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, $w = p\Delta V$ .