Lesson 4 of 13 · 7 min
First law of thermodynamics
NCERT §11.5
At lunch, the workshop kettle boils. Riya wonders where the energy goes: the water makes steam that takes up far more room, pushing the air aside, but it also stays at 100 °C.
The lesson in notes
In short
Let ΔQ be the heat supplied to the system by its surroundings, ΔW the work done by the system on its surroundings, and ΔU the change in its internal energy.
First law: ΔQ = ΔU + ΔW. The heat put in partly raises the internal energy and partly goes out as work. It is conservation of energy for a system that exchanges energy with its surroundings.
Written as ΔQ − ΔW = ΔU: a gas can go from (P₁, V₁) to (P₂, V₂) by many routes, e.g. first at constant pressure to (P₁, V₂) and then at constant volume, or the other way round. ΔQ and ΔW generally depend on the route, but ΔQ − ΔW = ΔU does not, because U is a state variable.
If a process has ΔU = 0 (an ideal gas expanding isothermally, say), then ΔQ = ΔW: all the heat supplied is used as work on the surroundings.
Work by a gas pushing a piston at constant pressure P: force = P × area and area × displacement = volume change, so ΔW = PΔV. Then ΔQ = ΔU + PΔV.
Worked example, boiling 1 g of water: latent heat 2256 J/g, so ΔQ = 2256 J. Under atmospheric pressure this gram fills 1 cm³ as liquid and 1671 cm³ as vapour.
Work against the atmosphere: ΔW = P(V_g − V_l) = 1.013 × 10⁵ × (1671 × 10⁻⁶) = 169.2 J. So ΔU = 2256 − 169.2 = 2086.8 J.
Most of the heat supplied in boiling raises the water's internal energy; only a small part pushes back the atmosphere.
Sign convention: Q > 0 when heat is added to the system, Q < 0 when heat is taken out; W > 0 when the system does work, W < 0 when work is done on it.