The first law of thermodynamics

Internal Energy of a System

Key Equations

Internal energy of an ideal monatomic gas:

$$U=\frac{3}{2}nRT=\frac{3}{2}N{k}_{B}T$$

Variables: $U$ = internal energy, $n$ = number of moles, $R$ = universal gas constant ($8.314 \text{J/(mol}\cdot\text{K)}$), $T$ = absolute temperature in K, $N$ = number of atoms, ${k}_{B}$ = Boltzmann constant ($1.38\times {10}^{-23} \text{J/K}$)

SI unit: joule (J)

ProportionalityInternal energy is directly proportional to absolute temperature. Doubling the temperature doubles the internal energy.

Reference sheet status: On the reference sheet.

Every object is made of particles in constant motion. The internal energy of a system is the sum of the kinetic energy of the objects that make up the system and the potential energy of the configuration of those objects. This is the total microscopic energy — it has nothing to do with the motion of the system’s center of mass.

For an ideal monatomic gas, two simplifications apply. First, the atoms do not interact through conservative forces, so the gas has no internal potential energy. Second, internal structure of individual atoms is ignored.

The internal energy of an ideal monatomic gas is therefore just the sum of the translational kinetic energies of all its atoms. Because the average kinetic energy per atom is $\frac{3}{2}{k}_{B}T$, multiplying by the number of atoms gives the total internal energy.

Changes to a system’s internal energy can alter the internal structure and behavior of that system. For example, heating a gas increases the average speed of its molecules. Crucially, these changes occur without changing the motion of the system’s center of mass. A sealed container of gas sitting on a table can gain internal energy through heating while its center of mass stays perfectly still.

MisconceptionStudents often confuse internal energy with temperature. Temperature measures the average kinetic energy per particle. Internal energy is the total energy of all particles, so it depends on both temperature and the number of particles.

Exam TipA large cold system can have more internal energy than a small hot system.

The First Law of Thermodynamics

Key Equations

First law of thermodynamics:

$$\Delta U=Q+W$$

Variables: $\Delta U$ = change in internal energy of the system, $Q$ = energy transferred into the system by heating ($Q>0$ when energy enters; $Q<0$ when energy leaves), $W$ = work done on the system

SI unit: joule (J)

Reference sheet status: On the reference sheet.

Work done on a system at constant or average external pressure:

$$W=-P\Delta V$$

Variables: $W$ = work done on the system, $P$ = external pressure (Pa), $\Delta V$ = change in volume (${\text{m}}^{3}$)

SI unit: joule (J)

Rearrangements:

$$P=-\frac{W}{\Delta V}$$

$$\Delta V=-\frac{W}{P}$$

ProportionalityThe magnitude of work is directly proportional to both pressure and the change in volume.

Reference sheet status: On the reference sheet.

Sign convention for $W=-P\Delta V$:

The negative sign carries essential physical meaning. When the gas expands, $\Delta V$ is positive, so $W=-P\Delta V$ is negative. Negative $W$ means the system does work on its surroundings and loses energy. When the gas is compressed, $\Delta V$ is negative, so $W$ is positive. Positive $W$ means the surroundings do work on the system and add energy to it. If you drop the negative sign, expansion and compression give the wrong energy flow direction.

The first law of thermodynamics is a restatement of conservation of energy that accounts for energy transferred into or out of a system by work, heating, or cooling. The equation $\Delta U=Q+W$ states that the change in internal energy equals the net energy added to the system. Energy enters through positive $Q$ (heating) or positive $W$ (compression). Energy leaves through negative $Q$ (cooling) or negative $W$ (expansion).

Two important cases follow directly. For an isolated system — one where no energy crosses the boundary by any means — the total energy is constant, so $\Delta U=0$. For a closed system — one where matter cannot enter or leave but energy can — the change in internal energy is the sum of energy transferred by heating and work done on the system.

Examiner InsightAP FRQs frequently test sign interpretation. A question might give you $Q$ and $\Delta V$ and ask whether the gas heats up. You must track signs through $W=-P\Delta V$ and then $\Delta U=Q+W$.

Exam TipAlways state the sign convention before calculating.

Worked Example: Applying the First Law

Scenario

An ideal gas in a cylinder receives 500 J of energy by heating while expanding against a constant external pressure of $1.0\times {10}^{5} \text{Pa}$. The volume increases by $2.0\times {10}^{-3} {\text{m}}^{3}$. Determine the change in internal energy of the gas.

Finding the work done on the gas:

Equation used

$$W=-P\Delta V$$

Given

$$P=1.0\times {10}^{5} \text{Pa}$$

$$\Delta V=2.0\times {10}^{-3} {\text{m}}^{3}$$

Working — substitution

$$W=-(1.0\times {10}^{5})(2.0\times {10}^{-3})$$

$$W=-200 \text{J}$$

The negative sign confirms the gas loses energy by doing work on its surroundings during expansion.

