Question

If \( f \) is the number of degrees of freedom of a molecule of a gas and ratio of molar specific heats of a gas, \( \gamma = 1 + \frac{2}{f} \) where \( \gamma = \frac{C_p}{C_v} \). The ratio of \( \gamma \) for monoatomic gas to \( \gamma \) for (rigid) diatomic gas is

• A. \( \frac{25}{21} \)
• B. \( \frac{35}{15} \)
• C. \( \frac{21}{25} \)
• D. \( \frac{15}{35} \)

Hint:

For ideal gases, the ratio of specific heats \( \gamma = 1 + \frac{2}{f} \), where \( f \) is the number of degrees of freedom. For monoatomic gases, \( f = 3 \), and for diatomic gases, \( f = 5 \).

Answer 1

The Correct Option is A

Step 1: Formula for \( \gamma \).
The ratio of specific heats \( \gamma \) for a gas is related to the number of degrees of freedom \( f \) by: \[ \gamma = 1 + \frac{2}{f} \] Step 2: Calculate \( \gamma \) for monoatomic gas.
For a monoatomic gas, the number of degrees of freedom \( f = 3 \), since it has 3 translational degrees of freedom. Therefore: \[ \gamma_{\text{monoatomic}} = 1 + \frac{2}{3} = \frac{5}{3} \] Step 3: Calculate \( \gamma \) for diatomic gas.
For a rigid diatomic gas, the number of degrees of freedom \( f = 5 \) (3 translational + 2 rotational). Therefore: \[ \gamma_{\text{diatomic}} = 1 + \frac{2}{5} = \frac{7}{5} \] Step 4: Calculate the ratio of \( \gamma \) values.
The ratio of \( \gamma \) for the monoatomic gas to the diatomic gas is: \[ \frac{\gamma_{\text{monoatomic}}}{\gamma_{\text{diatomic}}} = \frac{\frac{5}{3}}{\frac{7}{5}} = \frac{5}{3} \times \frac{5}{7} = \frac{25}{21} \] Step 5: Conclusion.
Thus, the ratio of \( \gamma \) for the monoatomic gas to the diatomic gas is \( \frac{25}{21} \), which corresponds to option (A).