Question:medium

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 Concept:
When a gas is "suddenly" compressed or expanded, there is no time for heat exchange between the system and its surroundings. Such a process is considered an adiabatic process. In an adiabatic process, the pressure and volume follow a specific power-law relationship.
Step 2: Key Formula or Approach:
For an adiabatic process: \[ PV^{\gamma} = \text{constant} \] Or, \( P_1 V_1^{\gamma} = P_2 V_2^{\gamma} \).
Step 3: Detailed Explanation:
1. Identify the initial states: \( P_1 = P_0 \) and \( V_1 = V_0 \).
2. Identify the final states: \( P_2 = ? \) and \( V_2 = \frac{V_0}{4} \).
3. Substitute these into the adiabatic equation: \[ P_0 (V_0)^{\gamma} = P_2 \left( \frac{V_0}{4} \right)^{\gamma} \]
4. Solve for \( P_2 \): \[ P_2 = \frac{P_0 (V_0)^{\gamma}}{(\frac{V_0}{4})^{\gamma}} \] \[ P_2 = P_0 \left( \frac{V_0}{V_0/4} \right)^{\gamma} \] \[ P_2 = P_0 (4)^{\gamma} \]
Step 4: Final Answer
The final pressure of the gas is \( P_0(4)^{\gamma} \).
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