Isothermal vs Adiabatic: The Difference

The sign convention, before anything else

Every number on this page depends on the convention, and it is the opposite of what many textbooks print, so it goes first.

The AP Physics 2 equation sheet's Thermal Physics panel prints

and the symbol list on that same panel defines the letters: $Q$ is the energy transferred to a system by heating, and $W$ is the work done on a system. Essential knowledge 9.4.B.1.ii and 9.4.B.1.iii are worded to match.

So compression, which makes $\Delta V$ negative, makes $W$ positive: work goes into the gas. Expansion makes $W$ negative. Both $Q$ and $W$ are positive when energy enters. If a source writes $\Delta U = Q - W$, its $W$ is the work done by the gas, and every sign on this page would need reversing to match it. Use the sheet's version on the exam. Heat vs work works through the translation.

The distinction, stated once

Both of these processes are named for a quantity being held at zero, and they are named for different quantities in the same equation.

Essential knowledge 9.4.B.3 lists the special cases together: constant volume (isovolumetric), constant temperature (isothermal), and constant pressure (isobaric), as well as processes where no energy is transferred to or from the system through thermal processes (adiabatic).

Read the two definitions side by side and the difference is structural.

• Isothermal fixes the temperature. For an ideal gas, 9.4.A.1.ii gives $U = \frac{3}{2}nRT$, so a fixed temperature means a fixed internal energy: $\Delta U = 0$. The first law then reads $0 = Q + W$, so $Q = -W$.
• Adiabatic fixes the heating at zero. That is 9.4.B.3's own wording, no energy transferred to or from the system through thermal processes: $Q = 0$. The first law then reads $\Delta U = W$.

So isothermal zeroes the left-hand side of $\Delta U = Q + W$, and adiabatic zeroes the first term on the right. Neither process zeroes the work, and that is the point: in both of them work is being done, and the difference is only where that energy ends up.

In an isothermal compression the work you put in leaves again as cooling, and the gas is exactly as warm at the end as at the start. In an adiabatic compression the work you put in has nowhere to go, so it stays as internal energy and the gas gets hotter. Same push, different destination.

One clause in the isothermal case is doing quiet work. $\Delta U = 0$ follows from constant temperature only because the internal energy of an ideal gas depends on temperature alone, which is 9.4.A.1.i and 9.4.A.1.ii: the atoms in an ideal gas do not interact via conservative forces and the internal structure is not considered, so an ideal gas does not have internal potential energy. Drop the ideal-gas model and constant temperature stops guaranteeing constant internal energy.

Isothermal against adiabatic, row by row

Row three is the one to memorise, because it is the only row a question can hinge on that you cannot recover from the others.

Row nine is worth stating plainly rather than working around. The AP Physics 2 sheet gives you $PV = nRT$, so you can compute anything you like along an isotherm. It prints no equation relating $P$ and $V$ along an adiabat, and the CED introduces no exponent for one, so an adiabatic process is handled through the first law and through qualitative reasoning about the direction of the temperature change. If a question needs the final pressure of an adiabatic compression, it has to give you enough to get there.

Row eight can be derived rather than recalled, which is safer. 9.4.B.2.i says lines of constant temperature on a PV diagram are called isotherms. An adiabatic compression raises the temperature, so its final state must sit on a higher isotherm than its start. A curve that climbs across isotherms is steeper than the isotherm it crosses. That is the whole argument, and it needs no formula.

The case that separates them: the same compression, two different outcomes

Take $0.50\ \text{mol}$ of a monatomic ideal gas at $300\ \text{K}$ and compress it, putting $250\ \text{J}$ of work into it. Do it twice, once with the cylinder sitting in a large water bath at $300\ \text{K}$ and once with the cylinder insulated.

In the water bath, isothermally. The bath holds the gas at $300\ \text{K}$, so $\Delta T = 0$ and by 9.4.A.1.ii $\Delta U = 0$. The first law gives $Q = -W = -250\ \text{J}$: exactly $250\ \text{J}$ leaves the gas by cooling into the bath. The gas ends at $300\ \text{K}$, the temperature it started at, and its internal energy is unchanged at $\frac{3}{2}(0.50)(8.31)(300) = 1870\ \text{J}$.

Insulated, adiabatically. No thermal process is available, so $Q = 0$ and the first law gives $\Delta U = W = +250\ \text{J}$. Using $\Delta U = \frac{3}{2}nR\Delta T$ with $\frac{3}{2}nR = 6.2325\ \text{J/K}$, the temperature rises by $250/6.2325 = 40.1\ \text{K}$, ending at $340\ \text{K}$.

Same gas, same amount of work, and a $40\ \text{K}$ difference in where it ends up. The only thing that changed is whether energy was allowed to leave through the walls.

Now reverse the direction, because the reversal is where the marks are. Let the same gas expand, doing $400\ \text{J}$ of work on its surroundings, so $W = -400\ \text{J}$.

• Isothermally, $\Delta U = 0$ still, so $Q = -W = +400\ \text{J}$. Energy flows in from the bath, and it does so precisely because the gas would otherwise cool.
• Adiabatically, $Q = 0$ still, so $\Delta U = W = -400\ \text{J}$ and the temperature falls by $400/6.2325 = 64.2\ \text{K}$, to $236\ \text{K}$.

