Question

An ideal gas is found to obey \( PV^{\frac{3}{2}} = \text{constant} \) during an adiabatic process. If such a gas initially at a temperature \( T \) is adiabatically compressed to \( \frac{1}{4} \)th of its volume, then its final temperature is:

• A. \( \sqrt{3T} \)
• B. \( \sqrt{2T} \)
• C. \( 2T \)
• D. \( 3T \)

Hint:

In an adiabatic process, the final temperature depends on the polytropic index and volume ratio. The temperature increases when volume is compressed.

Answer 1

The Correct Option is C

Step 1: Applying the Adiabatic 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: - \( \gamma \) is the polytropic index, given as \( \gamma = \frac{3}{2} \)
- \( T_1 \) is the initial temperature
- \( V_1 \) is the initial volume
- \( T_2 \) is the final temperature
- \( V_2 \) is the final volume, given as \( V_2 = \frac{V_1}{4} \) Step 2: Substituting Values Rewriting the equation: \[ T \cdot V_1^{\frac{3}{2} - 1} = T_2 \cdot V_2^{\frac{3}{2} - 1} \] Since \( \gamma -1 = \frac{1}{2} \), we get: \[ T \cdot V_1^{\frac{1}{2}} = T_2 \cdot \left(\frac{V_1}{4}\right)^{\frac{1}{2}} \] \[ T \cdot V_1^{\frac{1}{2}} = T_2 \cdot V_1^{\frac{1}{2}} \times \frac{1}{2} \] Canceling \( V_1^{\frac{1}{2}} \) from both sides: \[ T = T_2 \times \frac{1}{2} \] Step 3: Solving for \( T_2 \) \[ T_2 = 2T \] Thus, the correct answer is option (3).