Chemistry - Thermodynamics - II Question with Solution | TestHub
ChemistryThermodynamics - IIGibb's Free EnergyHard2 minQB
ChemistryHardmatching list
Match Column I (graph) with the physical quantity in Column II (area shown in the graph represents the magnitude of)
Column I | Column II |
|---|---|
P) S vs T, rising curve S = f(T), area shaded between T1 and T2
| i) q |
Q) T vs S, rising curve T = f(S), area shaded between S1 and S2
| ii) w |
R) P vs V, falling curve P = f(V), area shaded between V1 and V2
| iii) (Δ G)T |
S) V vs P, falling curve V = f(P), area shaded between P1 and P2
| iv) (Δ G)P |
Options:
Answer:
B
Solution:
(P) Area = ∫ S dT; at constant pressure Δ G = -∫ S dT, so the shaded area equals |(Δ G)P| — iv. (Q) Area = ∫ T dS = qrev — i. (R) Area = ∫ P dV = |w| — ii. (S) Area = ∫ V dP = (Δ G)T for an isothermal change — iii.
Stream:JEESubject:ChemistryTopic:Thermodynamics - IISubtopic:Gibb's Free Energy
⏱ 2mℹ️ Source: QB
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