Topic 9.4 First Law of Thermodynamics and PV Diagrams: apply the first law to track internal energy, heat and work, and read work as the area on a PV diagram.

What this topic is asking

The College Board (Topic 9.4) wants you to apply the first law of thermodynamics, the conservation of energy for a gas, tracking how heat and work change the internal energy. You must also read the work as an area on a PV diagram and recognize the four named processes.

The first law of thermodynamics

The first law is just conservation of energy applied to a gas: energy can enter as heat or leave as work, and whatever is left over changes the internal energy. For an ideal gas the internal energy is the total kinetic energy of the atoms, so $U \propto T$: raising the internal energy raises the temperature, and vice versa. Watch the sign convention. AP uses $\Delta U = Q - W_{by}$ (work by the gas), but some texts write $\Delta U = Q + W_{on}$. They are the same law; just be consistent.

Work is the area on a PV diagram

The PV diagram turns the work into geometry: trace the gas's path from its start to end state and the area beneath it is the work. This is why constant-volume (vertical) lines do no work (no area), and why the work depends on the path, not just the endpoints. A cyclic process, returning to its start, encloses an area equal to the net work done per cycle, which is how engines are analyzed.

The four named processes

Each standard process pins down one variable, which simplifies the first law:

• Isothermal ($\Delta T = 0$): internal energy is unchanged, $\Delta U = 0$, so $Q = W_{by}$. All heat added becomes work done.
• Isobaric (constant $P$): work is simply $W_{by} = P\,\Delta V$, the rectangular area.
• Isovolumetric (constant $V$): no work, $W_{by} = 0$, so $\Delta U = Q$. All heat goes into internal energy.
• Adiabatic ($Q = 0$): no heat exchanged, so $\Delta U = -W_{by}$. Compressing warms the gas; expanding cools it.

The strategic move in any first-law problem is to identify which quantity is fixed, use it to evaluate one term, then apply $\Delta U = Q - W_{by}$. Because internal energy is proportional to temperature, the change in temperature tells you $\Delta U$ directly. This energy accounting, built on conservation of energy from mechanics, is what lets you predict whether a gas heats or cools in any process, and it leads into the directionality of energy flow captured by entropy and the second law (Topic 9.6).

Try this

Q1. A gas does $300$ J of work while $300$ J of heat is added. State the change in its internal energy. [1 point]

• Cue. $\Delta U = Q - W_{by} = 300 - 300 = 0$ J (an isothermal process).

Q2. State the work done by a gas in an isovolumetric (constant-volume) process. [1 point]

• Cue. Zero, because $\Delta V = 0$ so there is no area under the path.