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Calculate the entropy of vaporisation (in J K⁻¹ mol⁻¹) of liquid X at 27 deg C, given: normal boiling point = 360 K, delta_vap_H at 360 K = 28.8 kJ/mol, Cp,m(X, liquid) = 136 J K⁻¹ mol⁻¹, Cp,m(X, gas) = 82 J K⁻¹ mol⁻¹. Use ln 6 = 1.69 and ln 5 = 1.61.
- 75.5
- 84.2
- 90.0
- 105.3
Correct answer: 75.5
Solution
To find delta_vap_S at 300 K, use a thermodynamic cycle: (i) heat liquid from 300 to 360 K: delta_S1 = 136*ln(360/300) = 136*ln(6/5) = 136*(1.69-1.61) = 136*0.08 = 10.88 J/K/mol. (ii) vaporise at 360 K: delta_S2 = 28800/360 = 80 J/K/mol. (iii) cool gas from 360 to 300 K: delta_S3 = 82*ln(300/360) = 82*ln(5/6) = 82*(-0.08) = -6.56 J/K/mol. Total delta_vap_S(300K) = 10.88 + 80 - 6.56 = 84.32 ~ 84.2 J/K/mol. If we assume the answer choices are approximate, 84.2 fits.
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