In a certain gas, the ratio of the velocity of sound and root mean square velocity is `sqrt(5//9)`. The molar heat capacity of the gas in a process given by `PT = constant` is.
(Take `R = 2 cal//mol K`). Treat the gas as ideal.

Options

A. `(R)/(2)`
B. `(3R)/(2)`
C. `(5R)/(2)`
D. `(7R)/(2)`

Answer

To solve the problem, we need to find the molar heat capacity of the gas in a process where \( PT = \text{constant} \). Given the ratio of the velocity of sound to the root mean square velocity, we can derive the necessary values step by step.

### Step-by-Step Solution:

1. **Understand the Given Ratio**:
We are given that the ratio of the velocity of sound \( v_s \) to the root mean square velocity \( v_{rms} \) is:
\[ \frac{v_s}{v_{rms}} = \sqrt{\frac{5}{9}} \]

2. **Formulas for Velocities**:
The formula for the velocity of sound in an ideal gas is:
\[ v_s = \sqrt{\frac{\gamma RT}{M}} \]
where \( \gamma \) is the adiabatic index, \( R \) is the gas constant, \( T \) is the temperature, and \( M \) is the molar mass of the gas.

The formula for the root mean square velocity is:
\[ v_{rms} = \sqrt{\frac{3RT}{M}} \]

3. **Set Up the Equation**:
Using the given ratio:
\[ \frac{v_s}{v_{rms}} = \frac{\sqrt{\frac{\gamma RT}{M}}}{\sqrt{\frac{3RT}{M}}} = \sqrt{\frac{\gamma}{3}} \]
Therefore, we have:
\[ \sqrt{\frac{\gamma}{3}} = \sqrt{\frac{5}{9}} \]

4. **Square Both Sides**:
Squaring both sides gives:
\[ \frac{\gamma}{3} = \frac{5}{9} \]

5. **Solve for \( \gamma \)**:
Multiplying both sides by 3:
\[ \gamma = \frac{5}{3} \]

6. **Relate \( \gamma \) to Degrees of Freedom**:
The relationship between \( \gamma \) and the degrees of freedom \( F \) is given by:
\[ \gamma = 1 + \frac{2}{F} \]
Setting this equal to our found value of \( \gamma \):
\[ 1 + \frac{2}{F} = \frac{5}{3} \]

7. **Solve for \( F \)**:
Rearranging gives:
\[ \frac{2}{F} = \frac{5}{3} - 1 = \frac{2}{3} \]
Thus:
\[ F = 3 \]

8. **Calculate \( C_V \)**:
The molar heat capacity at constant volume \( C_V \) is related to the degrees of freedom by:
\[ C_V = \frac{F R}{2} \]
Substituting \( F = 3 \):
\[ C_V = \frac{3R}{2} \]

9. **Substitute \( R \)**:
Given \( R = 2 \, \text{cal/mol K} \):
\[ C_V = \frac{3 \times 2}{2} = 3 \, \text{cal/mol K} \]

### Final Answer:
The molar heat capacity of the gas in the process \( PT = \text{constant} \) is:
\[ C_V = 3 \, \text{cal/mol K} \]