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← JEE Advanced 2025 · Paper 1
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JEE Advanced 2025 · Paper 1, Chemistry question 10

Molar volume (VmV_m) of a van der Waals gas can be calculated by expressing the van der Waals equation as a cubic equation with VmV_m as the variable. The ratio (in mol dm−3\mathrm{mol\,dm^{-3}}) of the coefficient of Vm2V_m^2 to the coefficient of VmV_m for a gas having van der Waals constants a=6.0 dm6 atm mol−2a = 6.0\ \mathrm{dm^6\,atm\,mol^{-2}} and b=0.060 dm3 mol−1b = 0.060\ \mathrm{dm^3\,mol^{-1}} at 300 K and 300 atm is _____.

Use: Universal gas constant (R) =0.082 dm3 atm mol−1 K−1= 0.082\ \mathrm{dm^3\,atm\,mol^{-1}\,K^{-1}}

Answer and worked solution

Official answer: -7.2

Cubic form. Expand (P+aVm2)(Vm−b)=RT\left(P+\frac a{V_m^2}\right)(V_m-b)=RT:
PVm3−(Pb+RT)Vm2+aVm−ab=0.PV_m^3-(Pb+RT)V_m^2+aV_m-ab=0.

Ratio of coefficients.
−(Pb+RT)a=−300×0.060+0.082×3006.0=−18+24.66=−7.1 mol dm−3.\frac{-(Pb+RT)}{a}=-\frac{300\times0.060+0.082\times300}{6.0}=-\frac{18+24.6}{6}=-7.1\ \text{mol dm}^{-3}.