For the adiabatic expansion of a perfect monoatomic gas, when volume increases by 24% , what is the percentage decrease in pressure ? Given : `(25/31)^(5//3)=0.7`

Options

A. `24%`
B. `40%`
C. `48%`
D. `71%`

Answer

To solve the problem of finding the percentage decrease in pressure during the adiabatic expansion of a perfect monoatomic gas when the volume increases by 24%, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Adiabatic Process**:
The relationship for an adiabatic process for a perfect gas is given by:
\[ P V^\gamma = \text{constant} \]
where \( P \) is the pressure, \( V \) is the volume, and \( \gamma \) (gamma) is the adiabatic constant.

2. **Determine the Value of Gamma**:
For a monoatomic gas, the value of \( \gamma \) is:
\[ \gamma = \frac{C_p}{C_v} = \frac{\frac{5R}{2}}{\frac{3R}{2}} = \frac{5}{3} \]

3. **Relate Changes in Pressure and Volume**:
From the relationship \( P V^\gamma = \text{constant} \), we can express the change in pressure in terms of the change in volume:
\[ \frac{dP}{P} = -\gamma \frac{dV}{V} \]
This implies:
\[ \frac{dP}{P} = -\frac{5}{3} \frac{dV}{V} \]

4. **Calculate the Change in Volume**:
Given that the volume increases by 24%, we can express this as:
\[ \frac{dV}{V} = \frac{24}{100} = 0.24 \]

5. **Substitute the Values into the Equation**:
Now substituting \( \frac{dV}{V} \) into the equation for \( \frac{dP}{P} \):
\[ \frac{dP}{P} = -\frac{5}{3} \times 0.24 \]

6. **Calculate the Change in Pressure**:
Performing the multiplication:
\[ \frac{dP}{P} = -\frac{5 \times 0.24}{3} = -\frac{1.2}{3} = -0.4 \]

7. **Convert to Percentage**:
To find the percentage decrease in pressure, we multiply by 100:
\[ \text{Percentage decrease in pressure} = -0.4 \times 100 = -40\% \]
The negative sign indicates a decrease in pressure.

### Final Answer:
The percentage decrease in pressure is **40%**.