Work Done On Or By A Gas

You're staring at a piston. Maybe it's just a diagram in a textbook that refuses to make sense. Here's the thing nobody tells you upfront: the gas inside doesn't care about your intuition. Maybe it's in a syringe you're pulling back. Maybe it's in a lawnmower engine. It only cares about pressure, volume, and whether the boundary moves.

What Is Work Done On or By a Gas

Work in thermodynamics isn't about effort. Now, it's not about sweat. It's a precise mechanical quantity: pressure times change in volume. That's it. When a gas expands, it pushes outward against whatever contains it — a piston, a balloon, the atmosphere itself. That push over a distance is work done by the gas. When something compresses the gas — you pushing the plunger, the atmosphere crushing a can — work is done on the gas.

The sign convention trips everyone up. Physics and chemistry historically picked opposite conventions. In physics, work done by the system is positive. On the flip side, the gas spends its energy. Now, in chemistry, work done on the system is positive. The gas gains energy. Neither is wrong. Plus, you just have to know which convention your textbook, professor, or simulation uses. Pick one. Stay consistent. Write it on a sticky note if you have to.

The integral you'll actually use

For a reversible process — quasistatic, no friction, the gas always in equilibrium — the work is the area under the curve on a PV diagram. Which means real expansions happen fast. ∫ P dV from initial volume to final volume. But real processes aren't always reversible. Which means simple. And the actual work is always ∫ P_ext dV — external pressure, not gas pressure. Even so, if pressure is constant, it collapses to PΔV. Real pistons have friction. Then the integral doesn't equal the actual work. That distinction matters more than most introductions admit.

Why It Matters

Engines. Refrigerators. Your lungs. So naturally, every heat engine that powers civilization — steam turbines, internal combustion, even the tiny Stirling engine on a desk toy — runs on a cycle where gas does net work over a loop. The area inside that loop on a PV diagram? That's the net work output per cycle. Practically speaking, no area, no net work. No engine.

Refrigerators and heat pumps flip the script. The diaphragm contracts, chest volume increases, lung pressure drops below atmospheric, air rushes in. Work done on the gas (negative in physics convention) during inhalation. Your lungs? You do work on the gas to move heat from cold to hot. So the compressor in your fridge is doing work on refrigerant gas right now. Exhalation is mostly passive — elastic recoil does work by the gas.

Get the sign wrong on a first law problem and your internal energy change flips sign. Think about it: your temperature prediction goes the wrong way. Practically speaking, this isn't pedantry. Your efficiency calculation becomes nonsense. It's the difference between an engine that runs and one that seizes.

How It Works

Constant pressure — the simplest case

Isobaric process. Pressure stays fixed while volume changes. Happens when the external pressure is constant and the process is slow enough for the gas to match it. A piston with a fixed weight on top. A balloon expanding in the atmosphere. Work = P(V₂ - V₁). Positive if V₂ > V₁ (expansion), negative if compression. The PV diagram is a horizontal line. The area under it is a rectangle. You can see the answer before you calculate.

Constant temperature — isothermal

Ideal gas, constant T. But expansion work is positive. On a PV diagram, it's a hyperbola. The natural log appears because pressure isn't constant — it drops as volume grows. Day to day, compression work is negative. But the magnitude is the same for equal volume ratios. Plug into the integral: W = nRT ln(V₂/V₁). That's why pV = nRT so P = nRT/V. The area under the curve from V₁ to V₂ — that's your work.

Here's what textbooks sometimes skip: for a real gas, the isotherm isn't a perfect hyperbola. Think about it: intermolecular forces matter. The van der Waals equation gives a different curve. The integral still works — you just integrate the real equation of state. But in an intro course? Also, ideal gas. Always ideal gas unless told otherwise.

Adiabatic — no heat exchange

Fast processes. 4. In practice, temperature drops during expansion. For an ideal gas, ΔU = nCᵥΔT. In practice, for diatomic gases like N₂ and O₂, γ ≈ 1. Temperature rises during compression. Q = 0 so ΔU = W (physics sign convention). No time for heat to flow. Because of that, for monatomic like helium or argon, γ ≈ 1. Because of that, the adiabat on a PV diagram is steeper than an isotherm — PV^γ = constant where γ = Cₚ/Cᵥ. The work equals the change in internal energy. Consider this: insulated containers. 67.

