NEET UG CHEMISTRY MCQ: THERMODYNAMICS

1 Hard

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Application +4 | -1

The magnitude of work done by one mole of a Van der Waals gas during its isothermal reversible expansion from volume

V_1

V_2

at temperature

T

\mathrm{K}

, is

View answer and explanation

Correct option: C

R T \ln \left(\frac{V_2-b}{V_1-b}\right)+a\left(\frac{1}{V_2}-\frac{1}{V_1}\right)

Option C (

R T \ln \left(\frac{{V_2-b}}{{V_1-b}}\right)+a\left(\frac{{1}}{{V_2}}-\frac{{1}}{{V_1}}\right)

) is correct because it represents the work done by a Van der Waals gas, accounting for both the volume correction (b) and the intermolecular attraction (a).

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