Question

$ {C_p} - {C_v} = R $ . This $ R $ is:
(A). Change in $ KE $
(B). Change in rotational energy
(C). Work done which system can do, on expanding the gas per mol per degree increase in temperature
(D). All are correct

Answer 1

Hint :The energy of an ideal gas is proportional to its absolute temperature. Using the expression for work done and ideal gas equation, we can deduce a relationship with universal gas constant $ R $ .

Complete Step By Step Answer:
It is given that, $ {C_p} - {C_v} = R $ . Here, $ {C_p} $ is heat capacity at constant pressure, $ {C_v} $ is the heat capacity at constant volume and $ R $ is universal gas constant.
(A) Kinetic energy of an ideal gas is proportional to absolute temperature. Therefore, change in kinetic energy cannot be equal to $ R $ , which is a constant.
(B) Rotational energy of an ideal gas is proportional to absolute temperature. Therefore, change in rotational energy cannot be equal to $ R $ , which is a constant.
(C) Work done by system on expansion of gas, $ w = p\Delta V $
From the ideal gas equation, $ pV = nRT $
Thus, $ \Delta (pV) = \Delta (nRT) $
For expansion of gas, pressure, $ p $ is constant and volume, $ V $ is increasing. For one mole of gas, $ n = 1 $ and for per degree increase in temperature, $ \Delta T = 1 $ .
Simplifying the above expression, we get:
 $ p\Delta V = nR\Delta T $
Substituting the values of $ n $ and $ \Delta T $ in this expression:
 $ p\Delta V = 1 \times R \times 1 $
 $ \Rightarrow w = p\Delta V = R $
Hence, option (C) is correct.

Note :
In this question, the given expression, $ {C_p} - {C_v} = R $ implies that the gas is behaving ideally. This is the reason we used the ideal gas equation. Then, we assumed that the gas is expanding at a constant pressure and therefore, we expressed the work done as, $ w = p\Delta V $ .