Consider a fully adiabatic piston-cylinder arrangement as shown in the figure. The piston is massless and cross-sectional area of the cylinder is 𝐴. The fluid inside the cylinder is air (considered as a perfect gas), with γ 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 𝐿₁. The initial pressure of the air inside the cylinder is 𝑃₁ ≫ 𝑃₀, where 𝑃₀ is the atmospheric pressure. The stop S1 is instantaneously removed and the piston moves to the position 𝐿₂, where the equilibrium pressure of air inside the cylinder is 𝑃₂ ≫ 𝑃₀.

What is the work done by the piston on the atmosphere during this process?

The correct option is 𝑃₀𝐴(𝐿₂ − 𝐿₁).

Analysis of the Figure:
From the provided diagram, we can identify the following labels and parameters:
1. The cylinder contains air at pressure 𝑃₁, and the piston is at the "Initial position of the piston" at a distance 𝐿₁ from the left wall.
2. A stop S₁ prevents the piston from moving further right initially.
3. When the stop S₁ is removed, the piston moves to the "Final position of the piston" at a distance 𝐿₂, where it is stopped by S₂.
4. The space outside the cylinder is labeled "Atmosphere, pressure 𝑃₀".

Step-by-Step Explanation:
The work done by the piston on the atmosphere is the work done against the constant external atmospheric pressure 𝑃₀.
The constant force exerted by the atmosphere on the piston of cross-sectional area 𝐴 is:

$$F_{\text{atm}}=P_{0}A$$

During the process, the piston moves from position 𝐿₁ to position 𝐿₂. The displacement of the piston is:

$$\Delta x=L_2-L_1$$

Since the atmospheric pressure remains constant at 𝑃₀ throughout this movement, the work done by the piston to push back the atmosphere is the product of the atmospheric force and the displacement:

$$W=F_{\text{atm}}\times \Delta x$$

Substituting the force and displacement into the equation, we get:

$$W=P_{0}A(L_2-L_1)$$

Thus, the work done by the piston on the atmosphere during this process is indeed 𝑃₀𝐴(𝐿₂ − 𝐿₁).