The normal boiling point of a liquid A is 350 K. $Δ H_{v a p}$ at normal boiling point is 35 KJ/mole. Pick out the correct statement(s). (Assume $Δ H_{v a p}$ to be independent of pressure).

Δ S_{v a p o r i s a t i o n}

> 100 J/K mole at 350 K and 0.5 atm

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Δ S_{v a p o r i s a t i o n}

< 100 J/K mole at 350 K and 0.5 atm

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Δ S_{v a p o r i s a t i o n}

< 100 J/K mole at 350 K and 2 atm

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Δ S_{v a p o r i s a t i o n}

= 100 J/K mole at 350 K and 2 atm

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Solution

The correct option is A $Δ S_{v a p o r i s a t i o n}$ > 100 J/K mole at 350 K and 0.5 atm
The equilibrium at the boiling point is
$A (1) ⇋ A (g); Δ_{v a p} H = 35 k J / m o l$
$Δ_{v a p} S = \frac{Δ_{v a p} H}{350 K} = 100 J / K m o l$
At 350 K and 0.5 atm
The boiling point $(T_{v a p})$ will be lower at the decreased pressure.
Since $T_{v a p} < 35 K, Δ_{v a p} S > 100 J / K m o l$

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