Chemical Thermodynamics · Lesson 2 of 10

Pressure–Volume Work and Gas Expansion

“Use pistons and pressure–volume graphs to understand how the route of expansion controls work.”

Learning Objectives

• Derive the sign and magnitude of pressure–volume work. • Interpret one-step, finite-step, and reversible work on a pressure–volume diagram. • Derive the reversible isothermal ideal-gas work expression. • Compare free expansion with expansion against nonzero external pressure. • Apply the first law under isothermal, adiabatic, and constant-volume conditions.

A gas pushes against its surroundings

A gas can transfer energy by pushing a piston outward. The gas is then doing work, so the work term in the chemical convention is negative. If the surroundings push the piston inward, they do work on the gas and the term is positive. To calculate the energy transfer, we must know the pressure opposing the motion, not just the gas pressure at the beginning.

Consider a piston of area $A$ moving a small distance $l$ . The magnitude of the volume change is $A l$ . External pressure $p_{ext}$ produces a force of magnitude $p_{ext} A$ . Multiplying force by displacement gives the magnitude of the work. Attaching the correct direction gives the following expression for a constant external pressure.

Constant external pressureLaTeX

w = - p_{ext} Δ V = - p_{ext} (V_{f} - V_{i})

V_i and V_f are initial and final volumes. Expansion has positive ΔV and negative w; compression has negative ΔV and positive w.

The pressure must be compatible with the volume unit. A pascal multiplied by a cubic metre is a joule. Useful conversions are $1 L bar = 100 J$ and $1 L atm = 101.325 J$ . These conversions matter because the first law cannot combine work in litre-atmospheres with heat in joules without converting one of them.

Example — Compression against a constant load

Problem
A gas is compressed from 5.0 L to 2.0 L against a constant external pressure of 2.0 bar. Find w.

1.Calculate the signed volume change: $Δ V = 2.0 - 5.0 = - 3.0 L$ .
2.Substitute the external pressure: $w = - (2.0) (- 3.0) = + 6.0 L bar$ .
3.Convert units: $w = + 6.0 (100) = + 600 J$ .
4.The positive result agrees with the physical picture: the surroundings transfer energy to the compressed gas.

Work is the area associated with a path

If the opposing pressure changes in several finite steps, calculate the work for each step and add the results. A pressure–volume graph makes this sum visible: each step contributes a rectangle with height equal to external pressure and width equal to the volume change. Different routes between the same volumes can enclose different areas and therefore transfer different amounts of work.

Finite steps and continuous pathsLaTeX

w = - i \sum p_{ext, i} Δ V_{i}, w = - \int_{V_{i}}^{V_{f}} p_{ext} d V

The integral is the continuous limit of the sum. Use external pressure unless reversibility permits replacing it with the system pressure.

Reading work from a pressure–volume path— Rectangles add in a stepped process. Their continuous limit becomes the area under a reversible pressure curve. The paths are schematic comparisons.

A reversible process is an ideal limiting process that could be reversed by an infinitesimal change in the driving conditions. During reversible compression, the opposing pressure exceeds the gas pressure only infinitesimally; during reversible expansion it is infinitesimally smaller. The process passes through a continuous sequence of equilibrium states. Moving slowly helps, but friction or a finite temperature difference can still make a slow process irreversible.

Reversible pressure–volume workLaTeX

w_{rev} = - \int_{V_{i}}^{V_{f}} p d V

In this limit the external pressure can be replaced by the equilibrium gas pressure p. For expansion between the same states, the reversible route delivers the greatest magnitude of expansion work.

An isothermal ideal gas

For an ideal gas at constant temperature, $p = n R T / V$ . Substituting this into the reversible work integral makes the inverse relationship between pressure and volume part of the calculation. Since n, R, and T are constant, they can be taken outside the integral. Integrating $1/ V$ gives a natural logarithm.

Isothermal reversible workLaTeX

w_{rev} = - n R T \int_{V_{i}}^{V_{f}} \frac{d V}{V} = - n R T ln \frac{V _{f}}{V _{i}} = - 2.303 n R T lo g_{10} \frac{V _{f}}{V _{i}}

Use absolute temperature in kelvin and a dimensionless volume ratio. For an ideal gas U depends only on temperature, so an isothermal change has ΔU = 0 and q = −w.

Example — The reversible expansion calculation

Problem
One mole of ideal gas expands reversibly and isothermally at 298 K from 2.0 L to 10.0 L. Find w and q.

