First slide

2nd Law of Thermodynaimcs

Question

ΔS for vapourization of 900 g water (in KJ/ K) is [ ΔH_vap=40KJ/mol]

Easy

Solution

$\Delta S=\frac{\Delta H_{vapourization}}{B.P}=\frac{40}{373}KJ/mole$

$For\;\;\;18gm ---- \frac{40}{373}KJ$

$900gm ----\;\;?$

$=\frac{900}{18}\times \frac{40}{373}=\frac{50\times 40}{373}$

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