16.2 The first law of thermodynamics
- Syllabus
- 9702–2028–2029
- Topic
- 16.2
- Level
- A2
Learning objectives
16.2.1W = p∆V for the work done when the volume of a gas changes at constant• recall and use W = p∆V for the work done when the volume of a gas changes at constant pressure and understand the difference between the work done by the gas and the work done on the gas16.2.2The first law of thermodynamics ∆U = q + W expressed• recall and use the first law of thermodynamics ∆U = q + W expressed in terms of the increase in internal energy, the heating of the system (energy transferred to the system by heating) and the work done on the system
Constant-pressure volume change transfers work p times change in volume
| Quantity at constant pressure p | Formula | Expansion V_f>V_i | Compression V_f<V_i |
|---|---|---|---|
| work done by gas | W_by=p(V_f-V_i) | positive | negative |
| work done on gas | W_on=-p(V_f-V_i)=p(V_i-V_f) | negative | positive |
forceonpistonF=pAanddisplacementxgivesWby=Fx=pAx=pΔV
At p=1.01×10⁵ Pa, expansion by 5.20×10⁻⁵ m³ gives W_by=(1.01×10⁵)(5.20×10⁻⁵)=+5.25 J and W_on=-5.25 J.
Compression from 0.32 m³ to 0.18 m³ at 1.6×10⁵ Pa gives W_on=-p(V_f-V_i)=-(1.6×10⁵)(-0.14)=+2.2×10⁴ J.
Use p in Pa and volume in m³; Pa m³ = J. At constant volume, ΔV=0 so no pressure-volume work is done.
Always label W_by or W_on before assigning a sign. The simple product pΔV requires constant pressure; for changing pressure, work magnitude is the area under the p–V path.
The first law tracks heating and work done on a system
ΔU=q+W
| Symbol | Positive when | Negative when |
|---|---|---|
| q | energy is transferred to the system by heating | energy is transferred from the system by heating |
| W | work is done on the system | work is done by the system |
| ΔU | internal energy increases | internal energy decreases |
If 300 J enters by heating and the gas does 100 J of work, q=+300 J and W=-100 J. Therefore ΔU=300-100=+200 J.
| Process | Useful consequence |
|---|---|
| rapid insulated compression | q≈0, W>0, so ΔU>0 and temperature rises |
| constant-volume heating | W=0, so ΔU=q |
| complete cycle | final state equals initial state, so ΔU_cycle=0 and q_cycle=-W_cycle |
q=ΔU−WandW=ΔU−q
In this syllabus equation W means work done on the system. Do not insert positive work done by the gas. Heating and work are transfer routes, not energy stored in the system.