Consider two diatomic ideal gases A &B at some temperature T. Molecules of A rigid and have mass 2m. Molecules of B have vibrational modes in addition and have mass m.The ratio of the specific heats `( C_(v) ^(4) ` & ` C_(v) ^(B))` of gas A &B,if both the molecules have planar structure is `:`

Options

A. `5/7`
B. `5/6`
C. `6/7`
D. `3/4`

Explanation

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

### Step 1: Identify the degrees of freedom for each gas
- For gas A (rigid diatomic molecules):
- Since the molecules are diatomic and planar, they have 5 degrees of freedom (3 translational + 2 rotational).

- For gas B (diatomic molecules with vibrational modes):
- Gas B has 3 translational, 2 rotational, and 2 vibrational modes (since each vibrational mode contributes 1 degree of freedom). Therefore, the total degrees of freedom for gas B is 7.

### Step 2: Use the formula for molar heat capacity at constant volume
The molar heat capacity at constant volume \( C_{v} \) is given by 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:
\[ C_{v}^{A} = \frac{5}{2} R \]

- For gas B:
\[ C_{v}^{B} = \frac{7}{2} R \]

### Step 4: Find the ratio \( \frac{C_{v}^{A}}{C_{v}^{B}} \)
Now, we can find the ratio of the specific heats:
\[ \frac{C_{v}^{A}}{C_{v}^{B}} = \frac{\frac{5}{2} R}{\frac{7}{2} R} \]
The \( R \) and \( \frac{1}{2} \) cancel out:
\[ \frac{C_{v}^{A}}{C_{v}^{B}} = \frac{5}{7} \]

### Final Answer
The ratio of the specific heats \( C_{v}^{A} : C_{v}^{B} \) is:
\[ \frac{5}{7} \]

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