UPSC Combined Geo-Scientist, Mains · 2018 · Mains Paper II
Q1(e). Joule–Thomson cooling
Question
In a Joule–Thomson experiment the gas temperature decreases. What is the sign of the Joule–Thomson coefficient? Give reasons.
Answer in brief
The coefficient is positive: μ_(JT)=(∂ T/∂ p)H>0.
Key points
- Throttling through a porous plug is approximately isenthalpic when the apparatus is insulated and changes in bulk kinetic energy are negligible.
- Pressure decreases across the plug, dp<0. Cooling means dT<0, so their ratio (∂ T/∂ p)H is positive.
- Thermodynamically, μ_(JT)=[T(∂ Vₘ/∂ T)ₚ-Vₘ]/C_(p,m). Ideal gases have zero coefficient and do not cool on ideal throttling; real-gas interactions determine the sign.
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.
CoversJoule–Thomson coolingReal-gas thermodynamics
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