW =ΔWₐᵦ +ΔWᵦ₍

     =(0.05-0.02)*200x10³ + 0 joules

     =6000 J

2625*J =5000 +6000 =11000

→J =11000/2625 =4.19 J/cal.

   



  13. A gas is taken along the path AB as shown in Figure (26-E8). If 70 cal of heat is extracted from the gas in the process, calculate the change in the internal energy of the system.
The figure for Q-13
   


Answer:  Work done by the system,

ΔW =-(250-100)x10⁻⁶*{200 +½(500-200)x10³ J

→ΔW = -150*350*10⁻³ J = -52.5 J

ΔQ = -70 cal = -70*4.186 J

So, ΔU =ΔQ -ΔW =-70*4.186-(-52.5) J

           =241 J =Change in internal energy.   




 

  14. The internal energy of a gas is given by U = 1.5pV. It expands from 100 cm³ to 200 cm³ against a constant pressure of 1.0x10⁵ Pa. Calculate the heat absorbed by the gas in the process.   


Answer:  ΔQ =ΔU+ΔW

ΔW =pV =1.0x10⁵*(200-100)x10⁻⁶ J

=10 J

Change in internal energy

ΔU =1.5*1.0x10⁵*(200-100)x10⁻⁶ J 

=15 J

Heat absorbed ΔQ =10 J +15 J =25 J.





 

  15. A gas is enclosed in a cylindrical vessel fitted with a frictionless piston. The gas is slowly heated for some time. During the process, 10 J of heat is supplied, and the piston is found to move out 10 cm. Find the increase in the internal energy of the gas. The area of the cross-section of the cylinder = 4 cm² and the atmospheric pressure = 100 kPa.    


Answer: Work done by the gas =p*dV

→ΔW=p*dV=100x10³*(4x10⁻⁴*10/100)

=4 J.
Heat given, ΔQ = 10 J.
Change in internal energy =ΔU =?
From the first law of thermodynamics,  

ΔQ =ΔU +ΔW

→10 = ΔU + 4

→ΔU = 10 - 4 = 6 J.   





 

  16. A gas is initially at a pressure of 100 kPa, and its volume is 2.0 m³. Its pressure is kept constant, and the volume is changed from 2.0 m³ to 2.5 m³. Its volume is now kept constant, and the pressure is increased from 100 kPa to 200 kPa. The gas is brought back to its initial state, the pressure varying linearly with its volume. (a) Whether heat is supplied to or extracted from the gas in the complete cycle? (b) How much heat was supplied or extracted?   


Answer:  At initial state A,

p = 100 kPa, V = 2.0 m³

At 2nd state B, p=100 kPa, V= 2.5 m³.

Hence dV =0.5 m³
Diagram for Q-16

Hence work done (for process AB)

= p*dV 

=(100 kPa)*(0.5 m³)

=100x1000*0.5 J

=50000 J 

At the state C, p=200 kPa, V=2.5 m³

So for process BC, dV = 0. Thus, the work done = zero. 

For the process CA,

dV = -0.5 m³. Work done = area under CA and V axis.

=-(100x1000*0.5+½*100x1000*0.5) J

=-1.5*50000 J

=-75000 J

Net work done in the process ABCA,

ΔW =50000+0-75000 =-25000 J

For a cyclic process, ΔU =0.

Hence from,

ΔQ =ΔU+ΔW

→ΔQ = 0 -25000 J =-25000 J.


(a) A negative sign shows that heat is extracted from the gas in the complete cycle.


(b) The amount of heat extracted =25000 J.     




 

  17. Consider the cyclic process ABCA, shown in Figure (26-E9), performed on a sample of 2.0 moles of an ideal gas. A total of 1200 J of heat is withdrawn from the sample in the process. Find the work done by the gas during the part BC.
The figure for Q-17
   

Answer:  For the complete cycle, ΔQ = -1200 J, ΔU = 0. From the first law of thermodynamics,

ΔQ =ΔU+ΔW

→-1200 = 0+ ΔW

→ΔW = -1200 J = work done in the whole cycle.

During the process of CA, the volume does not change. So no work is done in this process.

During the process AB, the temperature is proportional to the volume; hence, here the pressure must be constant. 

Now pV = nRT
At A, pVₐ = nR(300)
At B, pVᵦ = nR(500)
Hence, pVᵦ-pVₐ =nR(500-300)
→p(Vᵦ-Vₐ) =nR*200
→p*𐊅V =2*8.3*200
→Work done in the process AB=3320 J. 

Now the total work done in the process AB, BC, and CA = 3320 +Wᵦ₍+0
=Wᵦ₍+3320 J.
But the work done in the whole cycle =-1200 J. Hence,
Wᵦ₍ +3320 =-1200
→Wᵦ₍ =-3320-1200 =4520 J. 

