JEE Advanced 2018 Paper 2 · Chemistry · Mole Concept and Stoichiometry · hard

Two moles of a van der Waals gas (a=0.40 Pa m^6 mol^-2, b=4.0×10^-5 m^3 mol^-1) expand reversibly and isothermally at 300 K from 5.0 L to 10.0 L.…

Numerical answer

Two moles of a van der Waals gas (

a=0.40\ \mathrm{Pa\,m^6\,mol^{-2}}

b=4.0\times10^{-5}\ \mathrm{m^3\,mol^{-1}}

) expand reversibly and isothermally at

300\ \mathrm{K}

from

5.0\ \mathrm{L}

10.0\ \mathrm{L}

. Using

R=8.314\ \mathrm{J\,K^{-1}\,mol^{-1}}

, the magnitude of the work done by the gas, in

\mathrm{kJ}

, is ______. Give the answer correct to two decimal places.

Show answer and solution

Answer: 3.34

For the van der Waals equation

P=\dfrac{nRT}{V-nb}-\dfrac{an^2}{V^2}

, the reversible work done by the gas is

w=\int_{V_1}^{V_2}P\,dV=nRT\ln\dfrac{V_2-nb}{V_1-nb}+an^2\left(\dfrac{1}{V_2}-\dfrac{1}{V_1}\right)

. Here

nb=2\times4.0\times10^{-5}=8.0\times10^{-5}\ \mathrm{m^3}

V_1=5.0\times10^{-3}\ \mathrm{m^3}

and

V_2=1.0\times10^{-2}\ \mathrm{m^3}

. The first term is

2\times8.314\times300\times\ln\dfrac{9.92\times10^{-3}}{4.92\times10^{-3}}=4988.4\times0.7014=3499\ \mathrm{J}

. The attraction term is

0.40\times4\times(100-200)=-160\ \mathrm{J}

(the attractive forces reduce the work obtained). Hence

w=3499-160=3338\ \mathrm{J}\approx3.34\ \mathrm{kJ}

. For comparison an ideal gas would give

nRT\ln2=3.46\ \mathrm{kJ}

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