Finding the change in internal energy:

Equation used

$$\Delta U=Q+W$$

Given

$$Q=+500 \text{J}$$

$$W=-200 \text{J}$$

Working — substitution

$$\Delta U=500+(-200)$$

$$\Delta U=+300 \text{J}$$

Answer

$$\Delta U=300 \text{J}$$

Interpretation

The gas gained 500 J by heating but spent 200 J doing work as it expanded. The net gain in internal energy is 300 J, so the temperature of the gas increases.

PV Diagrams and Thermodynamic Work

Understanding how pressure and volume change together is central to thermodynamics. PV diagrams are graphical representations used to represent thermodynamic processes, with pressure on the vertical axis and volume on the horizontal axis.

Each point on a PV diagram represents a specific state of the gas defined by its pressure and volume. A process appears as a path connecting an initial state to a final state.

Lines of constant temperature on a PV diagram are called isotherms. Because $PV=nRT$, an isotherm is a hyperbola — as volume increases, pressure decreases so that the product $PV$ stays constant. Higher temperatures correspond to isotherms further from the origin.

The absolute value of the work done on a gas equals the area underneath the curve on a PV diagram. For expansion (the gas moves rightward on the diagram), the gas does work on its surroundings, so $W$ on the system is negative. For compression (leftward), the surroundings do work on the gas, so $W$ is positive. Because the area depends on the shape of the path, the work done depends on how the process occurs — not just on the initial and final states.

MisconceptionStudents often treat work as a state function, calculating it from endpoints alone. Work depends on the path taken between states because the area under different curves differs even when the endpoints are the same.

Exam TipOn PV diagrams, always identify the path before calculating work.

Reading a PV diagram:

Special Thermodynamic Processes

Four special-case thermodynamic processes appear repeatedly on the AP exam. Each places a specific constraint on the system, which simplifies the first law.

In an isovolumetric process, the volume does not change. No work is done because $W=-P\Delta V=0$. All energy transferred by heating goes directly into changing the internal energy. On a PV diagram, this appears as a vertical line.

In an isobaric process, the pressure stays constant. Work is calculated directly from $W=-P\Delta V$. On a PV diagram, this is a horizontal line. Both $Q$ and $W$ can be nonzero.

In an isothermal process, the temperature remains constant. For an ideal gas, internal energy depends only on temperature, so $\Delta U=0$. The first law then gives $Q=-W$: any energy added by heating is entirely spent on work done by the gas (or vice versa). On a PV diagram, this process follows an isotherm — a hyperbola.

In an adiabatic process, no energy is transferred by heating or cooling ($Q=0$). The first law simplifies to $\Delta U=W$. If the gas is compressed adiabatically, the work done on it increases its internal energy and temperature. If it expands adiabatically, it does work at the expense of internal energy, and the temperature drops. On a PV diagram, an adiabatic curve is steeper than an isotherm through the same point.

Examiner InsightFRQs frequently present a PV diagram and ask students to rank processes by work done, temperature change, or heat transferred. Use the table above as a quick reference for which quantities are zero.

Exam TipStart every thermodynamic analysis by identifying the constraint ($\Delta V=0$, $P$ = const, $T$ = const, or $Q=0$).

Worked Example: Identifying a Process and Applying the First Law

Scenario

An ideal monatomic gas at a constant pressure of $1.5\times {10}^{5} \text{Pa}$ is heated, and its volume increases from $4.0\times {10}^{-3} {\text{m}}^{3}$ to $6.0\times {10}^{-3} {\text{m}}^{3}$. The gas receives 750 J of energy by heating. Determine the change in internal energy.

This is an isobaric process ($P$ = constant).

Finding the work done on the gas:

Equation used

$$W=-P\Delta V$$

Given

$$P=1.5\times {10}^{5} \text{Pa}$$

$$\Delta V=6.0\times {10}^{-3}-4.0\times {10}^{-3}=2.0\times {10}^{-3} {\text{m}}^{3}$$

Working

$$W=-(1.5\times {10}^{5})(2.0\times {10}^{-3})$$

$$W=-300 \text{J}$$

The gas expands, so $W$ is negative — the gas does 300 J of work on its surroundings.

Finding the change in internal energy:

Equation used

$$\Delta U=Q+W$$

Given

$$Q=+750 \text{J}$$

$$W=-300 \text{J}$$

Working

$$\Delta U=750+(-300)$$

$$\Delta U=+450 \text{J}$$

Answer

$$\Delta U=450 \text{J}$$

Interpretation

Of the 750 J added by heating, 300 J was spent doing work during the expansion. The remaining 450 J increased the internal energy, raising the gas temperature.