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Entropy of Phase Change

If Hvapourisa...

Question

If $△ H_{v a p o u r i s a t i o n}$ of substance $X (l)$ (molar mass = 30 g/mol) is 300 J/g at its boiling point 300 K, molar entropy change for reversible condensation process is:

30 J / m o l K^{- 1}

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- 60 J / m o l K^{- 1}

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- 30 J / m o l K^{- 1}

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None of these

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Solution

The correct option is C $- 30 J / m o l K^{- 1}$
Vapourisation and condensation are reversible process,
Enthalpy of vapourisation = $300 J / g$
Enthalpy of condensation = $- 300 J / g$
Molar enthalpy of condensation = $- 300 \times 30 g / m o l$
Molar entropy, $△ S = \frac{△ H_{c o n d e n s a t i o n}}{T}$
$△ S = \frac{- 300 \times 30}{300}$
$= - 30 J / m o l K^{- 1}$

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If △Hvapourisation of substance X(l) (molar mass = 30 g/mol) is 300 J/g at its boiling point 300 K, molar entropy change for reversible condensation process is:

The correct option is C −30 J/molK−1Vapourisation and condensation are reversible process, Enthalpy of vapourisation = 300 J/g Enthalpy of condensation = −300 J/g Molar enthalpy of condensation = −300×30 g/mol Molar entropy, △S=△HcondensationT △S=−300×30300 =−30 J/molK−1

If $△ H_{v a p o u r i s a t i o n}$ of substance $X (l)$ (molar mass = 30 g/mol) is 300 J/g at its boiling point 300 K, molar entropy change for reversible condensation process is: