The enthalpy of vaporization of a liquid is 30 kJ ${\rm{mo}}{{\rm{l}}^{ - 1}}$ and entropy of vaporization is 75 J ${\rm{mo}}{{\rm{l}}^{ - 1}}$. The boiling point of liquid at 1 atm is :

NEET UG CHEMISTRY MCQ: THERMODYNAMICS

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The enthalpy of vaporization of a liquid is 30 kJ

\mathrm{mo}{\mathrm{l}^{ - 1}}

and entropy of vaporization is 75 J

\mathrm{mo}{\mathrm{l}^{ - 1}}

. The boiling point of liquid at 1 atm is :

View answer and explanation

Correct option: B

400 K

Option B (400 K) is correct. It is calculated using the Clausius-Clapeyron equation, where boiling point is determined by the enthalpy and entropy of vaporization.

More Questions from THERMODYNAMICS

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The internal energy of one mole of a gas is: An ideal gas is allowed to expand both reversibly and irreversibly in an isolated system. If

{T_i}

is the initial temperature and

{T_f}

is the final temperature, which of the following statements is correct? For which of the process, ∆S is negative? The work done by a weightless piston in causing an expansion ∆V (at constant temperature), when the opposing pressure, P is variable, is given by: Assertion :- At constant pressure of the change

\mathrm{H}_2 \mathrm{O}(\mathrm{s}) \rightarrow \mathrm{H}_2 \mathrm{O}(\mathrm{g})

work done is negative.

Reason :- During phase transition work done is always negative. For the reaction,

\mathrm{X}_2 \mathrm{O_4(l}) \longrightarrow 2 \mathrm{XO}_2(\mathrm{g}), \Delta U=2.1 \mathrm{~kcal}, \Delta S=20 \mathrm{~cal} \mathrm{~K}^{-1}

300 \mathrm{~K}

. Hence,

\Delta G