Chemistry - Thermodynamics - II Question with Solution | TestHub
Match Column I with Column II
Column I | Column II |
|---|---|
P) Fusion at melting point | i) Δ G = 0 |
Q) Vaporization at boiling point | ii) Δ G < 0 |
R) Condensation at triple point | iii) Δ S > 0 |
S) Melting at normal boiling point | iv) Δ H ≈ Δ U |
Options:
Answer:
Solution:
(P) At the melting point solid and liquid are in equilibrium, Δ G = 0; the solid-liquid volume change is tiny, so Δ H ≈ Δ U — i,iv. (Q) At the boiling point liquid and vapour are in equilibrium, Δ G = 0; liquid → gas has Δ S > 0 — i,iii. (R) At the triple point all three phases coexist in equilibrium, so condensation there also has Δ G = 0 — i only (gas → liquid has Δ S < 0 and a large Δ V, so no other option fits). (S) Melting at the normal boiling point means T is far above the melting point, so melting is spontaneous, Δ G < 0; solid → liquid has Δ S > 0 and Δ V tiny, so Δ H ≈ Δ U — ii,iii,iv.
