Certain perfect gas is found to obey `pV^(3//2)` = constant during adiabatic process. If such a gas at intial temperature T, is adiabatically compressed to half the initial volume, its final temperature will be

Options

A. `sqrt(2)` T
B. 2 T
C. `2 sqrt(2) T `
D. 4 T

Answer

To solve the problem, we need to determine the final temperature of a perfect gas that obeys the relation \( pV^{3/2} = \text{constant} \) during an adiabatic process, given that the gas is compressed to half its initial volume.

### Step-by-Step Solution:

1. **Understand the Given Relation**:
We start with the relation for the adiabatic process:
\[ pV^{3/2} = \text{constant} \]

2. **Use the Ideal Gas Law**:
The ideal gas law states:
\[ pV = nRT \]
where \( n \) is the number of moles, \( R \) is the gas constant, and \( T \) is the temperature.

3. **Express Pressure in Terms of Volume and Temperature**:
Rearranging the ideal gas law gives:
\[ p = \frac{nRT}{V} \]

4. **Substitute Pressure into the Adiabatic Relation**:
Substitute \( p \) into the adiabatic relation:
\[ \frac{nRT}{V} V^{3/2} = \text{constant} \]
This simplifies to:
\[ nRT V^{1/2} = \text{constant} \]

5. **Establish a Relationship Between Temperature and Volume**:
From the equation above, we can express the relationship:
\[ T V^{1/2} = \text{constant} \]
Therefore, we can say:
\[ T \propto \frac{1}{V^{1/2}} \]

6. **Relate Initial and Final Temperatures and Volumes**:
Let \( T_i \) be the initial temperature and \( V_i \) be the initial volume. Let \( T_f \) be the final temperature and \( V_f \) be the final volume. We know:
\[ \frac{T_f}{T_i} = \frac{V_i^{1/2}}{V_f^{1/2}} \]

7. **Substitute the Known Values**:
Given that the initial volume \( V_i = V \) and the final volume \( V_f = \frac{V}{2} \):
\[ \frac{T_f}{T} = \frac{V^{1/2}}{(V/2)^{1/2}} = \frac{V^{1/2}}{(V^{1/2} \cdot (1/\sqrt{2}))} = \sqrt{2} \]

8. **Calculate the Final Temperature**:
Rearranging gives:
\[ T_f = T \cdot \sqrt{2} \]

### Final Answer:
The final temperature \( T_f \) after adiabatic compression to half the initial volume is:
\[ T_f = \sqrt{2} T \]