 


 

  18. Figure (26-E10) shows the variation in the internal energy U with the volume V of 2.0 mole of an ideal gas in the cyclic process abcda. The temperatures of the gas at b and c are 500 K and 300 K, respectively. Calculate the heat absorbed by the gas during the process.
The figure for Q-18
   


Answer:  During the cyclic process, 𐊅U = 0. Hence 𐊅Q = 𐊅W. In the given picture, the volume during the processes bc and da is constant. Hence, no work is done during these processes. During the processes ab and cd, the temperature is constant, and these are isothermal processes.

The temperature at a and b is T = 500 K.

The temperature at c and d is T' = 300 K.

Work done during an isothermal process is given as W = nRT*ln(V₂/V₁).

The work done during the process ab is W= nRT*ln(2V₀/V₀) =nRT*ln(2), 

Similarly, the work done during the process cd, W' = nRT'*ln(V₀/2V₀) =nRT'*ln(1/2).

So the total work done during the cyclic process abcda, 𐊅W = W+W'

→𐊅W =nRT*ln(2) +nRT'*ln(1/2)

=nRT*ln(2) +nRT'*ln(1) -nRT'*ln(2)

=nR*ln(2){T-T'} +nRT'*0
{Since ln(1) =0}
=nR*ln(2){500-300}
=2*8.3*0.693*200
=2300 J.

As we have seen in the beginning, here 𐊅Q =𐊅W

The total heat absorbed during this cyclic process =𐊅Q =𐊅W =2300 J.

  


 


  19. Find the change in the internal energy of 2 kg of water as it is heated from 0°C to 4°C. The specific heat capacity of water is 4200 J/kg-K and its densities at 0°C and 4°C are 999.9 kg/m³ and 1000 kg/m³, respectively. Atmospheric pressure = 10⁵ Pa.   


Answer:  𐊅Q = ms*𐊅T

=2*4200*4 =33600 J.

Pressure, p = 10⁵ Pa.

Change in volume, 

𐊅V =2/1000-2/999.9 =2.0x10⁻⁷ m³

Hence 𐊅W =p*𐊅V =10⁵*2.0x10⁻⁷ J

               =0.02 J.

From the first law of thermodynamics, 𐊅Q =𐊅W+𐊅U

→𐊅U =𐊅Q -𐊅W =(33600 -0.02) J.

= The change in the internal energy. 


 


  20. Calculate the increase in the internal energy of 10 g of water when it is heated from 0°C to 100°C and converted into steam at 100 kPa. The density of steam = 0.6 kg/m³. Specific heat capacity of water = 4200 J/kg-°C and the latent heat of vaporization of water = 2.5x10⁶ J/kg.   


Answer:  Heat given to water in the process 𐊅Q =Heat given to water to raise the temperature from 0°C to 100°C + Heat given to convert the water into steam

=ms*𐊅T +mL

=(10/1000)*4200*100 +(10/1000)*2.5x10⁶

=4200 +25000 

=29200 J

The density of water varies with temperature, but the density of steam is much less. Hence, the change in the volume of water from 0°C to 100°C is negligible in comparison to the steam. Taking the density of water =1000 kg/m³, volume =(10/1000)/1000 m³ 

=1x10⁻⁵ m³ 

The density of steam at 100°C (given) =0.6 kg/m³, volume =(10/1000)/0.6 m³

=0.01666 m³

Chang in volume, 𐊅V =(0.01666 -1x10⁻⁵) m³

Pressure, p = 100 kPa =1x10⁵ Pa

The work done, 𐊅W =p*𐊅V

=1x10⁵*(0.01666-1x10⁻⁵) J

=1665 J.

If 𐊅U = change in internal energy, then from the first law of thermodynamics,

𐊅Q =𐊅W +𐊅U

→𐊅U =𐊅Q -𐊅W =29200-1665 = 27535 J =2.75x10⁴ J.       



 

 

 

  21. Figure (26E-11) shows a cylindrical tube of volume V with adiabatic walls containing an ideal gas. The internal energy of this ideal gas is given by 1.5nRT. The tube is divided into two equal parts by a fixed diathermic wall. Initially, the pressure and the temperature are p₁, T₁ on the left and p₂, T₂ on the right. The system is left for sufficient time so that the temperature becomes equal on the two sides. (a) How much work has been done by the gas on the left part? (b) Find the final pressure on the two sides. (c) Find the final equilibrium temperature. (d) How much heat has flown from the gas on the right to the gas on the left?
The figure for Q-21
    


Answer:  (a) Since the diathermic wall is fixed, there is no change in the volumes of the gases on either side. So zero work is done by the gas on the left part.   


Let us first answer part (c)
(c) Let the final temperature on both parts = T. n₁ =Number of moles on the left part. n₂ =Number of moles on the right part. 
  