1.The volume ratio is $V_{f} / V_{i} = 10.0/2.0 = 5.0$ . Its natural logarithm is approximately 1.609.
2.Using $R = 8.314 J mo l^{- 1} K^{- 1}$ , $w = - (1) (8.314) (298) ln 5 \approx - 3.99 kJ$ .
3.The ideal gas has $Δ U = 0$ because its temperature does not change.
4.From $0 = q + w$ , the gas absorbs $q = + 3.99 kJ$ . This heat replaces the energy transferred out as work.

Free expansion and useful limits

Free expansion means expansion against a vacuum. The external pressure is zero, so no pressure–volume work is transferred. It is irreversible: gas does not spontaneously gather back into its original smaller region. If an ideal gas starts and ends at the same temperature, its internal-energy change is zero as well; the first law then gives zero net heat transfer.

Example — Two expansions from the same starting volume

Problem
An ideal gas initially occupies 2.0 L at 5.0 atm. Compare its isothermal expansion to 10.0 L into (a) a vacuum and (b) a constant external pressure of 1.0 atm.

1.For (a), $p_{ext} = 0$ . Therefore $w = 0$ even though the volume increases.
2.The temperature is constant and the gas is ideal, so $Δ U = 0$ . Thus $q = 0$ in (a).
3.For (b), $w = - (1.0) (10.0 - 2.0) = - 8.0 L atm \approx - 811 J$ .
4.Again $Δ U = 0$ , so $q = + 811 J$ . The gas must receive heat while it does work.
5.This comparison shows how q and w depend on process conditions. The final gas pressure is $5.0 (2.0/10.0) = 1.0 atm$ , consistent with mechanical equilibrium against the load in (b).

For a rigid closed vessel, $Δ V = 0$ and pressure–volume work is zero. If no other work occurs, $Δ U = q_{V}$ . In an adiabatic process $q = 0$ , so $Δ U = w_{ad}$ . An expanding adiabatic ideal gas usually cools because it supplies work from its own internal energy. Isothermal and adiabatic therefore impose different conditions.

Condition	What is zero?	First-law consequence
Rigid, only pressure–volume work	$w = 0$	$Δ U = q_{V}$
Adiabatic	$q = 0$	$Δ U = w$
Isothermal ideal gas	$Δ U = 0$	$q = - w$
Free expansion	$p_{ext} = 0$ , so pressure–volume work is zero	Heat and ΔU still depend on the remaining conditions

Common mistake

Do not use $p V = n R T$ to replace external pressure in an irreversible-work formula. The gas may not even have one uniform equilibrium pressure during the change. Also, free expansion is intrinsically irreversible; zero work does not make a process reversible.

Quick check

A gas expands against a positive constant external pressure. What is the sign of w?

Quick check

Which pressure belongs in the irreversible boundary-work integral?

Quick check

A gas freely expands into a vacuum. What is its pressure–volume work?

Quick check

For a reversible isothermal ideal-gas expansion, which statement is correct?

Quick check

A gas is compressed from 4 L to 1 L against 1 bar. What is w?

Quick check

Which feature is necessary for the ideal reversible limit?

Practice Problems

Derive $w = - p_{ext} Δ V$ using piston area and displacement, explaining the negative sign.
Calculate the work when a gas expands from 3.0 L to 7.0 L against 1.5 bar.
A gas expands from 2 L to 4 L against 3 bar, then from 4 L to 8 L against 1 bar. Find its total work and compare with a single expansion from 2 L to 8 L against 1 bar.
Calculate the reversible isothermal expansion work for 0.50 mol of ideal gas at 300 K when its volume doubles.
Explain why no heat is absorbed in an isothermal ideal-gas free expansion even though the gas occupies more space.
A closed adiabatic system does 200 J of work. Determine its internal-energy change and describe a possible temperature response.

Key Takeaways

Pressure–volume work uses the external opposing pressure and the signed volume change. Work is path dependent and can be interpreted through areas on pressure–volume graphs. Reversible isothermal ideal-gas work is $- n R T ln (V_{f} / V_{i})$ . A vacuum gives zero boundary work; adiabatic, isothermal, and constant-volume conditions impose different restrictions.

Previous · Lesson 1

Systems, States, and the First Law

Next · Lesson 3

Enthalpy, Heat Capacity, and Calorimetry

A gas pushes against its surroundings

Work is the area associated with a path

An isothermal ideal gas

Free expansion and useful limits

Quiz

Practice Problems

Key Takeaways