Questions OBJECTIVE - II and Answer Laws of Thermodynamics HC Verma Part II

Q#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). 

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

W = ∫pdV

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 increases, T will also increase. Option (c) is true.

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

Q#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). 

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 the 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.

Q#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) 

The heat given ΔQ to the system and the work done ΔW by the system depends 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 the 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 the thermodynamics,

ΔQ = ΔU + ΔW

ΔU = ΔQ - ΔW will be the same in both processes. Option (d) is true.

Q#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. 

(a) ΔU₁ + ΔU₂ = 0.

(b) ΔU₁ - ΔU₂ = 0.

(c) ΔQ - ΔW = 0.

(d) ΔQ + ΔW = 0.   

Answer:  (a), (c). 

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.  

Q#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). 

From the first law of thermodynamics,

ΔQ = ΔU + ΔW

Given that, ΔW = -ΔU, hence,

ΔQ = 0. So the heat exchanged to 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.