Question

Consider two ideal diatomic gases A and B at some temperature T. Molecules of the gas A are rigid and have a mass of m. Molecules of the gas B have an additional vibration mode and have a mass $m/4$ . The ratio of molar specific heat at constant volume of gas A and B is;

• A. $\raisebox{1ex}{$7$}\!\left/ \!\raisebox{-1ex}{$9$}\right.$
• B. $\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$9$}\right.$
• C. $\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$5$}\right.$
• D. $\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$7$}\right.$

Answer 1

The correct option is D

$\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$7$}\right.$

Explanation for correct option:

D $\frac{\mathbf{5}}{\mathbf{7}}$

1. It is given that diatomic gas A has molecules of mass m and another diatomic gas B has mass $\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$4$}\right.$ with an additional vibration mode.
2. We know that the specific heat at constant volume for a diatomic gas is determined as follows: $\left(C_V\right)=\frac{1}{2}\times f\times R$ where $f$ is the degree of freedom and $R$ is gas constant.
3. For gas A, degree of freedom, $f=3+2=5$ (for diatomic gas).
4. So molar specific heat for gas A, ${\left(C_v\right)}_A=\frac{1}{2}\times 5\times R=\frac{5R}{2}$
5. For gas B, degree of freedom, $f=3+2+2=7$ as it has additional vibration mode.
6. So molar specific heat for gas B, ${\left(C_v\right)}_B=\frac{1}{2}\times 7\times R=\frac{7R}{2}$
7. Now, the ratio of molar specific heat at constant volume of gas A and B is: $$\begin{gathered} \frac{{\left(C_V\right)}_A}{{\left(C_V\right)}_B}=\frac{5R}{2}\times \frac{2}{7R} \\ \frac{{\left(C_V\right)}_A}{{\left(C_V\right)}_B}=\frac{5}{7} \end{gathered}$$
8. Therefore, the ratio of molar specific heat at constant volume of gas A and B is $\frac{5}{7}$.

Hence option D is correct, the ratio of molar specific heat at constant volume of gas A and B is $\frac{5}{7}$.