Thermodynamic Systems (HL) | HL IB Physics Revision Notes 2025

Change in Internal Energy

The change in the internal energy $∆ U$ of an object is intrinsically related to a change in its temperature $∆ T$

$∆ U \propto ∆ T$

When a container containing gas molecules is heated up, the molecules begin to move around faster, increasing their kinetic energy
- In a solid, where the molecules are tightly packed, molecules begin to vibrate more as they are heated
- Molecules in liquids and solids have both kinetic and potential energy because they are close together and bound by intermolecular forces

However, the molecules in an ideal gas are assumed to have no intermolecular forces
- This means they do not possess potential energy, only kinetic energy

Change in internal energy, downloadable AS & A Level Physics revision notes

As the container is heated up, the gas molecules move faster with higher kinetic energy and therefore higher internal energy

The (change in) internal energy of an ideal gas is equal to:

$∆ U = \frac{3}{2} N k_{B} ∆ T$

Where
- $∆ U$ = change in internal energy (J)
- $k_{B}$ = Boltzmann constant
- $∆ T$ = change in temperature (K)
- $N$ = number of particles

Another form of this equation related to the translational kinetic energy of the particles is

$∆ U = \frac{3}{2} n R ∆ T$

Where:
- $n$ = number of moles of gas (mol)
- $R$ = molar gas constant

Worked example

A student suggests that, when an ideal gas is heated from 50°C to 150°C, the internal energy of the gas is tripled.

State and explain whether the student’s suggestion is correct.

Answer:

The change in internal energy of an ideal gas is directly proportional to its change in temperature

$∆ U \propto ∆ T$

The temperature change is the thermodynamic temperature i.e. Kelvin
The temperature change in degrees (from 50°C to 150°C) increases by three times
The temperature change in Kelvin is:

50°C + 273.15 = 323.15 K

150°C + 273.15 = 423.15 K

$\frac{423.15}{323.15} = 1.3$

The temperature change, in Kelvin, does not increase by three times, therefore, neither does the internal energy
Hence, the student is incorrect

Worked example

An ideal gas expands at constant pressure. The following data are available:

amount of gas = 126 mol

initial temperature of gas = −23.0°C

final temperature of gas = +27.0°C

Determine the change in internal energy of the gas during this expansion.

Answer:

The change in internal energy of a gas is equal to

$∆ U = \frac{3}{2} n R ∆ T$

Where
- Amount of gas, $n$ = 126 mol
- Gas constant, $R$ = 8.31 J mol⁻¹ K⁻¹
- Change in temperature, $∆ T$ = 27 − (−23) = 50°C

$∆ U = \frac{3}{2} \times 126 \times 8.31 \times 50$

$∆ U = 7.85 \times 10^{4} J$

Exam Tip

If an exam question about an ideal gas asks for the total internal energy, remember that this is equal to the total kinetic energy since an ideal gas has zero potential energy

Work Done by a Gas

When a gas expands, it does work on its surroundings by exerting pressure on the walls of the container it's in
For a gas inside a piston, the force exerted by the gas pushes the piston outwards
- As a result, work is done on the piston by the gas

Alternatively, if an external force is applied to the piston, the gas will be compressed
- In this case, work is done on the gas by the piston

Work done by gas, downloadable AS & A Level Physics revision notes

The expansion of the gas does work on the piston by exerting a force over a distance, s

Assuming the volume of gas is kept at constant pressure, this means the force F exerted by the gas on the piston is equal to:

$p = \frac{F}{A} \Rightarrow F = p A$

Where A = cross-sectional area of the cylinder (m²)

The definition of work done is:

$W = F s$

Where s = displacement in the direction of force (m)

The displacement s of the gas multiplied by the cross-sectional area A is equal to the increase in volume ΔV of the gas:

$W = p A s$

This gives the equation for the work done when the volume of a gas changes at constant pressure:

$W = p ∆ V$

Where:
- W = work done (J)
- p = pressure of the gas (Pa)
- ΔV = change in the volume of the gas (m³)

This equation assumes that the surrounding pressure does not change as the gas expands
- This is true if the gas is expanding against the pressure of the atmosphere, which changes very slowly

p-V diagrams

Pressure-volume (p-V) diagrams are often used to represent changes in the state of a gas in thermodynamic processes

2-4-1-positive-and-negative-work-done-convention

Positive or negative work done depends on whether the gas is compressed or expanded

When a gas expands (at constant pressure) work done is positive
- Volume increases +ΔV
- Work is done by the gas +W

When a gas is compressed (at constant pressure) work done W is negative
- Volume decreases −ΔV
- Work is done on the gas −W

When both the volume and pressure of gas changes

The work done can be determined from the area under a p-V diagram

2-4-1-area-under-a-pv-diagram-changing-pressure-and-volume

Worked example

When a balloon is inflated, its rubber walls push against the air around it.

Calculate the work done when the balloon is blown up from 0.015 m³ to 0.030 m³.

Atmospheric pressure = 1.0 × 10⁵ Pa.

Answer:

The work done by a gas is equal to

$W = p ∆ V$

Where the change in volume is

ΔV = final volume − initial volume = 0.030 − 0.015 = 0.015 m³

Therefore, work done is

W = (1.0 × 10⁵) × 0.015 = 1500 J

Worked example

An ideal gas is compressed, as shown on the graph below.

2-4-1-area-under-a-pv-diagram-worked-example2-4-1-area-under-a-pv-diagram-worked-example

(a)

For this change, state and explain whether work is done on the gas or by the gas

(b)

Determine the value of the work done and state whether it is positive or negative

Answer:

(a)

The volume decreases, therefore, work is done on the gas

(b)

The work done is equal to the area under the p-V diagram

2-4-1-area-under-a-pv-diagram-worked-example-ma

$W = \frac{1}{2} (600 \times 10^{3}) (400 \times 10^{- 6}) + (300 \times 10^{3}) (400 \times 10^{- 6})$

W = −240 J

When volume decreases, the work done is negative

Exam Tip

The pressure p in the work done by a gas equation is not the pressure of the gas but the pressure of the surroundings. This is because when a gas expands, it does work on the surroundings.

Thermodynamic Systems (HL) (HL IB Physics)

Revision Note

Change in Internal Energy

Worked example

Worked example

Exam Tip

Work Done by a Gas

p-V diagrams

Worked example

Worked example

Exam Tip

Author: Katie M