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 efficiency of the engine is

Options

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

Explanation

To find the efficiency of a Carnot engine using a diatomic ideal gas during an adiabatic expansion from volume \( V \) to \( 32V \), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Carnot Efficiency Formula**:
The efficiency \( \eta \) of a Carnot engine is given by the formula:
\[ \eta = 1 - \frac{T_2}{T_1} \]
where \( T_1 \) is the temperature of the hot reservoir and \( T_2 \) is the temperature of the cold reservoir.

2. **Using the Adiabatic Process Relation**:
For an adiabatic process involving an ideal gas, the relationship between temperature and volume is given by:
\[ T V^{\gamma - 1} = \text{constant} \]
where \( \gamma \) (gamma) is the heat capacity ratio. For a diatomic gas, \( \gamma = \frac{7}{5} = 1.4 \).

3. **Setting Up the Equation**:
Let \( V_1 = V \) and \( V_2 = 32V \). According to the adiabatic relation:
\[ T_1 V_1^{\gamma - 1} = T_2 V_2^{\gamma - 1} \]
Substituting the values:
\[ T_1 V^{\gamma - 1} = T_2 (32V)^{\gamma - 1} \]

4. **Simplifying the Equation**:
We can simplify this equation:
\[ T_1 V^{\gamma - 1} = T_2 (32^{\gamma - 1} V^{\gamma - 1}) \]
Dividing both sides by \( V^{\gamma - 1} \):
\[ T_1 = T_2 \cdot 32^{\gamma - 1} \]

5. **Substituting the Value of \( \gamma \)**:
Now substituting \( \gamma = \frac{7}{5} \):
\[ T_1 = T_2 \cdot 32^{\frac{7}{5} - 1} \]
Simplifying \( \frac{7}{5} - 1 = \frac{2}{5} \):
\[ T_1 = T_2 \cdot 32^{\frac{2}{5}} \]

6. **Calculating \( 32^{\frac{2}{5}} \)**:
We know \( 32 = 2^5 \), so:
\[ 32^{\frac{2}{5}} = (2^5)^{\frac{2}{5}} = 2^2 = 4 \]
Therefore:
\[ T_1 = 4 T_2 \]

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

8. **Final Answer**:
Thus, the efficiency of the engine is:
\[ \eta = 0.75 \]