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Question

(B) For a diatomic gas,the degrees of freedom $f_1 = 5$.The molar specific heat capacity at constant volume is $C_{v_1} = \frac{f_1 R}{2} = \frac{5 R}

Vedclass · Accepted Answer

$\frac{3 C}{7}$

Vedclass · Answer

(B) For a diatomic gas,the degrees of freedom $f_1 = 5$.The molar specific heat capacity at constant volume is $C_{v_1} = \frac{f_1 R}{2} = \frac{5 R}{2}$.The molar specific heat capacity at constant pressure is $C_{p_1} = C_{v_1} + R = \frac{5 R}{2} + R = \frac{7 R}{2} = C$.For a monoatomic gas,the degrees of freedom $f_2 = 3$.The molar specific heat capacity at constant volume is $C_{v_2} = \frac{f_2 R}{2} = \frac{3 R}{2}$.Now,taking the ratio $\frac{C}{C_{v_2}} = \frac{\frac{7 R}{2}}{\frac{3 R}{2}} = \frac{7}{3}$.Therefore,$C_{v_2} = \frac{3 C}{7}$.

The molar specific heat capacity of a diatomic gas at constant pressure is $C$. The molar specific heat capacity of a monoatomic gas at constant volume is

Similar Questions

The ratio of the specific heats $\frac{C_{p}}{C_{v}}=\gamma$ in terms of degrees of freedom $n$ is given by

Obtain the equation for $\gamma = \frac{C_P}{C_V}$ in terms of the degree of freedom $f$.

According to the law of equipartition of energy, the molar specific heat of a diatomic gas at constant volume, where the molecule has one additional vibrational mode, is:

The molar specific heat at constant pressure for a monoatomic gas is

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