Q.What will be the work done on an ideal gas enclosed in a cylinder, when it is compressed by a constant external pressure, $p_{ext}$ in a single step as shown in Fig. 6.2. Explain graphically.

The key idea is the First Law of Thermodynamics applied to mechanical work: $W = -\int p_{\text{ext}} \, dV$, where the sign convention is work done on the gas positive.

1. Since $p_{\text{ext}}$ is constant, the integral simplifies:

   $W = -p_{\text{ext}} \int_{V_1}^{V_2} dV = -p_{\text{ext}} (V_2 - V_1)$.

2. Because the gas is compressed, $V_2 < V_1$, so $(V_2 - V_1)$ is negative. The negative sign in the formula then makes the work positive, confirming work is done on the gas. …

The First Law of Thermodynamics tells us that work is a path-dependent quantity — it depends on how the volume change happens, not just the start and end states. For a gas, the work done on the gas is given by

where $p_{\text{gas}}$ is the pressure of the gas at each instant. The negative sign ensures that when volume decreases ($dV < 0$), work on the gas comes out positive.

But here’s the key: in a single-step compression against a constant external pressure $p_{\text{ext}}$, the gas does not control the process — the surroundings do. The gas is forced to compress until its pressure equals $p_{\text{ext}}$ (or until the piston stops). During the entire compression, the external pressure remains fixed at $p_{\text{ext}}$, so the work done on the gas by the surroundings is simply:

Why no integral? Because $p_{\text{ext}}$ is constant, the integral becomes $p_{\text{ext}} \times (-\Delta V)$, and since $V_2 < V_1$, the result is positive — work is done on the gas.

Now let’s walk through the reasoning step by step.

1. Identify the process. The gas is compressed from $V_1$ to $V_2$ in a single step. The external pressure $p_{\text{ext}}$ is constant throughout. This is an irreversible compression because the gas pressure is not equal to $p_{\text{ext}}$ during the process — it jumps suddenly.

2. Work formula for constant external pressure. When the external pressure is constant, the work done by the surroundings on the gas is:

Here $\Delta V = V_2 - V_1$ is negative, so $V_1 - V_2$ is positive.

3. Why not use gas pressure? In an irreversible process, the gas pressure is not uniform or well-defined during the compression — it may vary with position and time. The only pressure we can reliably use is the external one, because that’s what the surroundings actually apply. The formula $W = -\int p_{\text{gas}} dV$ is valid only if the gas is in equilibrium throughout (a reversible process). For a single-step irreversible compression, we must use $p_{\text{ext}}$. …