Chemistry - Thermodynamics - II Question with Solution | TestHub
Match Column I (Process) with Column II (Feature)
Column I | Column II |
|---|---|
P) Vaporization of a liquid at its normal boiling point | i) Δ G = 0, Δ S > 0 |
Q) Freezing of a supercooled liquid | ii) Δ G < 0, Δ S < 0 |
R) Condensation of a vapour above its boiling point | iii) Δ G > 0 (non-spontaneous) |
S) Sublimation at the sublimation point | iv) Δ G = 0, solid in direct equilibrium with vapour |
Options:
Answer:
Solution:
At the normal boiling point the two phases are at equilibrium and liquid → vapour gains entropy (P – i). A supercooled liquid freezes spontaneously, Δ G < 0, with entropy loss (Q – ii). Above the boiling point the vapour is the stable phase, so condensation is non-spontaneous, Δ G > 0 (R – iii). At the sublimation point solid and vapour are in direct equilibrium, Δ G = 0 (S – iv).
