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

Answer

To solve the problem, we need to find the efficiency of a Carnot engine using a diatomic ideal gas as the working substance, given that the volume of the gas increases from \( V \) to \( 32V \) during the adiabatic expansion.

### Step-by-Step Solution:

1. **Understand the relationship during adiabatic expansion**:
The relationship between temperature and volume for an adiabatic process is given by:
\[ T V^{\gamma - 1} = \text{constant} \]
where \( \gamma = \frac{C_p}{C_v} \). For a diatomic ideal gas, \( \gamma = \frac{7}{5} \).

2. **Set up the equation for the initial and final states**:
Let \( T_1 \) be the initial temperature and \( T_2 \) be the final temperature. The initial volume is \( V_1 = V \) and the final volume is \( V_2 = 32V \). Thus, we can write:
\[ T_1 V^{\frac{7}{5} - 1} = T_2 (32V)^{\frac{7}{5} - 1} \]
Simplifying gives:
\[ T_1 V^{\frac{2}{5}} = T_2 (32^{\frac{2}{5}} V^{\frac{2}{5}}) \]

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

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

5. **Determine the efficiency of the Carnot engine**:
The efficiency \( \eta \) of a Carnot engine is given by:
\[ \eta = 1 - \frac{T_2}{T_1} \]
Substituting \( T_1 = 4T_2 \) into the efficiency formula:
\[ \eta = 1 - \frac{T_2}{4T_2} = 1 - \frac{1}{4} = \frac{3}{4} \]

6. **Final result**:
Therefore, the efficiency of the Carnot engine is:
\[ \eta = 0.75 \]