One mole of a diatomic gas undergoes a thermodynamic process, whose process equation is `P prop V^(2)`. The molar specific heat of the gas is

Options

A. `(17R)/3`
B. `(17R)/6`
C. `(15R)/4`
D. `(15R)/8`

Answer

To find the molar specific heat of a diatomic gas undergoing a thermodynamic process described by the equation \( P \propto V^2 \), we can follow these steps:

### Step 1: Understand the relationship
The given relationship \( P \propto V^2 \) implies that \( P = k V^2 \) for some constant \( k \). This suggests that we can express the relationship in terms of a polytropic process.

### Step 2: Identify the polytropic process
In a polytropic process, the relationship can be expressed as:
\[ PV^n = \text{constant} \]
Comparing this with our equation \( P = k V^2 \), we can rewrite it as:
\[ P V^{-2} = k \]
This indicates that \( n = -2 \).

### Step 3: Use the formula for molar specific heat
The formula for the molar specific heat \( C \) in a polytropic process is given by:
\[ C = C_v + \frac{R}{1 - n} \]
where \( C_v \) is the molar specific heat at constant volume, and \( R \) is the universal gas constant.

### Step 4: Determine \( C_v \) for a diatomic gas
For a diatomic gas, the molar specific heat at constant volume \( C_v \) is:
\[ C_v = \frac{5R}{2} \]

### Step 5: Substitute values into the specific heat formula
Now substituting \( C_v \) and \( n \) into the specific heat formula:
\[ C = \frac{5R}{2} + \frac{R}{1 - (-2)} \]
This simplifies to:
\[ C = \frac{5R}{2} + \frac{R}{1 + 2} = \frac{5R}{2} + \frac{R}{3} \]

### Step 6: Find a common denominator and combine the terms
To combine \( \frac{5R}{2} \) and \( \frac{R}{3} \), we find a common denominator, which is 6:
\[ C = \frac{15R}{6} + \frac{2R}{6} = \frac{17R}{6} \]

### Final Answer
Thus, the molar specific heat of the gas is:
\[ C = \frac{17R}{6} \]