Question

A gaseous system, enclosed in an adiabatic container, is in equilibrium at pressure \( P_1 \) and volume \( V_1 \). Work is done on the system in a quasi-static manner due to which the pressure and volume change to \( P_2 \) and \( V_2 \), respectively, in the final equilibrium state. At every instant, the pressure and volume obey the condition \( PV^\gamma = C \), where \( \gamma = \frac{C_p}{C_v} \) and \( C \) is a constant. If the work done is zero, then identify the correct statement(s).

• A. \( P_2 V_2 = P_1 V_1 \)
• B. \( P_2 V_2 = \gamma P_1 V_1 \)
• C. \( P_2 V_2 = (\gamma + 1)P_1 V_1 \)
• D. \( P_2 V_2 = (\gamma - 1)P_1 V_1 \)

Hint:

In adiabatic processes, if the work done is zero, it means there is no change in \( PV \), implying that the process behaves as if \( PV = \text{constant}. \)

Answer 1

The Correct Option is A

Step 1: Given adiabatic condition.
The process follows \( PV^\gamma = C \). Normally, work is done in adiabatic processes, but here it is stated that the net work done is zero.
Step 2: Work done in adiabatic process.
For an adiabatic process, \[ W = \frac{P_1 V_1 - P_2 V_2}{\gamma - 1}. \] If \( W = 0 \), then \( P_1 V_1 = P_2 V_2. \)
Step 3: Final Answer.
Therefore, the correct statement is \( P_2 V_2 = P_1 V_1. \)