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JEE Advanced Chemistry Thermodynamics (C) 2023 JEE Advanced 2023 (Paper 2)

JEE Advanced Chemistry Question (2023) — Solution

Question

The entropy versus temperature plot for phases α  and β at 1 bar pressure is given. S T and S 0  are entropies of the phases at temperatures T and 0   K , respectively. The transition temperature for α to β phase change is 600   K and C p , β - C p , α = 1   J   mol - 1   K - 1 . Assume C p , β - C p , α is independent of temperature in the range of 200 to 700   K . C p , α and C p , β are heat capacities of α and β phases, respectively. The value of entropy change, S β - S α (in Jmol - 1   K - 1 ), at 300   K is [Use: ln   2 = 0 . 69 0. Given: S β - S α = 0  at 0   K ]

Options

  1. A. A
  2. B. B
  3. C. C
  4. D. D

Answer

A. A

Step-by-step solution

S = S 0 + ∫ C p d T T S α = S 0 + ∫ C p α d T T S β = S 0 + ∫ C p β d T T Now,  S β - S α = S 0 + ∫ C p β - C p α d T T Given   C p β - C p α = 1 S β - S α = lnT + C  at any temperature of  T S β - S α T 2 - S β - S α T 1 = lnT 2 - lnT 1 T 2 = 600 K   ,   T 1 = 300   K   and   S β - S α = 1   at   600   K 1 - S β - S α 300 = ln 600   - ln 300 ⇒ S β - S α = 1 - 0 . 69 = 0 . 31  

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