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

△ S_{v a p o r i s a t i o n} > 100 J / K m o l e

at 350 K and 0.5 atm

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△ S_{v a p o r i s a t i o n} < 100 J / K m o l e

at 350 K and 0.5 atm

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△ S_{v a p o r i s a t i o n} < 100 J / K m o l e

at 350 K and 2 atm

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△ S_{v a p o r i s a t i o n} = 100 J / K m o l e

at 350 K and 2 atm

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Solution

The correct options are
A $△ S_{v a p o r i s a t i o n} > 100 J / K m o l e$ at 350 K and 0.5 atm
D $△ S_{v a p o r i s a t i o n} < 100 J / K m o l e$ at 350 K and 2 atm
Solution:- (A) and (C)
Equillibrium at the boiling point-
$A_{(l)} ⟶ B_{(g)}; {Δ H}_{v a p .} = 35 K J / m o l$
As we know that,
${Δ S}_{v a p .} = \frac{{Δ H}_{v a p .}}{T_{v a p .}}$
$T_{v a p .} = 350 K (Given)$
$∴ {Δ S}_{v a p .} = \frac{35}{350} = 0.1 K J {m o l}^{- 1} K^{- 1} = 100 J {m o l}^{- 1} K^{- 1}$
At $350 K$ and $0.5 a t m$ -
The boiling point $(T_{v a p .})$ will be lower at decreased pressure.
$∴ T_{v a p .} < 350 K$
Hence, ${Δ S}_{v a p .} > 100 J {m o l}^{- 1} K^{- 1}$
At $350 K$ and $0.5 a t m$ -
The boiling point $(T_{v a p .})$ will be higher at increased pressure.
$∴ T_{v a p .} > 350 K$
Hence, ${Δ S}_{v a p .} < 100 J {m o l}^{- 1} K^{- 1}$
Hence both $(A)$ and $(C)$ will be correct.

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