Question:
The initial pressure and volume of an ideal gas are \( P_0 \) and \( V_0 \). The final pressure of the gas when the gas is suddenly compressed to volume \( V_0/4 \) will be: (Given \( \gamma \) = ratio of specific heats at constant pressure and at constant volume)
The initial pressure and volume of an ideal gas are \( P_0 \) and \( V_0 \). The final pressure of the gas when the gas is suddenly compressed to volume \( V_0/4 \) will be: (Given \( \gamma \) = ratio of specific heats at constant pressure and at constant volume)
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In thermodynamic problems, words like "sudden" or "rapid" imply an adiabatic process (\( PV^\gamma = \text{const} \)), while "slow" implies an isothermal process (\( PV = \text{const} \)). \
Updated On: Apr 22, 2026
- \( P_0 \)
- \( 4P_0 \)
- \( P_0(4)^\gamma \)
- \( P_0(4)^{1/\gamma} \)
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The Correct Option is C
Solution and Explanation
Step 1: Understanding the Question:
The term "suddenly compressed" indicates that the process is adiabatic, as there is no time for heat exchange with the surroundings.
Step 2: Key Formula or Approach:
For an adiabatic process, the relationship between pressure and volume is: \[ PV^\gamma = \text{constant} \]
Step 3: Detailed Explanation:
Initial state: \( P_1 = P_0, \, V_1 = V_0 \)
Final state: \( P_2 = ?, \, V_2 = \frac{V_0}{4} \)
Applying the adiabatic equation: \[ P_1 V_1^\gamma = P_2 V_2^\gamma \] \[ P_0 V_0^\gamma = P_2 \left( \frac{V_0}{4} \right)^\gamma \] Solving for \( P_2 \): \[ P_2 = P_0 \left( \frac{V_0}{V_0/4} \right)^\gamma \] \[ P_2 = P_0 (4)^\gamma \]
Step 4: Final Answer:
The final pressure of the gas is \( P_0 (4)^\gamma \).
The term "suddenly compressed" indicates that the process is adiabatic, as there is no time for heat exchange with the surroundings.
Step 2: Key Formula or Approach:
For an adiabatic process, the relationship between pressure and volume is: \[ PV^\gamma = \text{constant} \]
Step 3: Detailed Explanation:
Initial state: \( P_1 = P_0, \, V_1 = V_0 \)
Final state: \( P_2 = ?, \, V_2 = \frac{V_0}{4} \)
Applying the adiabatic equation: \[ P_1 V_1^\gamma = P_2 V_2^\gamma \] \[ P_0 V_0^\gamma = P_2 \left( \frac{V_0}{4} \right)^\gamma \] Solving for \( P_2 \): \[ P_2 = P_0 \left( \frac{V_0}{V_0/4} \right)^\gamma \] \[ P_2 = P_0 (4)^\gamma \]
Step 4: Final Answer:
The final pressure of the gas is \( P_0 (4)^\gamma \).
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