ΔU = Q + W in AP Physics 2

ΔU = Q + W is the first law of thermodynamics in AP Physics 2: the change in a closed system's internal energy (ΔU) equals the energy transferred by heating (Q) plus the work done on the system (W). It's conservation of energy written for gases, where W = −PΔV for constant-pressure volume changes.

What is ΔU = Q + W?

ΔU = Q + W is conservation of energy dressed up for thermodynamics. It says a closed system's internal energy can only change in two ways. Energy can flow in or out by heating or cooling (that's Q), or the surroundings can do work on the system (that's W). Add those two transfers together and you get exactly how much the internal energy changed. Nothing appears or disappears.

The sign convention is where the points live. In the AP Physics 2 equation, W is work done ON the system. Compress a gas and W is positive, so compression alone raises internal energy. When the gas expands, the surroundings do negative work on it (W = −PΔV for constant or average pressure), so expansion alone lowers internal energy. Similarly, Q is positive when heat flows into the system and negative when it flows out. For an ideal gas, internal energy is just the total kinetic energy of the molecules, so ΔU is directly tied to temperature through U = 3/2 nRT. If ΔU goes up, temperature goes up. Period.

Why ΔU = Q + W matters in AP® Physics 2

This equation is the backbone of Topic 9.4 in Unit 9 (Thermodynamics) and directly supports learning objective AP Physics 2 Revised 9.4.B, describing the behavior of a system using thermodynamic processes. It also leans on 9.4.A, since you can't interpret ΔU without knowing internal energy is the sum of molecular kinetic energies (and that an ideal gas has no internal potential energy). Almost every thermodynamic process question on the exam, whether it's isothermal, adiabatic, isobaric, isochoric, or a full cycle, gets solved by tracking these three quantities and their signs. If you can bookkeep Q, W, and ΔU correctly, Unit 9 becomes accounting instead of mystery.

How ΔU = Q + W connects across the course

Work done on a system, W = −PΔV (Unit 9)

The W in ΔU = Q + W isn't a separate idea; it's calculated from pressure and volume change. The negative sign in W = −PΔV is what makes compression (negative ΔV) give positive work on the gas. The two equations are a matched set, and the exam loves making you use both in one problem.

P-V diagrams (Unit 9)

A P-V diagram is the first law drawn as a picture. The area under a process curve tells you the magnitude of the work, and the direction of travel tells you its sign. Reading W off the graph and plugging it into ΔU = Q + W is one of the most common moves in Unit 9.

Thermodynamic cycles (Unit 9)

In any complete cycle, the system returns to its starting state, so ΔU = 0. The first law then collapses to Q_net = −W_net. That single shortcut solves most cycle problems, like finding work done by a gas that absorbs 200 J and releases 150 J over a cycle (it does 50 J of work).

Internal energy of an ideal gas, U = 3/2 nRT (Unit 9)

This equation tells you what ΔU actually is for a monatomic ideal gas. Since U depends only on temperature, ΔU is your bridge between the first law and the gas's temperature change. ΔU positive means the gas got hotter, no exceptions.

Is ΔU = Q + W on the AP® Physics 2 exam?

Expect multiple-choice stems that give you two of the three quantities and ask for the third, like a gas whose internal energy rises 600 J while absorbing 400 J of heat (so 200 J of work was done on it). Sign tracking is the whole game. Multi-step problems chain two processes together and ask for net ΔU, which means summing Q and W across both steps with correct signs. Adiabatic questions test whether you know Q = 0, so ΔU comes entirely from work (adiabatic compression heats the gas). Cyclic questions test whether you know ΔU = 0 over a full cycle. No released FRQ has used the equation verbatim as a prompt, but it underpins the energy-reasoning FRQs in Unit 9, where you justify temperature changes from a P-V diagram using first-law logic.

ΔU = Q + W vs ΔU = Q − W

You may see the first law written as ΔU = Q − W in chemistry class or older textbooks. That version defines W as work done BY the gas. The AP Physics 2 equation sheet uses ΔU = Q + W, where W is work done ON the gas. They describe the same physics with opposite sign conventions. On the AP exam, always use W as work done on the system, so compression means positive W and expansion means negative W. Mixing the conventions flips your answer's sign and costs you the point.

Key things to remember about ΔU = Q + W

• ΔU = Q + W is the first law of thermodynamics, which is conservation of energy applied to a closed system that can exchange energy through heating and work.

• On the AP Physics 2 equation sheet, W means work done ON the system, so compressing a gas gives positive W and an expanding gas gives negative W.

• Q is positive when heat flows into the system and negative when heat flows out of it.

• For an ideal gas, ΔU depends only on temperature change because U = 3/2 nRT, so positive ΔU always means the gas got hotter.

• In an adiabatic process Q = 0, so any change in internal energy comes entirely from work done on or by the gas.

• Over a complete thermodynamic cycle ΔU = 0, so the net heat added equals the net work done by the system.

Frequently asked questions about ΔU = Q + W

What does ΔU = Q + W mean in AP Physics 2?

It's the first law of thermodynamics. The change in a system's internal energy (ΔU) equals the heat added to it (Q) plus the work done on it (W). It's just conservation of energy written for thermodynamic systems.

Is W work done on the gas or by the gas in ΔU = Q + W?

On the gas. The AP Physics 2 convention defines W as work done ON the system, calculated as W = −PΔV for constant pressure. Work done BY the gas is the negative of that.

Why does my chemistry book say ΔU = Q − W instead?

It's the same law with the opposite sign convention. In ΔU = Q − W, W means work done by the gas; in the AP Physics 2 version ΔU = Q + W, W means work done on the gas. Stick with the equation sheet's version on the exam.

Does adding heat to a gas always increase its internal energy?

No. If the gas expands while you heat it, it does work on its surroundings (W is negative), which can cancel or exceed Q. In an isothermal expansion, all the added heat goes into work and ΔU stays zero.

What is ΔU for a complete thermodynamic cycle?

Zero. Internal energy depends only on the state of the gas, so returning to the starting point means ΔU = 0. That forces the net heat added to equal the net work done by the system. For example, a cycle absorbing 200 J and releasing 150 J does 50 J of work.