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 solve the problem, we need to determine the molar specific heat of one mole of a diatomic gas undergoing a thermodynamic process described by the equation \( P \propto V^2 \). Here are the steps to arrive at the solution:

### Step 1: Understand the given relationship
The relationship given is \( P \propto V^2 \). This can be expressed mathematically as:
\[ P = C V^2 \]
where \( C \) is a constant.

### Step 2: Identify the type of process
The equation \( P V^n = \text{constant} \) represents a polytropic process, where \( n \) is the polytropic index. From our equation \( P = C V^2 \), we can rearrange it to fit the polytropic form:
\[ P V^{-2} = C \]
This indicates that \( n = -2 \).

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

### Step 4: Determine \( C_b \) for a diatomic gas
For a diatomic gas, the value of \( C_b \) is:
\[ C_b = \frac{5R}{2} \]

### Step 5: Substitute the values into the formula
Now we can substitute \( C_b \) 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 simplify
To add these fractions, we need a common denominator. The least common multiple of 2 and 3 is 6:
\[ C = \frac{15R}{6} + \frac{2R}{6} = \frac{17R}{6} \]

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