Pressure-Volume Work

Push a piston, inflate a balloon, or let a gas escape into a vacuum. Every one of these events moves a boundary and does work. Pressure-volume work is the currency by which gases trade energy with their surroundings. This happens whenever their volume changes. It is the first concrete term chemists learn to compute in the First Law of Thermodynamics. It also sets up everything that follows in thermochemistry. Understanding this work term matters for an important reason. It separates the energy a system exchanges as ordered mechanical work from disordered heat. That distinction is the foundation for defining enthalpy. It also helps in predicting whether a process can do useful work. And it allows comparing engines, batteries, and chemical reactions on the same energetic footing. This article derives the general expression for expansion and compression work. You will see how it reduces to a simple formula for constant external pressure. Then you will work two fully numeric examples. Finally you will see why the sign convention matters. You will also see where students most often go wrong.

Key takeaways

Pressure-volume work arises whenever a system's volume changes against an external pressure.
The general definition is $w = -\int p_{ext}\,dV$ , valid for any process, reversible or not.
Against a constant external pressure this collapses to $w = -p_{ext}\Delta V$ .
Expansion ( $\Delta V > 0$ ) makes $w$ negative: the system loses energy doing work on the surroundings.
Free expansion into a vacuum does zero work, because $p_{ext} = 0$ .
Key idea: work done by a gas depends on the external pressure it pushes against, not on the gas's own internal pressure.

The core insight

Pressure-volume work is force times distance rewritten in terms of pressure and volume; because the system does work against an external pressure, the work done on the system carries a minus sign, $w = -\int p_{ext}\,dV$ , and this single integral is the starting point for every constant-pressure, constant-volume, isothermal, or adiabatic calculation that follows in thermochemistry.

The physical picture

Picture a gas confined in a cylinder by a frictionless, weightless piston of cross-sectional area $A$ . The surroundings press on the piston with a fixed external pressure $p_{ext}$ , perhaps because a stack of weights sits on it, or because the atmosphere itself pushes down. If the gas expands, the piston moves outward through a distance $dx$ , sweeping out a volume $dV = A\,dx$ . The mechanical work done by the gas on the piston is force times distance:

dw_{by\,gas} = F\,dx = (p_{ext}A)\,dx = p_{ext}\,dV

Thermodynamics, however, tracks energy from the system's point of view. When the system does positive work on the surroundings, the system's own energy decreases. The convention adopted throughout modern physical chemistry (the "surroundings" or IUPAC convention baked into the First Law, $\Delta U = q + w$ ) defines $w$ as the work done on the system. Expansion work done by the gas therefore enters the system's energy balance with a negative sign:

dw = -p_{ext}\,dV

If instead the surroundings compress the gas ( $dV < 0$ ), $dw$ is positive: work flows into the system, raising its energy. The external pressure, not the internal gas pressure, controls how much work is exchanged, because it is the external pressure that actually supplies or absorbs the force at the moving boundary. This distinction becomes critical when comparing reversible and irreversible paths.

Figure

Piston pushing against a constant external pressure

Figure

Work as area under a p versus V curve

The core relations

dw = -p_{ext}\,dV

where: $dw$ is the infinitesimal work done on the system, $p_{ext}$ is the pressure the surroundings exert at the moving boundary, $dV$ is the infinitesimal volume change of the system.

w = -\int_{V_1}^{V_2} p_{ext}\,dV

where: $V_1, V_2$ are initial and final volumes; valid for expansion, compression, reversible, or irreversible paths, since it uses only $p_{ext}$ .

w = -p_{ext}(V_2 - V_1) = -p_{ext}\Delta V

where: this special case holds only when $p_{ext}$ is constant throughout the change, e.g. a piston opposed by a fixed atmospheric pressure.

\Delta U = q + w

where: $\Delta U$ is the change in internal energy, $q$ is heat absorbed by the system, $w$ is work done on the system (First Law, assumed known from the prerequisite article).

Derivation AFrom force on a piston to

w = -p_{ext}\Delta V

Goal. Derive the work done on a gas-filled cylinder when its piston moves against a constant external pressure, starting only from the mechanical definition of work.

Step 1: Define the geometry. Let the piston have cross-sectional area $A$ and let $x$ measure its position, increasing as the gas expands. The gas volume is $V = Ax$ , so a small displacement $dx$ corresponds to $dV = A\,dx$ .

