First Law of Thermodynamics
A short lesson with diagrams, a worked example, a problem to try, and a quick check. Then quiz yourself on 7 practice questions.
Energy is conserved: ΔU = Q + W. Q is heat added to the gas, and W is the work done on the gas. When a gas expands, it does work on its surroundings, so W is negative: W = −PΔV, minus the area under the P–V graph.
1. Work is area on a P–V graph
2. The four special processes
Cycles: after a full loop the gas is back where it started, so ΔU = 0. The net work is the area inside the loop. Clockwise on a P–V graph, the gas does net work (an engine), so it must absorb net heat.
An adiabatic expansion cools a gas (it does work with no heat coming in); an adiabatic compression heats it.
3. Example, then try it
A gas at 3.0 × 10⁵ Pa expands from 2.0 L to 4.0 L at constant pressure while absorbing 1500 J of heat. Find W and ΔU.
W = −PΔV = −(3.0 × 10⁵)(2.0 × 10⁻³) = −600 J.
ΔU = Q + W = 1500 − 600 = +900 J. The gas warms up.
A gas goes clockwise around the rectangle A → B → C → D → A on the P–V graph below. Find the net work done on the gas and the net heat it absorbs per cycle.
A → B (expands at high P): W = −(4.0 × 10⁵)(2.0 × 10⁻³) = −800 J.
B → C and D → A (constant volume): W = 0.
C → D (compressed at low P): W = +(1.0 × 10⁵)(2.0 × 10⁻³) = +200 J.
Net: W = −800 + 200 = −600 J (the enclosed area). Full cycle: ΔU = 0, so Q = −W = +600 J.
Answer: W = −600 J per cycle; the gas absorbs a net 600 J of heat.
4. Common mistakes
An ideal gas is quickly compressed in an insulated cylinder, so no heat flows. What happens to its temperature?
With Q = 0, ΔU = W. Compressing the gas does positive work on it, so its internal energy, and its temperature, increase.
Ready to practice?
7 questions on First Law of Thermodynamics, each with a hint and an explanation for every wrong answer.