JEE Advanced · 2025 · 18 May 2025 · Paper 1 · States of Matter

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}

Answer: -7.1

Solution

The van der Waals equation is:

(P + \frac{a}{V _{m}^{2}}) (V_{m} - b) = R T

Expanding and rearranging as a cubic equation:

P V_{m}^{3} - (b P + R T) V_{m}^{2} + a V_{m} - ab = 0

Coefficient of

V_{m}^{2}

- (b P + R T)

Coefficient of

V_{m}

a

Ratio =

\frac{- ( b P + R T )}{a}

Substituting values:

R T = 0.082 \times 300 = 24.6

b P = 0.060 \times 300 = 18.0

Ratio = - \frac{18.0 + 24.6}{6.0} = - \frac{42.6}{6.0} = - 7.10

JEE Advanced · 2025 · 18 May 2025 · Paper 1 · Numerical answer