Deduce an expression for molar specific heat of a monoatomic gas at constant volume.Find kinetic energy of 4000 cc of a gas at S.T.P.[ Given : Standard pressure is 1.013 × 105 N/m2 ]

Answer: \(C_V = \dfrac{3}{2}R\); kinetic energy \(\approx 607.8\) J.

(i) Molar specific heat at constant volume: A molecule of a monoatomic gas has only three translational degrees of freedom (along x, y and z). By the law of equipartition of energy, each degree of freedom has average energy \(\frac{1}{2}k_BT\), so the average energy per molecule is

The internal energy of one mole (\(N_A\) molecules) is

At constant volume no work is done, so all the heat supplied increases the internal energy. Hence

Also \(C_P = C_V + R = \dfrac{5}{2}R\) and \(\gamma = \dfrac{C_P}{C_V} = \dfrac{5}{3}\).

(ii) Numerical: \(V = 4000\) cc \(= 4\times10^{-3}\) m\(^3\), \(P = 1.013\times10^5\) N/m\(^2\).

From kinetic theory, \(P = \dfrac{2}{3}\dfrac{E}{V}\), so the total kinetic energy is