ThermodynamicsHard
Question
The normal boiling point of a liquid is 350 K and ΔHvap is 35 kJ/mol. Assume that ΔHvap is independent from temperature and pressure. The correct statement(s) is/are
Options
A.ΔSvap > 100 J/K-mol at 350 K and 0.5 atm.
B.ΔGvap < 0 at 350 K and 0.5 atm.
C.ΔSvap > 100 J/K-mol at 350 K and 2.0 atm.
D.ΔGvap > 0 at 350 K and 2.0 atm.
Solution
$\left( \Delta S_{\text{Vap}} \right)_{350\text{ K, 1 atm}} = \frac{35 \times 10^{3}}{350} = 100\frac{J}{Kmol}$
On increasing pressure at constant temperature entropy decreases.
$\left( \Delta G_{Vap} \right)_{\text{350 K, 1 atm}} = 0$
On increasing pressure at constant temperature energy increases.
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