What is the internal energy $({kJ})$ change occurs when $36 {~g}$ of ${H}_2 {O}(l)$ converted to ${H}_2 {O}({g})$ ? $\Delta {H}^{\circ}$ (vapourisation) $=$ $40.79 {~kJ} / {mol}$

NEET UG CHEMISTRY MCQ: THERMODYNAMICS

1 Hard

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Application +4 | -1

What is the internal energy

({kJ})

change occurs when

36 {~g}

{H}_2 {O}(l)

converted to

{H}_2 {O}({g})

\Delta {H}^{\circ}

(vapourisation)

=

40.79 {~kJ} / {mol}

View answer and explanation

Correct option: A

75.38

Option A (75.38) is correct because it is calculated by finding the change in internal energy using the enthalpy of vaporization and accounting for the work done during the phase change.

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