Define internal energy as the sum of molecular kinetic and potential energies, and apply the first law of thermodynamics to changes in a gas

What this dot point is asking

SEAB wants you to define the internal energy of a system as the sum of the random kinetic and potential energies of its molecules, and to apply the first law of thermodynamics to changes in a gas, using a consistent sign convention. The first law is conservation of energy applied to thermal systems.

The answer

Internal energy

The internal energy $U$ of a system is the sum of the random kinetic and potential energies of all its molecules. The kinetic part is due to molecular motion (translational, and for molecules also rotational and vibrational); the potential part is due to intermolecular forces.

For an ideal gas, there are no intermolecular forces, so the potential part is zero, and the internal energy is purely molecular kinetic energy. Since that kinetic energy depends only on temperature, the internal energy of an ideal gas depends only on its temperature:

This is why an isothermal change of an ideal gas has $\Delta U = 0$.

The first law of thermodynamics

The first law states that the increase in internal energy of a system equals the thermal energy supplied to it plus the work done on it:

with the sign convention used here:

• $Q > 0$: thermal energy is supplied to the system.
• $W > 0$: work is done on the system (for example, compression).
• $W < 0$: work is done by the system (for example, expansion).

Some texts write $\Delta U = Q - W_{\text{by}}$, where $W_{\text{by}}$ is the work done by the gas. State your convention to avoid ambiguity.

Work done by an expanding gas

When a gas at pressure $p$ expands by a small volume $\Delta V$ at constant pressure, the work done by the gas is:

On a pressure-volume graph, the work done is the area under the curve. An expanding gas does work on its surroundings; a compressed gas has work done on it.

Special processes

• Isothermal (constant temperature): $\Delta U = 0$ for an ideal gas, so $Q = -W$ (any heat in is matched by work out).
• Isobaric (constant pressure): $W_{\text{by}} = p\Delta V$; heat supplied both raises internal energy and does work.
• Adiabatic (no heat exchange, $Q = 0$): $\Delta U = W$, so compressing a gas adiabatically raises its temperature and expanding it lowers the temperature.
• Isochoric (constant volume): no work is done ($\Delta V = 0$), so $\Delta U = Q$.

Examples in context

Example 1. A diesel engine. In a diesel engine, air is compressed adiabatically and rapidly so little heat escapes. The work done on the air raises its internal energy and temperature high enough to ignite the injected fuel without a spark plug. This is the adiabatic case $\Delta U = W$ in action.

Example 2. A gas expanding against a piston. When a heated gas pushes a piston outward at constant pressure, the heat supplied splits into two parts: some raises the internal energy (temperature) and some does work $p\Delta V$ on the piston. This is why the heat needed to raise a gas's temperature at constant pressure exceeds that at constant volume.

Try this

Q1. Define the internal energy of a system and state what it depends on for an ideal gas. [2 marks]

• Cue. Sum of random molecular kinetic and potential energies; for an ideal gas it is purely kinetic and depends only on temperature.

Q2. A gas releases $80\ \text{J}$ of heat to the surroundings while $120\ \text{J}$ of work is done on it. Find the change in internal energy. [2 marks]

• Cue. $\Delta U = Q + W = (-80) + (+120) = +40\ \text{J}$.

Q3. Explain why the temperature of a gas rises when it is compressed adiabatically. [3 marks]

• Cue. $Q = 0$, so $\Delta U = W > 0$; all the work done on the gas increases its internal energy, raising molecular kinetic energy and hence temperature.