First Law of Thermodynamics
A gas in a cylinder can be heated, and it can push a piston outward doing work. The first law of thermodynamics is simply energy conservation applied to exactly this situation — but getting the signs right on every term is where most marks in this topic are actually won or lost.
Energy conservation for a thermodynamic system
A gas trapped in a cylinder — the working fluid of an engine, in effect — can exchange energy with its surroundings in exactly two ways: it can be heated (or cooled), and it can have work done on it or do work on something else, typically by pushing a piston. The first law of thermodynamics simply insists that energy is conserved when both of these happen at once:
Here $Q$ is the energy transferred to the system by heating, $\Delta U$ is the resulting increase in the system's internal energy, and $W$ is the work done by the system on its surroundings. In words: whatever heat you put into a gas either raises its internal energy, or is spent doing work as the gas pushes outward, or some combination of both — there is nowhere else for the energy to go.
This is nothing more exotic than the conservation of energy, but applying it correctly to real thermodynamic processes depends entirely on tracking the direction of every energy transfer with the correct sign.
Every term in $Q = \Delta U + W$ has a sign, and mixing one up is the most common way marks are lost throughout this topic. The convention is:
- $Q$ is positive when heat flows into the system (the system is heated). $Q$ is negative when the system releases heat to its surroundings (the system is cooled).
- $W$ is positive when the system does work on its surroundings — for example, a gas expanding and pushing a piston outward. $W$ is negative when work is done on the system by its surroundings — for example, a gas being compressed by an external force pushing the piston in.
- $\Delta U$ is positive when internal energy — and hence temperature, for an ideal gas — increases. $\Delta U$ is negative when it falls.
Before substituting a single number into $Q = \Delta U + W$, always pause and assign the correct sign to each quantity based on the physical direction of the transfer described in the question. A right-magnitude, wrong-sign answer is still a wrong answer.
Applying the first law to the four standard processes
Rather than treating $Q = \Delta U + W$ as an abstract formula, its real power in this topic comes from applying it systematically to four standard ways a gas can change state — each one holding a different macroscopic variable fixed.
- Isothermal (constant temperature). The gas expands or is compressed slowly enough, and with enough thermal contact with its surroundings, that its temperature never changes. Since internal energy depends only on temperature for an ideal gas, $\Delta U = 0$, so the first law collapses to $Q = W$ exactly — all the heat supplied goes directly into the work done, and vice versa. The process follows $pV = \text{constant}$ (Boyle's law, at fixed $T$, following directly from $pV=nRT$).
- Adiabatic (no heat transfer). The gas is thermally insulated, or the change happens so rapidly that no heat has time to flow in or out, so $Q = 0$. The first law then gives $W = -\Delta U$: any work the gas does as it expands comes entirely at the expense of its own internal energy, so an adiabatically expanding gas necessarily cools down; an adiabatically compressed gas necessarily heats up, since work is now done on it. This process follows $pV^\gamma = \text{constant}$, where $\gamma$ is the ratio of specific heat capacities of the gas.
- Constant volume (isochoric). The gas is held in a rigid, fixed-volume container, so it cannot expand or be compressed — there is no boundary to push against, so no work is done, $W = 0$. The first law then gives $Q = \Delta U$ exactly: every joule of heat supplied goes directly into raising internal energy (and hence temperature).
- Constant pressure (isobaric). The gas expands or contracts against a constant external pressure $p$, doing work $W = p\Delta V$. The first law becomes $Q = \Delta U + p\Delta V$: heat supplied is shared between raising the internal energy and doing work against the constant external pressure.
| Process | Held constant | Resulting relationship |
|---|---|---|
| Isothermal | Temperature ($pV=\text{const}$) | $\Delta U = 0 \Rightarrow Q = W$ |
| Adiabatic | No heat transfer ($pV^\gamma=\text{const}$) | $Q = 0 \Rightarrow W = -\Delta U$ |
| Constant volume | Volume (no work possible) | $W = 0 \Rightarrow Q = \Delta U$ |
| Constant pressure | Pressure | $W = p\Delta V \Rightarrow Q = \Delta U + p\Delta V$ |
Notice these four cases aren't four separate laws — they're all the same first law, $Q = \Delta U + W$, with one term forced to zero (or given a specific form) by the physical constraint of the process. Recognising which variable a question is holding fixed is usually the key step in choosing which simplified version to use.
Key takeaways
- The first law of thermodynamics, $Q = \Delta U + W$, is a statement of energy conservation: heat supplied either raises internal energy or is used to do work, or both.
- $Q$ is positive for heat into the system; $W$ is positive for work done by the system on its surroundings — getting these signs wrong is the single most common source of lost marks in this topic.
- Isothermal: $\Delta U = 0$, so $Q = W$ exactly, following $pV = \text{constant}$.
- Adiabatic: $Q = 0$, so $W = -\Delta U$ exactly — an expanding adiabatic gas cools itself by doing work, following $pV^\gamma = \text{constant}$.
- Constant volume: $W = 0$, so $Q = \Delta U$ exactly — no work is possible with no volume change.
- Constant pressure: $W = p\Delta V$, so $Q = \Delta U + p\Delta V$.
- The equation of state $pV = nRT$ underlies all four processes and connects pressure, volume and temperature throughout.