Question

Consider a fully adiabatic piston-cylinder arrangement as shown in the figure. The piston is massless and the cross-sectional area of the cylinder is \(A\). The fluid inside the cylinder is air (considered as a perfect gas), with \(\gamma\) being the ratio of the specific heat at constant pressure to the specific heat at constant volume for air. The piston is initially located at a position \(L_1\). The initial pressure of the air inside the cylinder is \(P_1 \gg P_0\), where \(P_0\) is the atmospheric pressure. The stop \(S_1\) is instantaneously removed and the piston moves to the position \(L_2\), where the equilibrium pressure of air inside the cylinder is \(P_2 \gg P_0\). What is the work done by the piston on the atmosphere during this process?

• A. 0
• B. \(P_0 A(L_2 - L_1)\)
• C. \(P_1 A L_1 \ln\frac{L_1}{L_2}\)
• D. \(\frac{(P_2 L_2 - P_1 L_1)A}{(1 - \gamma)}\)

Hint:

Read thermodynamics questions very carefully. Distinguish between: - Work done by the gas inside the cylinder. - Work done on the atmosphere. - Net work done by the piston. The work done on the atmosphere is almost always \(P_{atm} \Delta V\), as atmospheric pressure is constant.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The question asks for the work done by the piston on the atmosphere. It does not ask for the work done by the gas inside the cylinder, nor the net work done by the piston. The work done on the atmosphere is the work required to push the atmosphere back as the piston moves.
Step 2: Key Formula or Approach:
Work done by a system against a constant external pressure is given by the formula: \[ W = P_{ext} \times \Delta V \] where \(P_{ext}\) is the constant external pressure and \(\Delta V\) is the change in volume of the system.
Step 3: Detailed Explanation:
1. The "system" doing the work on the atmosphere is the piston.
2. The external pressure that the piston works against is the constant atmospheric pressure, \(P_{ext} = P_0\).
3. The piston moves from an initial position \(L_1\) to a final position \(L_2\).
4. The initial volume occupied by the gas is \(V_1 = A \times L_1\). The final volume is \(V_2 = A \times L_2\).
5. The distance the piston moves is \((L_2 - L_1)\).
6. The volume displaced by the piston against the atmosphere is \(\Delta V = V_2 - V_1 = A L_2 - A L_1 = A(L_2 - L_1)\).
7. The work done by the piston on the atmosphere is therefore: \[ W_{\text{on atm}} = P_0 \times \Delta V = P_0 A(L_2 - L_1) \] Step 4: Final Answer:
The work done by the piston on the atmosphere is \(P_0 A(L_2 - L_1)\).
Step 5: Why This is Correct:
The question is very specific. The work done on the atmosphere depends only on the atmospheric pressure and the volume displaced, regardless of the process happening inside the cylinder. The other options represent different quantities:
- (A) 0 is incorrect as the piston moves.
- (C) This form, \(P_1 V_1 \ln(V_1/V_2)\), represents work done during an isothermal process, which is not the case here.
- (D) \(\frac{P_2 V_2 - P_1 V_1}{1-\gamma}\) represents the work done by the gas inside the cylinder during a reversible adiabatic process. The process here is irreversible (instantaneous removal of stop), but this formula still gives the change in internal energy of the gas, not the work on the atmosphere.