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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

  1. A-27.58 J/K/mol
  2. B+27.58 J/K/mol
  3. C-24.28 J/K/mol
  4. 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

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