JEE Challenger
More from States of Matter
ChemMedAdv 2025 • Paper 1

Ratio of Coefficients in Van der Waals Cubic Equation

Molar volume (VmV_{\text{m}}) of a van der Waals gas can be calculated by expressing the van der Waals equation as a cubic equation with VmV_{\text{m}} as the variable. The ratio (in mol dm−3\text{mol dm}^{-3}) of the coefficient of Vm2V_{\text{m}}^2 to the coefficient of VmV_{\text{m}} for a gas having van der Waals constants a=6.0 dm6 atm mol−2a = 6.0\text{ dm}^6\text{ atm mol}^{-2} and b=0.060 dm3 mol−1b = 0.060\text{ dm}^3\text{ mol}^{-1} at 300 K300\text{ K} and 300 atm300\text{ atm} is ______.

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

Official Numerical Answer-7.2 to -7

Topics & Concepts

States of MatterReal Gases

Step-by-Step Solution

The van der Waals equation for 1 mole of a real gas is given by: (P+aVm2)(Vm−b)=RT\left(P + \frac{a}{V_{\text{m}}^2}\right)(V_{\text{m}} - b) = RT

Expanding this expression gives: PVm−Pb+aVm−abVm2=RTP V_{\text{m}} - P b + \frac{a}{V_{\text{m}}} - \frac{ab}{V_{\text{m}}^2} = RT

Multiplying the entire equation by Vm2V_{\text{m}}^2 and rearranging into standard cubic polynomial form with respect to VmV_{\text{m}}: PVm3−(Pb+RT)Vm2+aVm−ab=0P V_{\text{m}}^3 - (P b + RT)V_{\text{m}}^2 + a V_{\text{m}} - ab = 0

From this cubic equation:

  • The coefficient of Vm2V_{\text{m}}^2 is: Coefficient of Vm2=−(Pb+RT)\text{Coefficient of } V_{\text{m}}^2 = -(Pb + RT)

  • The coefficient of VmV_{\text{m}} is: Coefficient of Vm=a\text{Coefficient of } V_{\text{m}} = a

Thus, the ratio of the coefficient of Vm2V_{\text{m}}^2 to the coefficient of VmV_{\text{m}} is: Ratio=−(Pb+RT)a\text{Ratio} = \frac{-(Pb + RT)}{a}

Given values:

  • P=300 atmP = 300\text{ atm}
  • T=300 KT = 300\text{ K}
  • R=0.082 dm3 atm mol−1 K−1R = 0.082\text{ dm}^3\text{ atm mol}^{-1}\text{ K}^{-1}
  • a=6.0 dm6 atm mol−2a = 6.0\text{ dm}^6\text{ atm mol}^{-2}
  • b=0.060 dm3 mol−1b = 0.060\text{ dm}^3\text{ mol}^{-1}

Calculating PbPb and RTRT: Pb=300 atm×0.060 dm3 mol−1=18.0 atm dm3 mol−1Pb = 300\text{ atm} \times 0.060\text{ dm}^3\text{ mol}^{-1} = 18.0\text{ atm dm}^3\text{ mol}^{-1} RT=0.082 dm3 atm mol−1 K−1×300 K=24.6 atm dm3 mol−1RT = 0.082\text{ dm}^3\text{ atm mol}^{-1}\text{ K}^{-1} \times 300\text{ K} = 24.6\text{ atm dm}^3\text{ mol}^{-1}

Substituting these into the numerator: Pb+RT=18.0+24.6=42.6 atm dm3 mol−1Pb + RT = 18.0 + 24.6 = 42.6\text{ atm dm}^3\text{ mol}^{-1}

Now, calculating the ratio: Ratio=−42.6 atm dm3 mol−16.0 dm6 atm mol−2=−7.1 mol dm−3\text{Ratio} = \frac{-42.6\text{ atm dm}^3\text{ mol}^{-1}}{6.0\text{ dm}^6\text{ atm mol}^{-2}} = -7.1\text{ mol dm}^{-3}

The ratio of the coefficient of Vm2V_{\text{m}}^2 to the coefficient of VmV_{\text{m}} is −7.1-7.1.

Ratio of Coefficients in Van der Waals Cubic Equation | Chemistry PYQ Solution - JEE Challenger