For a diatomic gas: Let \({\gamma }_{1}=\frac{{C}_{p}}{{C}_{v}}\) for a rigid molecule (without vibrational modes). Let …

ExplanationKey Concepts:

1. Degrees of Freedom:

   • A rigid diatomic molecule has 5 degrees of freedom (3 translational + 2 rotational).
   • A diatomic molecule with vibrational modes has additional vibrational degrees of freedom.

2. Specific Heats (\({C}_{v}\)​ and \({C}_{p}\)):

   • The internal energy is proportional to the degrees of freedom. For a rigid diatomic gas: \({C}_{v}=\frac{5R}{2}\)​
   • When vibrational modes are included, the total degrees of freedom increase. Each vibrational mode adds \(2R\), making: \({C}_{v}=\frac{5R}{2}+2R=\frac{9R}{2}\)​
   • \({C}_{p}\)​ is related to \({C}_{v}\) by: \({C}_{p}={C}_{v}+R\)

3. Adiabatic Index (\(\gamma\)γ):

   • The adiabatic index \(\gamma\)γ is defined as: \(\gamma =\frac{{C}_{p}}{{C}_{v}}=1+\frac{R}{{C}_{v}}\)​
   • For the rigid diatomic gas: \({\gamma }_{1}=\frac{{C}_{p}}{{C}_{v}}=\frac{\frac{7R}{2}}{\frac{5R}{2}}=\frac{7}{5}=1.4\)
   • For the diatomic gas with vibrational modes: \({\gamma }_{2}=\frac{{C}_{p}}{{C}_{v}}=\frac{\frac{11R}{2}}{\frac{9R}{2}}=\frac{11}{9}\approx 1.22\)

Step 2: Compare \({\gamma }_{1}\)​ and \({\gamma }_{2}\)​

Clearly, \({\gamma }_{2}<{\gamma }_{1}\) because the inclusion of vibrational modes increases \({C}_{v}\)​ more significantly than \({C}_{p}\), leading to a smaller \(\gamma\).

Final Answer:

(A) \({\gamma }_{2}<{\gamma }_{1}\)​