Molar volume ( V m ) of a van der Waals gas can be calculated by expressing the van der Waals equation as a cubic equation with V m as the variable. The ratio (in mol dm -3 ) of the coefficient of V m 2 to the coefficient of V m for a gas having van der Waals constants a=6.0 dm 6 atm mol -2 and b=0.060 dm 3 mol -1 at 300 K and 300 atm is . Use: Universal gas constant (R)=0.082 dm 3 atm mol -1 K -1

Chemistry · States of Matter - Gaseous State

JEE Advanced 2025 — Paper 1 — Question 39

Molar volume ( $V_{m}$ ) of a van der Waals gas can be calculated by expressing the

van der Waals equation as a cubic equation with $V_{m}$ as the variable. The ratio

(in $\mathrm{mol} \mathrm{dm}^{-3}$ ) of the coefficient of $V_{m}^{2}$ to the coefficient of

$V_{m}$ for a gas having van der Waals constants

$a=6.0 \mathrm{dm}^{6} \mathrm{~atm} \mathrm{~mol}^{-2}$ and

$b=0.060 \mathrm{dm}^{3} \mathrm{~mol}^{-1}$ at 300 K and 300 atm is $\qquad$ .

Use: Universal gas constant $(R)=0.082 \mathrm{dm}^{3} \mathrm{~atm} \mathrm{~mol}{ }^{-1} \mathrm{~K}^{-1}$

Numerical answer — enter this value.

$\quad\left(P+\frac{a}{V_{m}^{2}}\right)\left(V_{m}-b\right)=R T$

$P V_{m}-b P+\frac{a}{V_{m}}-\frac{a b}{V_{m}^{2}}-R T=0$

$\Rightarrow \quad P V_{m}^{2}-(b P+R T) V_{m}^{2}+a V_{m}-a b=0$

Coefficient of $\mathrm{V}_{\mathrm{m}}^{2}=-(\mathrm{bP}+\mathrm{RT})$

Coefficient of $\mathrm{V}_{\mathrm{m}}=\mathrm{a}$

Ratio $=-\frac{(\mathrm{bP}+\mathrm{RT})}{\mathrm{a}}=-\left[\frac{0.06 \times 300+24.6}{6}\right]=-7.1$ .

Answer key and solution verified before publishing.

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