The enthalpy of vapourization of a liquid is $30 k J m o l^{- 1}$ and entropy of vapourization is $75 J m o l^{- 1} K$ . The boiling point of the liquid at $1 a t m$ is:

250 K

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400 K

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450 K

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600 K

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Solution

The correct option is A $400 K$
For the equilibrium process:

$T = \frac{Δ H_{v a p}}{Δ S}$
$= \frac{30 \times 10^{3} J m o l^{- 1}}{75 J m o l^{- 1}}$
$T = 400 K$

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Similar questions

30 k J {m o l}^{- 1}

75 J {m o l}^{- 1} K^{- 1}

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75 J m o l^{- 1} K^{- 1}

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30 k J m o l^{- 1}

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The enthalpy of vaporisation of liquid is 30KJ/mol and entropy of vaporisation is 75J/mol/k. The Boiling point of the liquid at 1 atm is

Δ H = 30 k J m o l^{- 1}

Δ S = 75 J K^{- 1} m o l^{- 1}

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