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Question

The enthalpy of vapourization of a liquid is $$30 \,kJ \,mol^{-1}$$ and entropy of vapourization is $$75 \,J \,mol^{-1}K$$. The boiling point of the liquid at $$1 \,atm$$ is:


A
250K
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B
400K
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C
450K
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D
600K
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Solution

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

$$T = \dfrac{\Delta H_{vap}}{\Delta S}$$
$$= \dfrac{30 \times 10^3 \,J \,mol^{-1}}{75 \,J \,mol^{-1}}$$
$$T = 400 \,K$$

Chemistry

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