Prove the Mayer's relation : \(C_p - C_v = \dfrac{R}{J}\)

Answer: Using the first law for one mole of an ideal gas heated at constant volume and at constant pressure, the extra heat at constant pressure equals the work \(P\,dV = R\,dT\), giving \(C_p - C_v = R/J\).

Consider one mole of an ideal gas. Let its temperature rise by \(dT\).

At constant volume: no work is done, so all the heat supplied increases the internal energy: \(dQ_1 = C_v\,dT = dE\) … (1) (in heat units, \(dE\) being the increase in internal energy expressed in heat units).

At constant pressure: the gas also expands by \(dV\) and does work \(P\,dV\): \(dQ_2 = C_p\,dT = dE + \dfrac{P\,dV}{J}\) … (2)

The internal energy of an ideal gas depends only on temperature, so \(dE\) is the same in both cases for the same \(dT\).

Subtracting (1) from (2): \((C_p - C_v)\,dT = \dfrac{P\,dV}{J}\) … (3)

For one mole at constant pressure, \(PV = RT\) gives \(P\,dV = R\,dT\) … (4)

From (3) and (4): \((C_p - C_v)\,dT = \dfrac{R\,dT}{J}\), so

\(C_p - C_v = \dfrac{R}{J}\).

Here \(J\) is the mechanical equivalent of heat. If heat and work are in the same units (joules), \(C_p - C_v = R\). This is Mayer's relation; it shows \(C_p \gt C_v\).