A diatomic ideal gas is used in a Carnot engine as the working substance. If during the adiabatic expansion part of the cycle the volume of the gas increase from V to 32V, the efficenecy of the engine is

Options

A. `0.5`
B. `0.75`
C. 0.99`
D. `0.25`

Explanation

To solve the problem regarding the efficiency of a Carnot engine using a diatomic ideal gas, we will follow these steps:

### Step-by-Step Solution:

1. **Understand the Carnot Engine Efficiency Formula**:
The efficiency (η) of a Carnot engine is given by the formula:
\[ \eta = 1 - \frac{T_2}{T_1} \]
where \(T_1\) is the temperature of the heat source and \(T_2\) is the temperature of the heat sink.

2. **Use the Adiabatic Process Relation**:
For an adiabatic process, the relation between temperature and volume is given by:
\[ T_1 V_1^{\gamma - 1} = T_2 V_2^{\gamma - 1} \]
where \(\gamma\) (gamma) for a diatomic gas is \(\frac{7}{5}\).

3. **Set Initial and Final Volumes**:
Given that the volume changes from \(V\) to \(32V\), we can set:
- \(V_1 = V\)
- \(V_2 = 32V\)

4. **Substitute Values into the Adiabatic Relation**:
Substitute \(V_1\) and \(V_2\) into the adiabatic relation:
\[ T_1 V^{\frac{7}{5} - 1} = T_2 (32V)^{\frac{7}{5} - 1} \]
This simplifies to:
\[ T_1 V^{\frac{2}{5}} = T_2 (32^{\frac{2}{5}} V^{\frac{2}{5}}) \]

5. **Cancel Common Terms**:
Since \(V^{\frac{2}{5}}\) appears on both sides, we can cancel it out:
\[ T_1 = T_2 \cdot 32^{\frac{2}{5}} \]

6. **Calculate \(32^{\frac{2}{5}}\)**:
We can simplify \(32^{\frac{2}{5}}\):
\[ 32 = 2^5 \implies 32^{\frac{2}{5}} = (2^5)^{\frac{2}{5}} = 2^2 = 4 \]
Thus, we have:
\[ T_1 = 4T_2 \]

7. **Substitute Back into the Efficiency Formula**:
Now substitute \(T_1\) back into the efficiency formula:
\[ \eta = 1 - \frac{T_2}{4T_2} = 1 - \frac{1}{4} = \frac{3}{4} \]

8. **Final Efficiency Value**:
Therefore, the efficiency of the engine is:
\[ \eta = 0.75 \text{ or } 75\% \]

### Final Answer:
The efficiency of the Carnot engine is \(0.75\) or \(75\%\).