Q2(b). Expansion of an ideal gas against a constant pressure of 1 atm

Reading: the gas stays at 273.15 K (isothermal). The initial pressure is RT/V₁=10.0 atm and the final pressure RT/V₂=1.00 atm equals the opposing pressure, so the expansion stops at 22.4 L.

Given: n=1 mol, T=273.15 K, V₁=2.24 L, V₂=22.4 L, Pₑₓₜ=1 atm; 1 L atm = 101.325 J.

1. Work (against constant opposing pressure): w=-Pₑₓₜ(V₂-V₁)=-(1 atm)(22.4-2.24) L=-20.16 L atm w=-20.16×101.325 J, w=-2043 J (-2.04 kJ) The minus sign means the gas does work on the surroundings.
2. Δ U: for an ideal gas U depends only on T (no intermolecular forces); T is unchanged, so Δ U=0
3. Δ H: Δ H=Δ U+Δ(PV)=0+nRΔ T=0, Δ H=0
4. Heat: from the first law, q=Δ U-w=+2043 J taken in from the surroundings.
5. Δ S of the gas: S is a state function, so it is found along a reversible isothermal path between the same states: Δ S=nRln(V₂)/(V₁)=8.31410 Δ S_(sys)=+19.14J K⁻¹
6. Surroundings: they give up 2043 J reversibly at 273.15 K: Δ Sₛᵤᵣᵣ=-(2043)/(273.15)=-7.48 J K⁻¹. Total Δ S=+11.67 J K⁻¹>0: the expansion is spontaneous and irreversible.
7. Comparison: a reversible isothermal expansion over the same range does |wᵣₑᵥ|=RT10=5229 J. The irreversible path does less work; the shaded rectangle in the figure lies under the isotherm. Δ U, Δ H and Δ S_(sys) are the same for both paths; q and w are not.

Application. the gap between |wᵣₑᵥ|=5229 J and |w|=2043 J is the work lost by expanding against a fixed back-pressure. Compressors and turbines are run in several stages to approach the reversible limit.