Q.How will you calculate work done on an ideal gas in a compression, when change in pressure is carried out in infinite steps?

Concept: Reversible (quasi-static) compression – work done by integrating $P\,dV$ along the path.

When pressure changes in infinite steps, the process is reversible: the gas passes through a continuous sequence of equilibrium states. At every instant the external pressure equals the gas pressure $P$, so the work done on the gas during an infinitesimal compression $dV$ (where $dV < 0$) is $dW = -P\,dV$.

Integrating from initial volume $V_i$ to final volume $V_f$:

The integral depends on the process. For an isothermal process ($PV = nRT = \text{const}$):

For an adiabatic process ($PV^\gamma = \text{const}$): …

Why infinite steps matter

When you compress a gas in one violent stroke, the gas churns, develops pressure gradients, and dissipates energy as internal turbulence. The system is never in a well-defined equilibrium state during the process. But when compression happens through infinite infinitesimal steps—each so gentle that the gas has time to settle into uniform pressure and temperature before the next nudge—the process becomes quasi-static or reversible. At every instant the gas sits in an equilibrium state, and we can describe it with the usual state variables $P$, $V$, $T$.

This matters because only in a quasi-static process can we write the work as an integral of $P \, dV$ where $P$ is the gas's own pressure (not some external imposed pressure that differs wildly from the internal one).

The calculation

The work done on the gas during compression from volume $V_i$ to $V_f$ is

The negative sign accounts for the sign convention: when volume decreases ($dV < 0$), work is done on the gas ($W > 0$). The integral runs from initial to final volume, and $P$ inside the integral is the gas's pressure at each volume along the path.

To evaluate this integral, you need the path—the relationship between $P$ and $V$ during the process. For an ideal gas, $PV = nRT$, so the path is determined by how temperature changes (or is held constant, or relates to $V$).

Common cases

1. Isothermal process ($T = \text{constant}$)

For an ideal gas at fixed temperature, $P = \frac{nRT}{V}$. Substituting:

Since $V_f < V_i$ in compression, the logarithm is negative and $W > 0$, as expected.

2. Adiabatic process ($Q = 0$, no heat exchange)

For an adiabatic process, $PV^\gamma = \text{constant} = K$, where $\gamma = C_P/C_V$. Then $P = K V^{-\gamma}$:

$W = -\int_{V_i}^{V_f} K V^{-\gamma} \, dV = -K \left[\frac{V^{1-\gamma}}{1-\gamma}\right]_{V_i}^{V_f} = \frac{K}{γ-1}\left(V_f^{1-\gamma} - V_i^{1-\gamma}\right)$ …