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The First Law of Thermodynamics (AQA A Level Physics): Revision Note

Exam code: 7408

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Ashika

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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 space equals space straight capital delta U space plus thin space 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 space equals space p increment 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:

increment U space equals space Q thin space minus thin space W

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

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

Examiner Tips and Tricks

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

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Ashika

Author: Ashika

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Ashika graduated with a first-class Physics degree from Manchester University and, having worked as a software engineer, focused on Physics education, creating engaging content to help students across all levels. Now an experienced GCSE and A Level Physics and Maths tutor, Ashika helps to grow and improve our Physics resources.