Q.The pressure-volume work for an ideal gas can be calculated by using the expression $w = -\int_{V_i}^{V_f} p_{ex}\,dV$. The work can also be calculated from the pV-plot by using the area under the curve within the specified limits. When an ideal gas is compressed

Concept: Irreversible Expansion Work

The work done on or by a gas depends on the external pressure $p_{\text{ex}}$ against which the process occurs. For compression (volume decreases from $V_i$ to $V_f$, so $\Delta V < 0$), we compare reversible and irreversible paths.

Reversible compression: The external pressure is infinitesimally greater than the gas pressure at every instant, so $p_{\text{ex}} = p_{\text{gas}} = \frac{nRT}{V}$ throughout. The work is

Since $V_f < V_i$, this logarithm is negative, making $w_{\text{rev}}$ positive (work done on the gas).

Irreversible compression: The external pressure is a constant $p_{\text{ex}}$ (typically much larger than the initial gas pressure), giving

$w_{\text{irr}} = -p_{\text{ex}}(V_f - V_i) = p_{\text{ex}}(V_i - V_f).$ …

Why reversible work differs from irreversible work

The expression $w = -\int_{V_i}^{V_f} p_{\text{ex}}\,dV$ tells us that the work done on or by a gas depends entirely on the external pressure $p_{\text{ex}}$ pushing against the piston at each instant, not on the gas's own pressure $p$. The integral sums up $p_{\text{ex}}\,dV$ over the path, which geometrically is the area under the $p_{\text{ex}}$ versus $V$ curve.

In a reversible process the system is always infinitesimally close to equilibrium, so $p_{\text{ex}} = p$ at every moment. The gas's own equation of state—say $p = \frac{nRT}{V}$ for an ideal gas at constant temperature—dictates the external pressure step by step. On a $pV$ diagram this traces the smooth isotherm (or adiabat, or whatever the constraint), and the area under that curve is the reversible work.

In an irreversible process the external pressure is fixed at some constant value $p_{\text{ex}}$ that does not match the gas pressure throughout. For compression, $p_{\text{ex}}$ must exceed the gas pressure to drive the volume down; the piston moves in a single stroke against this constant opposing force. The $p_{\text{ex}}$ versus $V$ graph is a horizontal line at height $p_{\text{ex}}$, and the area under that rectangle is the irreversible work.

Because the reversible path hugs the gas's own (higher, for compression) pressure curve while the irreversible path sits at a lower constant $p_{\text{ex}}$, the area—and hence the magnitude of work—differs.

Step-by-step comparison for compression ($V_i \to V_f$, $V_f < V_i$)

1. Sign convention and direction.

   Compression means $\Delta V = V_f - V_i < 0$. The integral $w = -\int_{V_i}^{V_f} p_{\text{ex}}\,dV$ will yield a positive number (work done on the gas) because $dV < 0$ and the minus sign flips it.

2. Reversible compression.

   At every instant $p_{\text{ex}} = p_{\text{gas}}$. For an ideal gas at constant temperature,

so

Since $V_i > V_f$, the logarithm is positive and $w_{\text{rev}} > 0$. On the $pV$ diagram the area under the hyperbola $p = nRT/V$ from $V_i$ to $V_f$ gives this work.

3. Irreversible compression. The external pressure is held constant at some $p_{\text{ex}}$ (chosen large enough to compress the gas). Then

This is the area of a rectangle of height $p_{\text{ex}}$ and width $V_i - V_f$.

4. Comparing the areas.

   The reversible curve $p = nRT/V$ is a decreasing function: at the start ($V = V_i$) the pressure is lower, and as volume shrinks toward $V_f$ the pressure climbs. The constant $p_{\text{ex}}$ must lie above the gas pressure at $V_i$ to initiate compression, but it will lie below the gas pressure near $V_f$ (where the gas is highly compressed). The reversible path therefore sweeps out a larger area under its curve than the flat irreversible rectangle.

5. Conclusion on magnitudes.

   Both works are positive (compression), but

   $w_{\text{rev}} > w_{\text{irrev}}.$ …