Explain the thermodynamics of adiabatic process to obtain adiabatic constant for rigid diatomic molecule of a gas?

Answer: \(\gamma = \dfrac{C_p}{C_v} = \dfrac{7}{5} = 1.4\)

In an adiabatic process no heat enters or leaves the gas (\(dQ = 0\)), so by the first law \(dU = -dW\): work done by the gas comes from its internal energy. The process obeys \(PV^{\gamma} = \text{constant}\), where the adiabatic constant \(\gamma = C_p/C_v\).

A rigid diatomic molecule has 3 translational and 2 rotational degrees of freedom (no vibration), i.e. \(f = 5\). By the law of equipartition, each degree of freedom carries \(\frac{1}{2}k_BT\) per molecule, so for one mole

By Mayer’s relation \(C_p = C_v + R = \dfrac{7}{2}R\). Hence