Consider, two ideal diatomic gases A and B at some temperature T. Molecules of the gas A are rigid, and have a mass m. Molecules of the gas B have an additional vibrations mode, and have a mass `m/4`. The ratio of the specific heats (`C_(V)^(A)` and `C_(V)^(B)`) of gas A and B, respectively is:

Options

A. `7:9`
B. `3:5`
C. `5:9`
D. `5:7`

Answer

To find the ratio of the specific heats \(C_V^A\) and \(C_V^B\) for the two ideal diatomic gases A and B, we will follow these steps:

### Step 1: Identify the degrees of freedom for each gas
- For gas A (rigid diatomic gas):
- The degrees of freedom \(f_A\) = 5 (3 translational + 2 rotational).

- For gas B (diatomic gas with an additional vibrational mode):
- The degrees of freedom \(f_B\) = 6 (3 translational + 2 rotational + 1 vibrational).

### Step 2: Use the formula for molar specific heat at constant volume
The molar specific heat at constant volume \(C_V\) for a gas can be calculated using the formula:
\[ C_V = \frac{f}{2} R \]
where \(f\) is the degrees of freedom and \(R\) is the universal gas constant.

### Step 3: Calculate \(C_V^A\) and \(C_V^B\)
- For gas A:
\[ C_V^A = \frac{f_A}{2} R = \frac{5}{2} R \]

- For gas B:
\[ C_V^B = \frac{f_B}{2} R = \frac{6}{2} R = 3R \]

### Step 4: Find the ratio of the specific heats
Now we can find the ratio of the specific heats \(C_V^A\) to \(C_V^B\):
\[ \frac{C_V^A}{C_V^B} = \frac{\frac{5}{2} R}{3R} = \frac{5}{2} \cdot \frac{1}{3} = \frac{5}{6} \]

### Final Answer
The ratio of the specific heats \(C_V^A\) and \(C_V^B\) is:
\[ \frac{C_V^A}{C_V^B} = \frac{5}{6} \]

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