For a certain process, pressure of diatomic gas varies according to the relation `P = aV^2`, where a is constant. What is the molar heat capacity of the gas for this process ?

Options

A. `(17R)/6`
B. `(6R)/(17)`
C. `(13R)/6`
D. `(16R)/7`

Answer

To solve the problem, we need to find the molar heat capacity of a diatomic gas for a process where the pressure varies with volume according to the relation \( P = aV^2 \).

### Step-by-Step Solution:

1. **Identify the given relation**:
The pressure of the gas is given by the equation:
\[ P = aV^2 \]
where \( a \) is a constant.

2. **Rearrange the equation**:
We can express this as:
\[ \frac{P}{V^2} = a \]
or equivalently:
\[ PV^{-2} = \text{constant} \]

3. **Identify the process type**:
The equation \( PV^n = \text{constant} \) indicates that this is a polytropic process. Here, we can identify \( n \) as:
\[ n = -2 \]

4. **Determine the degrees of freedom**:
For a diatomic gas, the degrees of freedom \( F \) is 5 (3 translational and 2 rotational). Thus, the molar heat capacity at constant volume \( C_v \) is given by:
\[ C_v = \frac{5}{2} R \]

5. **Use the formula for molar heat capacity**:
The molar heat capacity \( C \) for a polytropic process can be calculated using the formula:
\[ C = C_v + \frac{R}{1 - n} \]
Substituting \( C_v \) and \( n \):
\[ C = \frac{5}{2} R + \frac{R}{1 - (-2)} \]

6. **Calculate \( C \)**:
Simplifying the equation:
\[ C = \frac{5}{2} R + \frac{R}{1 + 2} = \frac{5}{2} R + \frac{R}{3} \]
To combine these, we need a common denominator:
\[ C = \frac{5}{2} R + \frac{1}{3} R = \frac{15}{6} R + \frac{2}{6} R = \frac{17}{6} R \]

7. **Final result**:
Thus, the molar heat capacity of the gas for this process is:
\[ C = \frac{17}{6} R \]