Step 2: Write the mechanical work. The external surroundings exert a force $F_{ext} = p_{ext}A$ on the piston face (Newton's third law pairs this with the force the gas exerts on the piston). The work done by the gas as the piston moves outward by $dx$ is

dw_{by\,gas} = F_{ext}\,dx = p_{ext}A\,dx = p_{ext}\,dV

Step 3: Switch to the system's energy bookkeeping. The First Law tracks work done on the system. Since work done by the gas on the surroundings is energy leaving the system,

dw = -dw_{by\,gas} = -p_{ext}\,dV

Step 4: Integrate over the whole process. For an arbitrary path from $V_1$ to $V_2$ , sum (integrate) these infinitesimal contributions:

w = -\int_{V_1}^{V_2} p_{ext}\,dV

This holds even if $p_{ext}$ varies with $V$ along the path — it is the fully general expansion-work integral.

Step 5: Specialize to constant external pressure. If a single fixed weight or the constant atmosphere supplies $p_{ext}$ , it can be pulled outside the integral:

w = -p_{ext}\int_{V_1}^{V_2} dV = -p_{ext}(V_2 - V_1)

Step 6: Box the result.

\boxed{w = -p_{ext}\Delta V}

Check the signs: if the gas expands, $\Delta V > 0$ , so $w < 0$ — energy leaves the system as work, exactly as expected for a gas pushing back a piston. If the gas is compressed, $\Delta V < 0$ , so $w > 0$ — the surroundings do work on the gas, increasing its energy. If the gas expands into a vacuum, $p_{ext} = 0$ , so $w = 0$ regardless of how much the volume changes — this is "free expansion," and no work crosses the boundary because there is nothing to push against.

Worked examples

Example 1 — Foundational: gas expanding against constant atmospheric pressure. A sample of gas in a cylinder expands from $V_1 = 2.00\ \text{L}$ to $V_2 = 6.50\ \text{L}$ at constant external pressure $p_{ext} = 1.00\ \text{atm}$ . Find the work done on the gas, in joules. Step 1. Compute $\Delta V$ :

\Delta V = V_2 - V_1 = 6.50\ \text{L} - 2.00\ \text{L} = 4.50\ \text{L}

Step 2. Apply the constant-pressure formula:

w = -p_{ext}\Delta V = -(1.00\ \text{atm})(4.50\ \text{L})

w = -4.50\ \text{L}\,\cdot\,\text{atm}

Step 3. Convert to joules using $1\ \text{L}\,\cdot\,\text{atm} = 101.325\ \text{J}$ :

w = -4.50 \times 101.325\ \text{J} = -455.96\ \text{J} \approx -456\ \text{J}

Sanity check: the gas expanded, so it did work on the surroundings; the system's energy from work alone should decrease, matching the negative sign. The magnitude (~456 J) is a modest, chemically reasonable value for a few liters of expansion at one atmosphere.

Example 2 — Applied, exam-level: two-stage compression with heat exchange, finding $\Delta U$ . A chemist compresses 1.00 mol of an ideal gas from $V_1 = 20.0\ \text{L}$ to $V_2 = 10.0\ \text{L}$ at $298\ \text{K}$ in two stages, each against a different constant external pressure, and measures that the whole process releases $q = -350\ \text{J}$ of heat to the surroundings. Stage A compresses the gas from $20.0\ \text{L}$ to $15.0\ \text{L}$ against $p_{ext,A} = 1.50\ \text{atm}$ ; stage B compresses it from $15.0\ \text{L}$ to $10.0\ \text{L}$ against $p_{ext,B} = 2.50\ \text{atm}$ . Find $w$ for the whole process and then $\Delta U$ . Step 1. Work in stage A:

w_A = -p_{ext,A}\Delta V_A = -(1.50\ \text{atm})(15.0 - 20.0)\ \text{L}

w_A = -(1.50\ \text{atm})(-5.00\ \text{L}) = 7.50\ \text{L}\,\cdot\,\text{atm}

Step 2. Work in stage B:

w_B = -p_{ext,B}\Delta V_B = -(2.50\ \text{atm})(10.0 - 15.0)\ \text{L}

w_B = -(2.50\ \text{atm})(-5.00\ \text{L}) = 12.5\ \text{L}\,\cdot\,\text{atm}

Step 3. Total work, converted to joules:

w = w_A + w_B = 7.50 + 12.5 = 20.0\ \text{L}\,\cdot\,\text{atm}

w = 20.0 \times 101.325\ \text{J} = 2026.5\ \text{J} \approx +2.03\ \text{kJ}

Step 4. Apply the First Law to get $\Delta U$ :

