Question

The \(\gamma \left( = \frac{C_p}{C_v} \right)\) values of a rigid diatomic gas molecule and a monoatomic gas molecule are in the ratio

• A. 5 : 3
• B. 21 : 25
• C. 7 : 5
• D. 25 : 21
• E. 7 : 3

Hint:

Rigid diatomic means vibration degrees of freedom are not counted. Always remember \(\gamma_{mono} = 1.67\) and \(\gamma_{di} = 1.4\). The ratio \(1.4 / 1.67\) is clearly less than 1, helping you eliminate options A, C, D, and E.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The ratio of specific heats \(\gamma\) depends on the degrees of freedom (\(f\)) of the gas molecules.
For any gas, \(\gamma = 1 + \frac{2}{f}\).

Step 2: Key Formula or Approach:
1. Monoatomic gas: \(f = 3 \implies \gamma_m = 1 + \frac{2}{3} = \frac{5}{3}\).
2. Rigid Diatomic gas: \(f = 5 \implies \gamma_d = 1 + \frac{2}{5} = \frac{7}{5}\).

Step 3: Detailed Explanation:
We need to find the ratio of \(\gamma\) for diatomic to monoatomic:
\[ \text{Ratio} = \frac{\gamma_d}{\gamma_m} = \frac{7/5}{5/3} \]
\[ \text{Ratio} = \frac{7}{5} \times \frac{3}{5} \]
\[ \text{Ratio} = \frac{21}{25} \]

Step 4: Final Answer:
The ratio is 21 : 25.