Question

If $Cp - Cv = R$. Then $R$ is
(A) change in $KE$.
(B) change in rotational energy.
(C) work done which the system can do on expanding the gas per mol per degree increase in temperature.
(D) all correct.

Answer 1

Hint: If $Cp$ and $Cv$ are measured in the units of work and $R$ is also in the units of work (or energy) the relation between $Cp,{\text{ }}Cv$ and $R$ is,
$Cp - Cv = R$

Complete step by step answer:
The specific heat constants for constant pressure and constant volume processes are related to the gas constant for a given gas. This rather remarkable result has been derived from thermodynamic relations, which are based on observations of physical systems and processes.
$Cp = Cv + R$
From ideal gas, $PV = RT$ at temperature $T$ for one mol and for a small change in volume
$P\left( {V + \Delta V} \right) = R\left( {T + 1} \right)$
At temperature $\left( {T + 1} \right)$ for one mole
Therefore, $W = P\Delta V = R$
$R$ is work done which system can do on expanding the gas per mol per degree increase in temperature.

So, the correct answer is Option C .

Note: $Cp - Cv = R$
This is the reaction between the two specific heats of the gas and is true for all types of gases i.e. perfect gas or an ideal gas or real gas at moderate pressures. It is to be noted that in this formula the values of $Cp,{\text{ }}Cv$ and $R$ must be put in the same system of units. i.e. all these values must either put in $Joule/mole^\circ c$ or in $Cal/mol^\circ c$.
The relation between the specific heat of gas at constant volume and at constant pressure.
$Cp - Cv = \dfrac{R}{M}$
Where, $M$ is the mass of one mole of gas.