Now, p₁(V/2) =n₁RT₁, p₂(V/2)=n₂RT₂,
→n₁ =p₁V/2RT₁ and n₂ =p₂V/2RT₂
Total number of moles, n =n₁+n₂
=(p₁V/2RT₁ +p₂V/2RT₂)
Internal energy is given as 1.5nRT; hence the internal energy on the left part, U₁=1.5n₁RT₁,
On the right part, U₂ =1.5n₂RT₂
The total internal energy of the gas initially, U =U₁+U₂ 
=1.5n₁RT₁ +1.5n₂RT₂ 
=1.5(p₁V/2RT₁)RT₁+1.5(p₂V/2RT₂)RT₂  =0.75p₁V+0.75p₂V
=0.75V(p₁+p₂) -------- (i)
Internal energy when equilibrium is reached =1.5nRT
=1.5(p₁V/2RT₁ +p₂V/2RT₂)RT --(ii) 

Since no work is done by the gases on either part and no heat is transferred due to adiabatic walls, the internal energy will remain the same. Equating (i) and (ii),

1.5(p₁V/2RT₁ +p₂V/2RT₂)RT=0.75V(p₁+p₂) 
→0.75(p₁/2T₁+p₂/2T₂)T=0.75(p₁+p₂)
→T =(p₁+p₂)/(p₁/T₁+p₂/T₂)
=T₁T₂(p₁+p₂)/(p₁T₂+p₂T₁)
=T₁T₂(p₁+p₂)/𝜆 =Final equilibrium temperature.
where 𝜆 =p₁T₂+p₂T₁

Now part (b)  
Assume that the final pressure on the left part =p, on the right part =p'  
For the left part,
p₁(V/2)/T₁ =n₁R =p(V/2)/T 
→p =p₁T/T₁
→p =p₁T₁T₂(p₁+p₂)/𝜆T₁ 
{Putting the value of T)
→p =p₁T₂(p₁+p₂)/𝜆 

Similarly, for the right side, the final pressure p' =p₂T/T₂ 
→p' =p₂T₁T₂(p₁+p₂)/𝜆T₂
→p' =p₂T₁(p₁+p₂)/𝜆  

(d) Since no work is done by the gas on the right side, the change in the internal energy of the gas on the right side will be the heat flown from the right side.
The change in the internal energy of the gas on the right side =1.5n₂R(T₂-T) 

=1.5(p₂V/2RT₂)R{T₂-T₁T₂(p₁+p₂)/𝜆}    =(3/4)(p₂V){1-T₁(p₁+p₂)/(p₁T₂+p₂T₁)}
Putting the value of 𝜆 above 
=(3/4)(p₂V){(p₁T₂+p₂T₁-p₁T₁-p₂T₁)/𝜆} 
=(3/4)(p₂V){(p₁T₂-p₁T₁)/𝜆} 
=3p₁p₂(T₂-T₁)V/4𝜆    



  22. An adiabatic vessel of total volume V is divided into two equal parts by a conducting separator. The separator is fixed in this position. The part on the left contains one mole of an ideal gas (U = 1.5nRT) and the part on the right contains two moles of the same gas. Initially, the pressure on each side is p. The system is left for sufficient time so that a steady state is reached. Find (a) the work done by the gas on the left part during the process. (b) The temperature on the two sides in the beginning, (c) the final common temperature reached by the gases. (d) the heat given to the gas in the right part and (e) the increase in the internal energy of the gas in the left part.    


Answer:  (a) Since the separator is fixed in position, the volume of the gas does not change on the left part; hence the work done by the gas in the left part during the process is zero.  


(b) Let the temperatures in the beginning and at the left and right parts be T₁, T₂, respectively. 

So for the left side, p(V/2)=(1 mol)RT₁

T₁ =pV/(2 mol)R

For the right side, p(V/2)=(2 mol)RT₂

T₂ =pV/(4 mol)R


(c) Let the final temperature =T 

Initial internal energy, U =U₁+U₂

=1.5n₁RT₁+1.5n₂RT₂ 

=1.5*1*R*pV/2R +1.5*2*R*pV/4R

=3pV/4 +3pV/4

=6pV/4

=1.5pV  

The final internal energy =1.5(3 mol)RT

=(4.5 mol)RT    

Since no work is done by the gas as there is no volume change, the change in internal energy will be the heat given. Since the walls of the vessel are adiabatic, no heat is exchanged. Thus the internal energy does not change.

Equating the initial and final internal energies, we get,
(4.5 mol)RT =1.5pV
►T =1.5pV/(4.5 mol)R
      = pV/(3 mol)R


(d) The heat given to the gas in the right part = Increase in the internal energy of the right part,

=1.5n₂R(T-T₂)
=1.5*2*R(pV/3R -pV/4R)
=3(pV/3 -pV/4)
=3(pV/12)
= pV/4

(e) Since the walls are adiabatic, the heat given to the right part comes from the left part. Hence, for the left part, ΔQ = -pV/4
Since for the left part ΔW=0, ΔQ =ΔU
→ΔU= increase in the internal energy of the left part =ΔQ = -pV/4. 


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Part-II

Solutions - "Concepts of Physics" Part-II, by H C Verma

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- 24 - Kinetic Theory of Gases

Questions for Short Answer

OBJECTIVE - I

OBJECTIVE - II

EXERCISES - Q1 to Q10

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-62

DOCTOR EXTRA SOFT

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