That is the pair of results worth carrying into an exam. An adiabatic compression heats a gas and an adiabatic expansion cools it, with no heating or cooling of any kind involved in either. Meanwhile an isothermal compression pushes energy out as cooling and an isothermal expansion pulls energy in as heating, and the temperature never moves.

One restriction on the comparison. The two processes do not end at the same volume even though the same work was done, and the AP Physics 2 sheet prints nothing that would let you find the adiabatic one's final volume from the work alone. What the first law gives you cleanly is the energy accounting, and that is what the exam asks for.

Where each one comes from physically

The two processes are not just algebraic cases. They correspond to two opposite laboratory strategies, and knowing which is which lets you read the setup off a question.

Isothermal means slow and connected. Put the gas in thermal contact with something so large that its own temperature does not budge, and move the piston slowly enough that energy has time to flow in or out and keep pace. 9.3.A.1 supplies the condition, that two systems are in thermal contact if they may transfer energy by thermal processes, and 9.3.A.4 supplies the reason it works, that thermal equilibrium results when no net energy is transferred between systems at the same temperature. A slow process stays effectively at equilibrium with the bath throughout.

Adiabatic means fast or insulated. Either block the transfer with insulating walls or move the piston so quickly that there is no time for a thermal process to matter. Both give $Q = 0$ for the same reason: no energy actually crossed the boundary thermally.

The CED's own Unit 9 sample activities point at the demonstration. Activity 3 for Topic 9.4 asks students to demonstrate each of eight thermodynamic processes, listing them as isobaric, isovolumetric, isothermal, and adiabatic, each in both possible directions, and its worked example is adiabatic compression: cap a syringe and push on the plunger really hard, which the CED flags under the name fire syringe. Push hard and fast and the trapped air heats enough to be interesting, and no hot plate is anywhere near it.

That activity's framing is worth borrowing. There are eight processes, not four: each of the four named types runs in two directions, and the direction flips every sign. An isothermal expansion and an isothermal compression have opposite signs for both $Q$ and $W$; an adiabatic expansion and an adiabatic compression have opposite signs for both $W$ and $\Delta T$. Naming the type is half the answer, and naming the direction is the other half.

The step-by-step procedure for reading these processes off a diagram, including cycles built from them, is in the PV diagram guide.

When it costs a mark

• Saying an isothermal process has $Q = 0$. That is the adiabatic one. An isothermal process has $\Delta U = 0$, which forces $Q = -W$, and unless the work is zero the heating is not. Swapping these two is the single error this page exists to prevent.
• Saying an adiabatic process has constant temperature. Constant $Q$ at zero, not constant $T$. An adiabatic compression always heats an ideal gas, because $\Delta U = W$ and compression makes $W$ positive.
• Assuming no heating means no temperature change. It means the opposite here: with nowhere for the work to go, the temperature has to absorb all of it.
• Using $W = -P\Delta V$ along an isotherm. 9.4.B.1.iii specifies a constant or average external pressure. On an isotherm the pressure varies continuously with volume, so the work has to come from the area under the curve, which is 9.4.B.2.ii, and the CED gives you the absolute value from the area with the sign coming from the direction of travel.
• Reaching for an adiabatic exponent. The AP Physics 2 sheet prints no relation between $P$ and $V$ for an adiabatic process and the CED introduces none, so any exponent you remember is outside the course. Handle an adiabat through $\Delta U = W$.
• Applying $\Delta U = 0$ to a non-ideal system at constant temperature. The step from constant $T$ to constant $U$ uses 9.4.A.1.i, that an ideal gas does not have internal potential energy. The sheet's exam conventions also state that ideal gases are monatomic, which is what licenses $U = \frac{3}{2}nRT$ rather than a different coefficient.
• Forgetting that both terms are still nonzero in the isothermal case. A student who reports $Q = 0$ and $W = 0$ for an isothermal compression has described nothing happening at all, when in fact a full $250\ \text{J}$ went in one side and out the other.

When they coincide, and why that lulls you

There is exactly one situation where the two processes are indistinguishable, and it is the situation where nothing happens.

If the volume is held fixed then $W = 0$. An isothermal process at fixed volume has $\Delta U = 0$ and therefore $Q = 0$; an adiabatic process at fixed volume has $Q = 0$ and therefore $\Delta U = 0$. Both descriptions collapse onto the same state, and both are correct, because a gas that is neither compressed nor heated is simply sitting there.

That is the whole overlap, and it is worth noticing because of what it reveals: the two processes are distinguished by what happens to the work, and if there is no work there is no distinction. Everything on this page follows from having a piston that moves.

The more dangerous near-coincidence is in the words. Both names begin by telling you something is not changing, and English does not help you keep track of which. "Isothermal" contains the word for heat in its root and yet is the case where heating definitely happens. "Adiabatic" sounds like a statement about temperature and is a statement about the walls. The mapping is:

• isothermal, iso-thermal, same temperature, therefore $\Delta U = 0$;
• adiabatic, no passage through, therefore $Q = 0$.

A third overlap is more subtle and shows up on cycles. Over a complete cycle $\Delta U = 0$ no matter which processes it was built from, because the gas returns to its starting state and 9.4.A.1.ii makes $U$ a function of $n$ and $T$ alone. So a cycle obeys $Q = -W$ overall, the same relationship an isothermal step obeys, without any individual step being isothermal. Reading that relationship backwards, and concluding a cycle must have been isothermal, is a real error and the cycle worked examples in the PV diagram guide are where to practise avoiding it.