Pressure-Volume Work

Introduction

When a gas pushes a piston, it can transfer energy to the surroundings as pressure-volume work. Under the chemistry sign convention, expansion work is negative for the gas system. At constant external pressure, the calculation is w = −P extΔV, with units chosen so the pressure-volume product is energy.

Core explanation

Let a gas occupy a cylinder with a movable piston of area A. If the piston moves outward by a small distance dx against external pressure P ext, the opposing force is P ext A and the gas does work of magnitude P ext A dx. Since A dx is the volume increase dV, the work assigned to the gas system is δw = −P ext dV. If P ext remains constant, integration gives w = −P ext(V f − V i) = −P extΔV.

For expansion, V f > V i, so ΔV > 0 and w < 0: energy leaves the gas as it pushes the surroundings. For compression, ΔV < 0 and w > 0: the surroundings do work on the gas. The formula uses external pressure, which is the mechanical opposition at the boundary. It is not always equal to the gas's internal pressure during an irreversible change. Confusing the two can give the wrong work value.

The pressure-volume product has energy units. In SI, 1 Pa m³ = 1 J because Pa = N m⁻². If pressure is in kPa and volume in liters, 1 kPa L = 1 J. If pressure is in atmospheres and volume in liters, 1 L atm ≈ 101.325 J. Convert before combining work with heat or internal-energy terms expressed in joules or kilojoules.

Work depends on the path, not only on endpoints. An expansion against a large external pressure transfers more work than the same volume change against a small external pressure, provided the process can occur. For a changing P ext, use w = −∫P ext dV along the path. The rectangle formula applies only when P ext is constant. A pressure-volume diagram represents the magnitude of P–V work by an area under the relevant pressure path.

If a gas expands freely into a vacuum, P ext = 0 at the moving boundary and P–V work is zero in the simple model, even though ΔV is positive. Conversely, if a rigid vessel cannot change volume, ΔV = 0 and P–V work is zero even if pressure rises. Other work modes, such as electrical work, could still exist and must be tracked separately.

One should not infer q directly from w without the first law and a statement about ΔU. For an isothermal ideal-gas change, ΔU = 0 and q = −w, but that is a special material and temperature condition. For a general gas, heat flow and work together determine the internal-energy change.

Step-by-step reasoning

1. Define the gas as system and read V i and V f. 2. Compute signed ΔV = V f − V i. 3. Verify P ext is constant for the stated formula. 4. Calculate w = −P extΔV and convert units. 5. Add q if the problem asks for ΔU.

Visual explanation

Draw a cylinder with the piston at initial and final positions. Mark external pressure arrows pointing inward and a gas-expansion arrow outward. On a P ext-versus-V diagram, shade a rectangle of height P ext and width ΔV; label its magnitude w and its expansion sign negative.

Real-world analogy

Pushing a heavy door through a distance transfers more mechanical energy than pushing a light door through the same distance. External pressure is the resisting load per area; volume change plays the role of displacement. The analogy does not capture molecular pressure details but helps identify the work direction.

Real-world example

A chemical reaction generating gas in a piston-cylinder can lift a weight. Some released energy becomes mechanical work, so the measured heat may not equal ΔU. The piston load fixes an approximately constant external pressure in a simple laboratory model.

Why?

Why is P ext used rather than the final gas pressure? Work is determined by the opposing force at the moving boundary throughout the path. A single final internal pressure does not specify that force or the route.

Common misconception

“If volume increases, work must be positive because the gas worked hard.” In chemistry notation, work done by the gas is negative for the gas system because energy leaves it.

Worked example

A gas expands from 2.0 L to 5.0 L against a constant external pressure of 100 kPa. ΔV = +3.0 L. Thus w = −(100 kPa)(3.0 L) = −300 J, using 1 kPa L = 1 J. If the gas simultaneously absorbs 500 J as heat, ΔU = q + w = +500 − 300 = +200 J.

Quick check

1. What is P–V work if a rigid container's volume does not change? Answer: Zero, because ΔV = 0, regardless of internal pressure change.

Exam focus

Use external pressure, signed ΔV and consistent energy units. State whether the boundary is expanding or compressing before substituting. A variable-pressure path needs an integral or area, not the constant-pressure rectangle formula.

Advanced insight

In a reversible quasistatic process, the external pressure differs only infinitesimally from the system's pressure, allowing work to be written as −∫P system dV along an equilibrium path. This does not justify replacing P ext by a final system pressure for an irreversible expansion.

Summary

Pressure-volume work transfers energy as a boundary moves. At constant external pressure, w = −P extΔV: expansion is negative and compression positive for the gas. Work depends on the resisting pressure path and must be converted to energy units.

Practice questions

1. A gas compresses from 6.0 L to 4.0 L at P ext = 50 kPa. Find w. Answer: ΔV = −2.0 L, so w = −(50)(−2.0) = +100 J. 2. A gas expands by 1.5 L against 2.0 atm. Estimate w in joules. Answer: w = −3.0 L atm ≈ −304 J using 101.325 J per L atm. 3. Why is work zero for free expansion into a vacuum in this model? Answer: The external opposing pressure is zero, so −P extΔV = 0 despite a positive volume change.