In a thermodynamic process on an ideal diatomic gas, work done by the gas is `eta` times the heat supplied `(eta lt 1)`. The molar heat capacity of the gas for the process is

Options

A. `(5R)/(2)`
B. `5eta(R)/(2)`
C. `(5R)/(2(1-eta))`
D. `(5R(1-eta))/(2)`

Answer

To find the molar heat capacity of an ideal diatomic gas in a thermodynamic process where the work done by the gas is `eta` times the heat supplied (with `eta < 1`), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the First Law of Thermodynamics**:
The first law of thermodynamics states:
\[ dQ = dU + dW \]
where \(dQ\) is the heat supplied, \(dU\) is the change in internal energy, and \(dW\) is the work done by the gas.

2. **Express Work Done**:
According to the problem, the work done by the gas is given by:
\[ dW = \eta \cdot dQ \]

3. **Substitute Work Done into the First Law**:
Substitute \(dW\) into the first law:
\[ dQ = dU + \eta \cdot dQ \]

4. **Rearrange to Find Change in Internal Energy**:
Rearranging gives:
\[ dU = dQ - \eta \cdot dQ = (1 - \eta) dQ \]

5. **Relate Change in Internal Energy to Temperature Change**:
For an ideal gas, the change in internal energy can be expressed as:
\[ dU = n C_v dT \]
where \(C_v\) is the molar heat capacity at constant volume.

6. **Equate the Two Expressions for Change in Internal Energy**:
From the previous steps, we have:
\[ n C_v dT = (1 - \eta) dQ \]

7. **Express Heat Supplied**:
The heat supplied can be expressed as:
\[ dQ = n C dT \]
where \(C\) is the molar heat capacity for the process.

8. **Substitute \(dQ\) in the Equation**:
Substitute \(dQ\) into the equation:
\[ n C_v dT = (1 - \eta) (n C dT) \]

9. **Cancel \(n dT\) from Both Sides**:
Assuming \(n \neq 0\) and \(dT \neq 0\), we can cancel \(n dT\):
\[ C_v = (1 - \eta) C \]

10. **Solve for Molar Heat Capacity \(C\)**:
Rearranging gives:
\[ C = \frac{C_v}{1 - \eta} \]

11. **Substitute \(C_v\) for a Diatomic Gas**:
For a diatomic ideal gas, the molar heat capacity at constant volume is:
\[ C_v = \frac{5}{2} R \]
Therefore,
\[ C = \frac{\frac{5}{2} R}{1 - \eta} \]

### Final Result:
The molar heat capacity \(C\) for the process is:
\[ C = \frac{5R}{2(1 - \eta)} \]