Reversible Isothermal Gas Work

Introduction

For a reversible isothermal expansion, an ideal gas pushes against an external pressure kept infinitesimally below its own pressure. The pressure falls as volume rises, so constant-pressure work is not appropriate. Integrating along the path gives w rev = −nRT ln(V f/V i), with a negative value for expansion.

Core explanation

In a reversible quasistatic process, the gas passes through a continuous sequence of states very close to mechanical equilibrium. The opposing external pressure differs only infinitesimally from the gas pressure. The incremental chemistry-convention work is δw = −P dV. For an ideal gas, P = nRT/V. If n and T remain constant, substitute to obtain δw = −(nRT/V)dV. Integrating from V i to V f gives w rev = −nRT∫(dV/V) = −nRT ln(V f/V i).

The natural logarithm is dimensionless, so its argument must be a ratio of volumes in the same units. Use R in energy-compatible units, commonly 8.314 J mol⁻¹ K⁻¹, temperature in kelvin and n in moles. The resulting w is in joules. For expansion V f/V i > 1, ln is positive and w rev is negative. For compression the ratio is less than 1, ln is negative and w rev is positive.

An ideal gas's internal energy depends only on temperature. Isothermal means ΔT = 0, so ΔU = 0 for this ideal-gas model. The first law then gives q = −w. In reversible expansion, heat enters the gas to replace the energy it exports as work; in reversible compression, heat leaves as work enters. This does not mean heat and work are state functions. Another route between the same ideal-gas endpoints can have different q and w while preserving ΔU = 0.

For fixed initial and final volumes at one temperature, reversible expansion gives the greatest magnitude of work done by the gas among mechanically possible expansion paths under common constraints, because it pushes against the largest opposing pressure compatible with each small forward step. A one-step expansion against a lower constant external pressure gives a smaller magnitude. Reversible compression requires the least work input relative to more strongly driven compression paths. These comparisons assume the same endpoints and a suitable heat exchange that maintains temperature.

Reversibility is an ideal limit, not a claim that a real piston has zero friction and can be run perfectly backward without change. It is useful because it provides a well-defined path and a limiting work value. A sudden free expansion has P ext = 0 and w = 0, contrasting sharply with the reversible result for the same volume ratio.

This formula applies to an ideal gas with constant n and T and only P–V work under the stated reversible path. It should not be applied automatically to a reacting gas whose amount changes, a nonideal gas at high pressure, or a process with varying temperature.

Step-by-step reasoning

1. Confirm ideal gas, fixed n, fixed T and reversible path. 2. Write δw = −P dV and substitute P = nRT/V. 3. Integrate between initial and final volumes. 4. Evaluate the natural logarithm of V f/V i. 5. Use ΔU = 0 and q = −w only if the ideal isothermal condition holds.

Visual explanation

Draw a downward-curving isotherm P = nRT/V on a P–V graph. Shade the area under the curve from V i to V f. Add a lower horizontal P ext line for an irreversible expansion and shade its smaller rectangle; the two areas show different work magnitudes for different paths.

Real-world analogy

Lifting a load through many tiny adjustments can keep the opposing force close to the maximum manageable force at each stage. A single sudden move may use a much smaller load. The analogy captures the reversible limit's work difference, though molecular heat exchange is not represented.

Real-world example

An idealized piston engine can be analyzed with reversible isothermal stages to estimate limiting work. Real engines have friction and finite temperature differences, so measured work differs. The ideal calculation provides a benchmark rather than a promise of practical performance.

Why?

Why does a logarithm appear? Ideal-gas pressure at constant temperature is inversely proportional to volume. Summing the changing pressure force over volume requires integrating 1/V, whose antiderivative is ln V.

Common misconception

“Isothermal means no heat flows.” For an ideal gas, ΔU = 0, but nonzero work can be balanced by equal-and-opposite heat. Isothermal and adiabatic are different conditions.

Worked example

One mole of ideal gas expands reversibly and isothermally at 300 K from 2.0 L to 4.0 L. Use w = −(1 mol)(8.314 J mol⁻¹ K⁻¹)(300 K)ln(4.0/2.0). Since ln 2 ≈ 0.693, w ≈ −1729 J or −1.73 kJ. Because ΔU = 0, q ≈ +1.73 kJ. The heat input maintains the temperature while the gas does work.

Quick check

1. What is the sign of reversible isothermal work for compression? Answer: Positive for the gas system, because V f/V i < 1 makes the logarithm negative and the leading minus sign reverses it.

Exam focus

Use the natural logarithm, matching volume units in the ratio, kelvin temperature and energy-compatible R. State all assumptions. Do not substitute a single final pressure into the constant-pressure work formula for a reversible path.

Advanced insight

Reversible work is a limiting path integral, while ΔU is an endpoint difference. The area between reversible and irreversible P–V paths quantifies work lost to the choice of route under matched endpoints. Entropy production provides a broader measure of irreversibility in real processes.

Summary

For a reversible isothermal ideal-gas path, integrating −P dV gives w = −nRT ln(V f/V i). Expansion work is negative; compression work positive. The ideal gas has ΔU = 0, so q = −w, yet both transfers remain path-dependent.

Practice questions

1. Find the sign of w when V f = 3V i in reversible isothermal expansion. Answer: Negative, because ln 3 is positive and w = −nRT ln 3. 2. If an ideal gas expands freely into a vacuum to the same final volume, is its work the same? Answer: No. Free-expansion P–V work is zero, while reversible expansion gives negative work. 3. For the worked example, what is ΔU and why? Answer: Zero, because the ideal gas is isothermal and its internal energy depends only on temperature.