The First Law of Thermodynamics

The first law of thermodynamics is based on the principle of conservation of energy
- This applies to heating, cooling and work done
The first law can be expressed in different ways, depending on the sign convention used
For AQA, the first law of thermodynamics is therefore defined as:

$Q = Δ U + W$

Where:
- ΔU = increase in internal energy (J)
- Q = energy supplied to the system by heating (J)
- W = work done by the system (J)
The 'system' is a region of space containing a quantity of gas
The system is open if gas or vapour flows into and out of the region
- Examples include a gas expanding through the nozzle of an aerosol can or steam passing through a turbine
The system is closed if gas or vapour remains within the region, although the boundary can expand or contract with changes in the volume of the gas
- Examples include gas expanding in a cylinder by moving a piston or air in a balloon being heated

Open and closed systems

11-2-1-open-and-closed-system

An open system is where gas can flow in and out, whilst in a closed system, gas can only remain within the boundary

In both systems, work can 'cross' the boundary

The first law of thermodynamics applies to all situations, not just to gases
- There is an important sign convention used for this equation

Amount of heat transfer Q is:
- Positive if heat energy is added
- Negative if heat energy is removed
The change in internal energy ΔU is:
- Positive if the internal energy increases
- Negative if the internal energy decreases
The work done, W is:
- Positive if work is done by the gas (the gas expands)
- Negative if the work is done on the gas (the gas is compressed)

Worked example

The volume occupied by 1.00 mol of a liquid at 50°C is 2.4 × 10⁻⁵ m³. When the liquid is vaporised at an atmospheric pressure of 1.03 × 10⁵ Pa, the vapour occupies a volume of 5.9 × 10⁻² m³.

The latent heat to vaporise 1.00 mol of this liquid at 50°C at atmospheric pressure is 3.48 × 10⁴ J.

For this change of state, determine the increase in internal energy ΔU of the system.

Answer:

Step 1: List the known quantities

Thermal energy, Q = 3.48 × 10⁴ J
Atmospheric pressure, p = 1.03 × 10⁵ Pa
Initial volume = 2.4 × 10⁻⁵ m³
Final volume = 5.9 × 10⁻² m³

Step 2: Calculate the work done W

The work done by a gas at constant pressure is

$W = p ∆ V$

Where the change in volume is:

ΔV = final volume − initial volume = (5.9 × 10⁻²) − (2.4 × 10⁻⁵) = 0.059 m³

Since the volume of the gas increases (it expands), the work done is positive

W = (1.03 × 10⁵) × 0.059 = 6077 = 6.08 × 10³ J

W = 6.08 × 10³ J

Step 3: Substitute the values into the equation for the first law of thermodynamics

From the first law of thermodynamics:

$∆ U = Q - W$

ΔU = (3.48 × 10⁴) – (6.08 × 10³) = 28 720

Increase in internal energy: ΔU = 28 700 J (3 s.f.)

Examiner Tip

The sign convention is very important for AQA, make sure you understand how it is used from the worked example.

The First Law of Thermodynamics (AQA A Level Physics)

Revision Note