Question

Let \(\gamma_1\)be the ratio of molar specific heat at constant pressure and molar specific heat at constant volume of a monoatomic gas and \(\gamma_2\) be the similar ratio of diatomic gas. Considering the diatomic gas molecule as a rigid rotator, the ratio, \(\frac{\gamma_1}{\gamma_2}\) is :

• A. \(\frac{21}{25}\)
• B. \(\frac{25}{21}\)
• C. \(\frac{35}{27}\)
• D. \(\frac{27}{35}\)

Answer 1

The Correct Option is B

To solve this problem, we need to understand the concepts of molar specific heat capacities and the ratio \( \gamma \) (gamma). The gamma ratio for a gas is defined as:

\(\gamma = \frac{C_p}{C_v}\)

where:

• \(C_p\) is the molar specific heat at constant pressure.
• \(C_v\) is the molar specific heat at constant volume.

For a monoatomic gas:

• \(C_v = \frac{3}{2}R\)
• \(C_p = C_v + R = \frac{5}{2}R\)

Thus, for a monoatomic gas:

\(\gamma_1 = \frac{C_p}{C_v} = \frac{\frac{5}{2}R}{\frac{3}{2}R} = \frac{5}{3}\)

For a diatomic gas considered as a rigid rotator:

• \(C_v = \frac{5}{2}R\)
• \(C_p = C_v + R = \frac{7}{2}R\)

Thus, for a diatomic gas:

\(\gamma_2 = \frac{C_p}{C_v} = \frac{\frac{7}{2}R}{\frac{5}{2}R} = \frac{7}{5}\)

Now, we need to find the ratio of \( \gamma_1 \) to \( \gamma_2 \):

\(\frac{\gamma_1}{\gamma_2} = \frac{\frac{5}{3}}{\frac{7}{5}} = \frac{5}{3} \times \frac{5}{7} = \frac{25}{21}\)

Hence, the ratio \( \frac{\gamma_1}{\gamma_2} \) is \( \frac{25}{21} \), which is the correct answer.