\Delta U = q + w = (-350\ \text{J}) + (2026.5\ \text{J}) = 1676.5\ \text{J} \approx +1.68\ \text{kJ}

Sanity check: compression means work is done on the gas, so $w$ should be positive — it is. The gas is compressed and cools by releasing heat, yet the net internal energy still rises because the work input (2.03 kJ) outweighs the heat lost (0.35 kJ); this is physically consistent for a fast, irreversible compression. As a rough cross-check via multi-stage staircase work, note that a single-stage compression straight to 2.50 atm would give only $w = -(2.50)(10.0-20.0) = 25.0$ L·atm — larger, confirming that splitting into stages against smaller pressures first, then larger, extracts a smaller total work input than one abrupt jump to the highest pressure, exactly the trend expected en route to the reversible limit.

Limiting cases & common traps

Situation	Formula to use	Key point
Constant $p_{ext}$	$w = -p_{ext}\Delta V$	Simplest and most common exam case
Free expansion (vacuum)	$w = 0$	No matter how large $\Delta V$ is, $p_{ext}=0$
Varying $p_{ext}$ (reversible path)	$w = -\int p_{ext}\,dV$ with $p_{ext}\approx p_{gas}$	Leads to isothermal reversible work, covered next
Rigid container	$w = 0$	$\Delta V = 0$ regardless of pressure changes

The chemist behind it

The modern treatment of pressure-volume work descends from the work of Rudolf Clausius, who gave thermodynamics its precise mathematical form in the 1850s, including the clear separation of heat and work in the energy balance that we now call the First Law. His formulation of internal energy as a state function, and his insistence on tracking work done at a moving boundary via $p\,dV$ , set the notation and sign conventions still taught today.

Origins · 1 minRudolf Clausius and how this was discovered

Rudolf Clausius

1822-1888 · German physicist who formalized the First and Second Laws and coined the term "entropy"

Clausius reformulated Sadi Carnot's ideas on heat engines into rigorous mathematics, establishing $dU = dq - dw_{by\,system}$ (equivalent to today's $\Delta U = q+w$ convention) and showing that pressure-volume work is the mechanical term linking a gas's volume changes to its energy exchanges with the surroundings.

Ask the tutor: How does the work integral change if the external pressure varies continuously, as in a reversible isothermal expansion?

Why is there a minus sign in w = -∫p_ext dV?

The minus sign encodes the convention that $w$ is the work done on the system, consistent with $\Delta U = q + w$ . When a gas expands ( $dV>0$ ), it pushes the surroundings back and loses energy as work, so $w$ must come out negative. Without the minus sign, expansion would incorrectly appear to increase the system's energy. The sign is not arbitrary — it is forced by defining $w$ from the system's perspective rather than the surroundings'.

Does the formula w = -p_ext ΔV work for any expansion or compression?

It works whenever $p_{ext}$ is constant throughout the volume change, such as a piston loaded with a fixed weight or exposed to a constant atmosphere. If $p_{ext}$ varies during the process — for example in a reversible expansion where $p_{ext}$ is continuously adjusted to match the gas pressure — you must return to the integral $w = -\int p_{ext}\,dV$ and integrate properly, often yielding a logarithmic result for isothermal ideal gases.

What is the difference between the gas pressure and p_ext, and why does it matter?

The gas pressure $p$ describes the internal state of the system, obeying an equation of state like $pV=nRT$ . The external pressure $p_{ext}$ is whatever mechanical force per area acts at the moving boundary, set by the surroundings. They are equal only in the idealized reversible (quasi-static) limit. In any real, finite-rate process they generally differ, and only $p_{ext}$ appears in the work integral because it is what actually does the pushing.

How does this connect to enthalpy and the First Law?

Enthalpy is defined as $H = U + pV$ specifically so that, at constant pressure, $\Delta H = q_p$ , absorbing the $p\Delta V$ work term automatically. Since so many chemical reactions run open to the atmosphere at constant $p_{ext}$ , this article's constant-pressure work formula is the mechanical piece that, combined with the First Law, motivates why chemists prefer enthalpy over raw internal energy for everyday calorimetry.

What comes next after mastering this constant-pressure case?

The natural extension is comparing reversible and irreversible paths between the same two states, since they generally give different work even though $\Delta U$ depends only on the states. That comparison sets up the classic calculation of isothermal reversible expansion of an ideal gas, where $p_{ext}$ tracks $p_{gas} = nRT/V$ at every instant and the integral yields a logarithm rather than a simple product.