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`

Answer

To find the molar heat capacity \( C \) of a diatomic gas when it does work of \( \frac{Q}{4} \) upon receiving heat \( Q \), we can follow these steps:

### Step 1: Apply the First Law of Thermodynamics
The first law of thermodynamics states:
\[ Q = W + \Delta U \]
Where:
- \( Q \) is the heat supplied,
- \( W \) is the work done by the system,
- \( \Delta U \) is the change in internal energy.

### Step 2: Substitute the Given Values
We know that the work done \( W = \frac{Q}{4} \). Substituting this into the first law gives:
\[ \Delta U = Q - W = Q - \frac{Q}{4} = \frac{3Q}{4} \]

### Step 3: Relate Change in Internal Energy to Temperature Change
For a diatomic gas, the change in internal energy can also be expressed as:
\[ \Delta U = nR \Delta T \left(\frac{1}{\gamma - 1}\right) \]
Where:
- \( n \) is the number of moles,
- \( R \) is the universal gas constant,
- \( \Delta T \) is the change in temperature,
- \( \gamma \) is the specific heat ratio, which for a diatomic gas is \( \frac{7}{5} \).

### Step 4: Substitute \( \gamma \) into the Equation
Substituting \( \gamma = \frac{7}{5} \) into the equation gives:
\[ \Delta U = nR \Delta T \left(\frac{1}{\frac{7}{5} - 1}\right) = nR \Delta T \left(\frac{1}{\frac{2}{5}}\right) = \frac{5}{2} nR \Delta T \]

### Step 5: Set the Two Expressions for \( \Delta U \) Equal
Now we equate the two expressions for \( \Delta U \):
\[ \frac{3Q}{4} = \frac{5}{2} nR \Delta T \]

### Step 6: Solve for \( n \Delta T \)
Rearranging gives:
\[ nR \Delta T = \frac{3Q}{4} \cdot \frac{2}{5} = \frac{3Q}{10} \]

### Step 7: Relate Heat Supplied to Molar Heat Capacity
The heat supplied can also be expressed as:
\[ Q = nC \Delta T \]
Substituting \( n \Delta T \) from the previous step:
\[ Q = nC \Delta T \implies Q = nC \left(\frac{3Q}{10R}\right) \]

### Step 8: Solve for Molar Heat Capacity \( C \)
Cancelling \( Q \) from both sides (assuming \( Q \neq 0 \)):
\[ 1 = nC \left(\frac{3}{10R}\right) \implies C = \frac{10R}{3} \]

### Final Answer
Thus, the molar heat capacity \( C \) of the diatomic gas is:
\[ C = \frac{10R}{3} \]