Chemistry - Thermodynamics - II Question with Solution | TestHub
Match Column I (graph) with the physical quantity in Column II (the shaded area under the curve represents the magnitude of)
Column I | Column II |
|---|---|
P) P vs V, falling adiabatic curve P = f(V), area shaded between V1 and V2
| i) qrev |
Q) V vs P, falling isothermal curve V = f(P), area shaded between P1 and P2
| ii) (Δ G)T |
R) T vs S, rising curve T = f(S), area shaded between S1 and S2
| iii) (Δ G)P |
S) S vs T, rising curve S = f(T), area shaded between T1 and T2
| iv) w |
Options:
Answer:
Solution:
(P) Area under the P–V curve = ∫ P dV = |w| — iv. (Q) Area under the V–P curve = ∫ V dP = (Δ G)T for an isothermal change — ii. (R) Area under the T–S curve = ∫ T dS = qrev — i. (S) Area under the S–T curve = ∫ S dT = -(Δ G)P at constant pressure — iii. Trap: for an isothermal ideal-gas change ∫ P dV and ∫ V dP are numerically equal, but only ∫ V dP equals (Δ G)T — identify the axes before matching.
