The molar heat capacity in a process of a diatomic gas if it does a work of `Q/4` when a heat of `Q` is supplied to it is

Options

A. `(2)/(5) R`
B. `(5)/(2) R`
C. `(10)/(3) R`
D. `(6)/(7) R`

Explanation

To solve the problem, we need to determine the molar heat capacity \( C \) of a diatomic gas given that it does work \( W = \frac{Q}{4} \) when a heat \( Q \) is supplied to it.

### Step-by-Step Solution:

1. **Understand the First Law of Thermodynamics**:
The first law of thermodynamics states that:
\[ \Delta U = Q - W \]
where \( \Delta U \) is the change in internal energy, \( Q \) is the heat added to the system, and \( W \) is the work done by the system.

2. **Substitute the Given Values**:
From the problem, we know:
\[ W = \frac{Q}{4} \]
Therefore, substituting this into the first law gives:
\[ \Delta U = Q - \frac{Q}{4} = Q \left(1 - \frac{1}{4}\right) = Q \left(\frac{3}{4}\right) = \frac{3Q}{4} \]

3. **Relate Change in Internal Energy to Temperature Change**:
For a diatomic gas, the change in internal energy can also be expressed as:
\[ \Delta U = n C_V \Delta T \]
where \( C_V \) is the molar heat capacity at constant volume. For a diatomic gas, \( C_V = \frac{5}{2} R \).

4. **Express \( \Delta T \)**:
We can equate the two expressions for \( \Delta U \):
\[ n C_V \Delta T = \frac{3Q}{4} \]
Substituting \( C_V \):
\[ n \left(\frac{5}{2} R\right) \Delta T = \frac{3Q}{4} \]

5. **Solve for \( \Delta T \)**:
Rearranging gives:
\[ \Delta T = \frac{3Q}{4} \cdot \frac{2}{5nR} = \frac{3Q}{10nR} \]

6. **Calculate Molar Heat Capacity \( C \)**:
Molar heat capacity \( C \) is defined as:
\[ C = \frac{Q}{\Delta T} \]
Substituting for \( \Delta T \):
\[ C = \frac{Q}{\frac{3Q}{10nR}} = \frac{Q \cdot 10nR}{3Q} = \frac{10nR}{3} \]

7. **Final Expression for Molar Heat Capacity**:
Since we are looking for the molar heat capacity per mole, we can express it as:
\[ C = \frac{10}{3} R \]

### Conclusion:
The molar heat capacity of the diatomic gas in this process is:
\[ C = \frac{10}{3} R \]