Molar heat capacity of an ideal gas in the process `PV^(x)` = constant , is given by : `C = (R)/(gamma-1) + (R)/(1-x)`. An ideal diatomic gas with `C_(V) = (5R)/(2)` occupies a volume `V_(1)` at a pressure `P_(1)`. The gas undergoes a process in which the pressure is proportional to the volume. At the end of the process the rms speed of the gas molecules has doubled from its initial value.
The molar heat capacity of the gas in the given process is :-

Options

A. `3R`
B. `3.5R`
C. `4R`
D. `2.5R`

Explanation

To solve the problem, we need to determine the molar heat capacity of an ideal diatomic gas undergoing a specific process where the pressure is proportional to the volume. We will follow the steps outlined in the video transcript to arrive at the solution.

### Step-by-Step Solution:

1. **Understanding the Given Information:**
- We have a diatomic ideal gas with \( C_V = \frac{5R}{2} \).
- The gas occupies a volume \( V_1 \) at a pressure \( P_1 \).
- The process follows the relation \( PV^x = \text{constant} \).
- The pressure is proportional to the volume, which implies \( P = kV \) for some constant \( k \).

2. **Identifying the Value of \( x \):**
- From the relation \( P \propto V \), we can express it as \( P = kV \).
- This can be rearranged to \( PV^{-1} = k \), which indicates that \( x = -1 \) when comparing with the general form \( PV^x = \text{constant} \).

3. **Using the Molar Heat Capacity Formula:**
- The molar heat capacity \( C \) for the process is given by:
\[ C = \frac{R}{\gamma - 1} + \frac{R}{1 - x} \]
- For a diatomic gas, the adiabatic coefficient \( \gamma \) is given by:
\[ \gamma = \frac{C_P}{C_V} = \frac{C_V + R}{C_V} = \frac{\frac{5R}{2} + R}{\frac{5R}{2}} = \frac{\frac{7R}{2}}{\frac{5R}{2}} = \frac{7}{5} \]
- Therefore, \( \gamma - 1 = \frac{7}{5} - 1 = \frac{2}{5} \).

4. **Calculating \( C \):**
- Substitute \( \gamma \) into the heat capacity formula:
\[ C = \frac{R}{\frac{2}{5}} + \frac{R}{1 - (-1)} = \frac{R \cdot 5}{2} + \frac{R}{2} \]
- Simplifying this gives:
\[ C = \frac{5R}{2} + \frac{R}{2} = \frac{6R}{2} = 3R \]

5. **Final Result:**
- The molar heat capacity of the gas in the given process is \( C = 3R \).

### Conclusion:
The molar heat capacity of the ideal diatomic gas in the process where the pressure is proportional to the volume is \( 3R \).