JEE Advanced 2025 Paper 1 · Q10 · Thermodynamics
Express the van der Waals equation as a cubic equation in $V_m$ (molar volume). For a gas with $a = 6.0$ $dm^6$ atm $mol^{-2}$ and $b = 0.060$ $dm^3$ $mol^{-1}$ at 300 K and 300 atm, find the ratio (in mol/$dm^3$) of the coefficient of $V_m^2$ to the coefficient of $V_m$. $R = 0.082$ $dm^3$ atm $mol^{-1}$ $K^{-1}$.
Reveal answer + step-by-step solution
Correct answer:-7.10
Solution
vdW: $(P + a/V_m^2)(V_m - b) = R T$. Multiply through by $V_m^2$: $P V_m^3 - P b V_m^2 + a V_m - a b = R T V_m^2$. Rearrange to standard cubic in $V_m$: $P V_m^3 - (P b + R T) V_m^2 + a V_m - a b = 0$. Coefficient of $V_m^2$ = $-(P b + R T) = -(300 \times 0.060 + 0.082 \times 300) = -(18 + 24.6) = -42.6$. Coefficient of $V_m$ = $a = 6.0$. Ratio = $-42.6 / 6.0 = -7.10$. Within official acceptance band $[-7.2, -7.0]$.
Want to drill deeper? Open this question inside SolveKar AI and tap "Why?" on any step — the AI Mentor reads your full solution context and explains until it clicks. Download SolveKar AI →