The work formula: W = (P₂V₂ - P₁V₁)/(1 - γ) = nR(T₂ - T₁)/(1 - γ). Plus, derive it once. Then memorize it. Because of that, or keep the derivation handy. The key insight: adiabatic work comes entirely from internal energy. No heat reservoir to draw from or dump into.

Want to learn more? We recommend how many protons does strontium have and what is the oxidation number of nitrogen in no2 for further reading.

Free expansion — the zero-work trap

Joule expansion. Gas rushes into vacuum. No piston. No external pressure. P_ext = 0. Work = 0. Always. Even though volume changes dramatically. Even though pressure drops. That said, even though it feels like "something happened. Even so, " No boundary moved against a force. No work. For an ideal gas, temperature doesn't change either (Joule's law). For real gases, there's a slight temperature change — the Joule-Thomson effect — but that's a different process (throttling, constant enthalpy). Because of that, free expansion is irreversible. Entropy increases. But work? But zero. This shows up on exams specifically to catch people who think "volume changed so work happened.

Cyclic processes

Engine cycles. Carnot, Otto, Diesel, Brayton, Rankine. On the flip side, the gas goes through a series of processes and returns to its initial state. Net work = area enclosed by the cycle on the PV diagram. On the flip side, clockwise cycle = net work done by the gas (engine). Counterclockwise = net work done on the gas (refrigerator/heat pump). The first law over a full cycle: ΔU = 0 so Q_net = W_net. That said, heat in minus heat out equals work out. That's the entire thermodynamic basis for every heat engine ever built.

Common Mistakes

Using gas pressure instead of external pressure for irreversible processes. The external pressure is what the gas actually pushes against. The formula W = ∫ P_gas dV only works for reversible processes. If the piston slams outward against a lower constant external pressure, the gas pressure isn't well-defined during the chaos — pressure waves, turbulence, non-equilibrium. Think about it: this is the big one. But the work is still ∫ P_ext dV. Always.

Forgetting the sign convention mid-problem. You start with physics convention (W_by positive), then halfway through you're thinking chemistry (W_on positive) because the textbook chapter before used that. Your ΔU = Q - W becomes ΔU = Q + W. The numbers look right but the physics is backwards. Because of that, pick a convention. Write it at the top of your paper.

Misinterpreting the Adiabatic condition

Students often assume that "adiabatic" and "isothermal" are interchangeable because both involve no heat exchange ($Q=0$). In an adiabatic process, the system is thermally isolated. Because no heat can enter to compensate for the energy used to do work, the internal energy must come from the kinetic energy of the molecules themselves, causing the temperature to plummet. On top of that, this is a fatal error. In an isothermal process, the system is in contact with a massive heat reservoir, allowing it to exchange energy to maintain a constant temperature. If a problem says "adiabatic," do not use $PV = \text{constant}$; you must use $PV^\gamma = \text{constant}$.

Neglecting the distinction between State and Path functions

Remember that $U$, $P$, $V$, and $T$ are state functions. On the flip side, $Q$ and $W$ are path functions. Now, you can reach the same final state via an isothermal path or an adiabatic path, but the work done and the heat exchanged will be vastly different. Their values depend only on the current state of the system, not how the system got there. When solving cycle problems, never try to calculate $Q$ or $W$ for an individual process using a "shortcut" formula unless you are certain the process is reversible and follows a specific mathematical path.

Summary and Final Checklist

Thermodynamics is a discipline of bookkeeping. If you can track the energy (Internal Energy), the heat (Thermal energy transfer), and the work (Mechanical energy transfer), you can solve almost any problem in the field. Before you start calculating, ask yourself these four questions:

Is the process reversible or irreversible? (This determines if you use $P_{gas}$ or $P_{ext}$ for work).
Is the system isolated, adiabatic, or isothermal? (This determines your choice of equation: $PV^\gamma$ vs $PV$ vs $\Delta U=0$).
Is it a state function or a path function? (This determines if you can simply look at the initial and final states).
Which sign convention am I using? (This prevents the most common mathematical error in the field).

Master these distinctions, and the complexity of the PV diagram becomes a map rather than a maze. Thermodynamics isn't about memorizing a dozen different formulas; it is about understanding how energy moves, how it is transformed, and how it is conserved.