Syllabus Edition
First teaching 2023
First exams 2025
The First Law of Thermodynamics (Cambridge (CIE) A Level Physics)
Revision Note
The first law of thermodynamics
The first law of thermodynamics is based on the principle of conservation of energy
When energy is put into a gas by heating it or doing work on it, its internal energy must increase:
The increase in internal energy = Energy transferred by heating + Work done on the system
The first law of thermodynamics is therefore defined as:
Where:
ΔU = increase in internal energy (J)
q = energy transferred to the system by heating (J)
W = work done on the system (J)
The first law of thermodynamics applies to all situations, not just for gases
There is an important sign convention used for this equation
A positive value for internal energy (+ΔU) means:
The internal energy ΔU of the system increases
Heat q is added to the system
Work W is done on the system
A negative value for internal energy (−ΔU) means:
The internal energy ΔU of the system decreases
Heat q is taken away from the system
Work W is done by the system
Worked Example
The volume occupied by 1.00 mol of a liquid at 50 oC is 2.4 × 10-5 m3. When the liquid is vaporised at an atmospheric pressure of 1.03 × 105 Pa, the vapour has a volume of 5.9 × 10-2 m3. The latent heat to vaporise 1.00 mol of this liquid at 50 oC at atmospheric pressure is 3.48 × 104 J.
Determine for this change of state the increase in internal energy ΔU of the system.
Answer:
Step 1: Write down the first law of thermodynamics
Step 2: Write the value of heating q of the system
This is the latent heat, the heat required to vaporise the liquid = 3.48 × 104 J
Step 3: Calculate the work done W
ΔV = final volume − initial volume
ΔV = 5.9 × 10-2 − 2.4 × 10-5 = 0.058976 m3
p = atmospheric pressure = 1.03 × 105 Pa
W = work done
W = (1.03 × 105) × 0.058976 = 6074.528 = 6.07 × 103 J
Since the gas is expanding, this work done is negative
W = −6.07 × 103 J
Step 4: Substitute the values into the first law of thermodynamics
ΔU = 3.48 × 104 + (−6.07 × 103) = 28 730 = 29 000 J (2 s.f.)
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