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.)