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

Answer

To solve the problem, we need to determine the molar heat capacity of an ideal diatomic gas that undergoes a process where pressure is proportional to volume, and the root mean square (rms) speed of the gas molecules doubles from its initial value.

### Step-by-Step Solution:

1. **Understand the relationship between rms speed and temperature:**
The rms speed \( v_{rms} \) of gas molecules is given by the equation:
\[ v_{rms} = \sqrt{\frac{3RT}{M}} \]
where \( R \) is the universal gas constant, \( T \) is the absolute temperature, and \( M \) is the molar mass of the gas.

2. **Initial and final rms speeds:**
Let the initial rms speed be \( v_{rms1} \) and the final rms speed be \( v_{rms2} \). According to the problem, \( v_{rms2} = 2 v_{rms1} \).

3. **Relate rms speed to temperature:**
From the rms speed equation, we can write:
\[ v_{rms1} = \sqrt{\frac{3RT_1}{M}} \quad \text{and} \quad v_{rms2} = \sqrt{\frac{3RT_2}{M}} \]
Therefore, we have:
\[ 2 v_{rms1} = \sqrt{\frac{3RT_2}{M}} \]
Substituting \( v_{rms1} \):
\[ 2 \sqrt{\frac{3RT_1}{M}} = \sqrt{\frac{3RT_2}{M}} \]
Squaring both sides gives:
\[ 4 \frac{3RT_1}{M} = \frac{3RT_2}{M} \]
Simplifying this results in:
\[ 4T_1 = T_2 \quad \Rightarrow \quad T_2 = 4T_1 \]

4. **Using the relationship between pressure and volume:**
Given that pressure \( P \) is proportional to volume \( V \), we can write:
\[ P = kV \quad \text{(for some constant } k\text{)} \]
Using the ideal gas law \( PV = nRT \), we can express this as:
\[ kV^2 = nRT \]
Since \( P \propto V \), we can substitute \( T_1 \) and \( T_2 \) into this equation:
\[ kV_1^2 = nRT_1 \quad \text{and} \quad kV_2^2 = nRT_2 \]
Substituting \( T_2 = 4T_1 \):
\[ kV_2^2 = nR(4T_1) \quad \Rightarrow \quad V_2^2 = 4 \frac{nRT_1}{k} \]
Since \( V_2^2 = 4V_1^2 \), we can conclude that:
\[ V_2 = 2V_1 \]

5. **Finding the molar heat capacity:**
The molar heat capacity \( C \) in a process can be expressed as:
\[ C = C_V + R \left( \frac{\Delta T}{\Delta T} \right) \]
Since \( C_V = \frac{5R}{2} \) for a diatomic gas, we need to find the heat capacity for the process where \( P \propto V \).

The relationship between heat capacities in this case can be derived from the first law of thermodynamics and the fact that:
\[ C = C_V + \frac{R}{\gamma - 1} \]
where \( \gamma \) is the ratio of specific heats. For a diatomic gas, \( \gamma = \frac{C_P}{C_V} = \frac{7R/2}{5R/2} = \frac{7}{5} \).

Using this, we can find:
\[ C = \frac{5R}{2} + R \left( \frac{1}{\frac{7}{5} - 1} \right) = \frac{5R}{2} + R \left( \frac{5}{2} \right) = \frac{5R}{2} + \frac{5R}{2} = 5R \]

### Final Answer:
The molar heat capacity of the gas in the given process is \( 5R \).