March 3, 2026

AP Physics 2: The First Law of Thermodynamics

« Back to AP Physics Guide / Unit 9 Part 1: Ideal Gas Law

Abstract illustration showing energy conservation: Heat entering a system, a piston doing work, and internal energy increasing.

The First Law of Thermodynamics is essentially the Law of Conservation of Energy, but specifically applied to thermal systems. It tells us that the total energy of a gas system only changes if heat is added/removed or if work is done on/by the gas.

1. The Energy Balance Equation

The internal energy (\Delta U) of a gas depends entirely on its temperature. If you add heat (Q) or do work (W), that energy has to go somewhere.

    \[\Delta U = Q + W\]

\Delta U: Change in Internal Energy | Q: Net Heat Transfer | W: Work done ON the gas.
⚠️ The Sign Convention: This is the biggest point of confusion. In AP Physics 2:
  • +Q: Heat is added to the system.
  • -Q: Heat is removed (exhausted).
  • +W: Work is done on the gas (compression).
  • -W: Work is done by the gas (expansion).

2. Work and the PV Diagram

On the AP exam, you rarely calculate work with complex calculus. Instead, you use the geometry of the PV diagram. Work is the area under the curve.

    \[W = -P \Delta V\]

If the gas expands (\Delta V is positive), the work done on the gas is negative.

3. Thermodynamic Cycles (Engines)

A cycle is a series of processes that returns the gas to its original state (P, V, T). Because the gas ends where it started, the total \Delta U for a full cycle is always zero.

A closed rectangular loop on a PV diagram. Arrows show the path: Isobaric expansion, Isochoric cooling, Isobaric compression, Isochoric heating.

Net Work: The work done during a full cycle is the area enclosed by the loop. Clockwise cycles represent heat engines (doing net work on the surroundings).

4. Heat Engines & Refrigerators

A Heat Engine takes heat from a hot reservoir (Q_H), does some work (W), and exhausts the rest to a cold reservoir (Q_C). A Refrigerator is just a heat engine running in reverse—it uses work to pull heat out of a cold area.

    \[e = \left| \frac{W}{Q_H} \right| = \frac{Q_H - Q_C}{Q_H}\]

Efficiency (e): How much of the heat you paid for actually turned into work.
Energy flow diagrams for a heat engine (heat moving from hot to cold while doing work) and a refrigerator (work being used to move heat from cold to hot).

Energy Flow: In an engine, work comes out. In a refrigerator, work must go in. Nature never lets Q flow from cold to hot for free!

The Carnot Limit (Ideal Efficiency)

The Second Law of Thermodynamics states that no engine can be 100% efficient. The maximum possible efficiency any engine can achieve is called the Carnot Efficiency, which depends only on the temperatures of the reservoirs.

    \[e_c = \frac{T_H - T_C}{T_H}\]

Note: Temperatures must be in Kelvin.
⚠️ AP Exam Tip: If a question asks why a real engine’s efficiency is lower than the calculated e_c, the answer is usually Entropy. Real processes are irreversible and always generate “lost” heat due to friction and turbulence.

5. Entropy & The Second Law

While the First Law is about quantity of energy, the Second Law is about quality and direction. It explains why heat never spontaneously flows from cold to hot.

What is Entropy?

Entropy (S) is a measure of the disorder or the number of possible microscopic states of a system. The Second Law states that for any spontaneous process, the total entropy of the universe must increase.

    \[\Delta S = \frac{Q}{T}\]

Entropy increases when heat is added to a system (Q > 0).
⚠️ The “Direction” Rule: Nature always moves toward a state of higher entropy (disorder). This is why energy spreads out and why 100% efficient engines are physically impossible.

6. Quick AP Practice

📚 AP Practice Problems

1. 500 \, \text{J} of heat is added to a gas, and the gas does 200 \, \text{J} of work expanding a piston. What is the change in internal energy?

Answer Using \Delta U = Q + W: Heat added is +500 \, \text{J}. Work done by the gas is -200 \, \text{J}. Therefore, \Delta U = 500 - 200 = <strong>+300 \, \text{J}</strong>.

2. In an isothermal expansion, the internal energy does not change (\Delta U = 0). If Q is positive, what must be true about W?

Answer If 0 = Q + W, then W = -Q. This means all the heat added to the system is immediately converted into work done by the gas.

🧠 Unit 9 Mastery Challenge

1. A sample of Argon gas and Neon gas are in a container at thermal equilibrium. The mass of an Argon atom is twice that of a Neon atom. What is the ratio of their average kinetic energies (K_{Ar} : K_{Ne})?

Check Answer The ratio is 1:1. “Thermal equilibrium” means they are at the same temperature. Since K_{avg} = \frac{3}{2}k_B T, the mass does not matter—the average kinetic energies are identical.

2. In a PV diagram, a gas moves from State A (P, V) to State B (P, 3V) at constant pressure. Is work being done on the gas or by the gas, and is the work positive or negative?

Check Answer The gas is expanding (\Delta V > 0), so work is done by the gas. In the AP Physics convention (\Delta U = Q + W), the work W done on the gas is negative.

3. During an adiabatic compression, the temperature of a gas increases. Where did the energy come from if no heat (Q=0) was added?

Check Answer From the First Law: \Delta U = Q + W. Since Q = 0, then \Delta U = W. The work done on the gas during compression was converted directly into internal energy, raising the temperature.

4. An ideal heat engine exhausts 400 \, \text{J} of heat for every 600 \, \text{J} of heat it absorbs. What is the efficiency of this engine?

Check Answer Efficiency e = \frac{W}{Q_H}. First, find Work: W = Q_H - Q_C = 600 - 400 = 200 \, \text{J}.
Then, e = \frac{200}{600} = \mathbf{33.3\%}.