The enthalpy of vaporisation of a liquid is $30 \mathrm{~kJ} \mathrm{~mol}^{-1}$ and entropy of vaporisation is $75 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$. Calculate boiling point of liquid at 1 atm .

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Aicharya · Accepted Answer

$\begin{aligned} & \Delta \mathrm{S}=\frac{\Delta \mathrm{H}}{\mathrm{T}}(	ext { At constant pressure of } 5 \mathrm{~atm}) \ & 75 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}=\frac{30 	imes 10^3 \mathrm{Jmol}^{-1}}{\mathrm{~T}} \ & \mathrm{~T}=\frac{30000 \mathrm{Jmol}^{-1}}{75 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}} \ & \mathrm{~T}=400 \mathrm{~K}\end{aligned}$

The enthalpy of vaporisation of a liquid is $30 \mathrm{~kJ} \mathrm{~mol}^{-1}$ and entropy of vaporisation…

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