B.4 · IB Physics
First law and p–V processes
An ideal monatomic gas in a cylinder goes from state 1 to state 2 by one of four processes. Follow the path on the p–V diagram, where the shaded area is the work done by the gas, and check the energy balance Q = ΔU + W.
Predict first
Start from the default settings (isobaric, n = 1.00 mol, V₁ = 20.0 and V₂ = 40.0 × 10⁻³ m³). How does the thermal energy Q absorbed compare with the work W done by the gas?
At constant pressure W = pΔV = nRΔT and ΔU = (3/2)nRΔT, so Q = ΔU + W = (5/2)nRΔT = 2.5 W. Here W ≈ 2490 J and Q ≈ 6230 J.
Check it in the model: Leave Process on Isobaric and compare the Thermal energy in Q and Work done by gas W readouts (about 6230 J and 2490 J).
Q (energy in) ΔU W (work by gas)
(Q in to gas, W done by gas)
ΔU = (3/2)nRΔT = (3/2)(pV₂ − pV₁)
isobaric W = pΔV · isothermal W = nRT ln(V₂/V₁)
adiabatic pV5/3 = constant, W = −ΔU
R = 8.31 J mol⁻¹ K⁻¹
Try this
- With the isothermal preset, compare the Q and W bars and the ΔU value. Then switch to adiabatic with the same volumes. Which quantity has become zero, and what has happened to the final temperature?
- Run the three expansions from 20 to 40 × 10⁻³ m³ (isobaric, isothermal, adiabatic) and compare W each time. Which has the largest shaded area, and why?
- Choose Isochoric and raise T₂. Why is W zero here, and how is Q related to ΔU?
Common error: thinking an isothermal change has no thermal energy transfer. At constant temperature ΔU = 0, so Q = W: a gas expanding isothermally must absorb energy equal to the work it does. It is the adiabatic process that has Q = 0, and there the temperature changes.