NTA Abhyas JEE Main2020ChemistryThermodynamics (C)Practice
For the process, 1 Ar (300 K, 1 bar) → 1 Ar (200 K, 10 bar), assuming ideal gas behaviour, the change in molar entropy is
Options
- A-27.58 J/K/mol
- B+27.58 J/K/mol
- C-24.28 J/K/mol
- D+24.28 J/K/mol
Correct answer
A. -27.58 J/K/mol
Step-by-step solution
Δ S = n C p ln T 2 T 1 - n R ln p 2 p 1 = 2.303 C p log T 2 T 1 - 2.303 R log p 2 p 1 For monoatomic gas like Ar, C p = 5 2 R = 5 × 8.314 2 = 20.8 Δ S = 2.303 × 20.8 log 200 300 - 2.303 × 8.314 log 10 1 = 2.303 × 20.8 log 2 3 - 2.303 × 8.314 = 47.9 × - 0.176 - 19.15 = - 8.43 - 19.15 = - 27.58 J/K/mol