UPSC Combined Geo-Scientist, Mains · 2018 · Mains Paper II
Q1(f). Gibbs and Helmholtz changes
Question
Show that ΔG=ΔA for isothermal expansion of a fixed amount of ideal gas.
Answer in brief
G-A=pV=nRT, which is unchanged when n and T are constant.
Key points
- The definitions are A=U-TS and G=H-TS=U+pV-TS, so G=A+pV.
- Therefore ΔG-ΔA=Δ(pV). For a fixed amount of ideal gas undergoing an isothermal change, this is nRΔT=0.
- Equivalently, ΔU=ΔH=0 and both free-energy changes equal -TΔS=-nRTln (V₂/V₁). This state-function equality does not require the actual expansion path to be reversible.
The full model answer
As written for the answer booklet: the diagram carries the working, the lines stay short. Read here only; it is not offered as a download.
CoversGibbsHelmholtz changesThermodynamics
All parts of Question 1
- 1(a) Equal molecular speeds5 marks
- 1(b) Fugacity under repulsion5 marks
- 1(c) Wetting of glass5 marks
- 1(d) FCC particle count5 marks
- 1(e) Joule–Thomson cooling5 marks
- 1(f) Gibbs and Helmholtz changes5 marks
- 1(g) Equilibrium-constant conventions5 marks
- 1(h) Very dilute hydrochloric acid5 marks
- 1(i) Pressure controlling boiling5 marks
- 1(j) Primary kinetic salt effect5 marks
- 1(k) Heat of spontaneous adsorption5 marks
- 1(l) Conductivity as an intensive property5 marks
- 1(m) Anthracene photodimerization5 marks
- 1(n) Particle-in-a-box eigenvalue5 marks
- 1(o) Second Balmer line5 marks
- 1(p) Normal-mode counts5 marks