JEE Advanced 2025 Paper 1, Chemistry Section 3 Q3: Deviation From Ideal Gas Behaviour

JEE Advanced2025Paper 1Chemistry Section 3

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 $^{- 1}$

Your answer

Solution

Start with the van der Waals equation and multiply through by $V_{m}^{2}$ to obtain a cubic in $V_{m}$ :

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

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

Reading off:

Coefficient of $V_{m}^{2}$ $= - (b P + R T)$ , coefficient of $V_{m}$ $= a$ .

$R T = 0.082 \times 300 = 24.6$ dm $^{3}$ atm mol $^{- 1}$ ; $b P = 0.060 \times 300 = 18.0$ dm $^{3}$ atm mol $^{- 1}$ .

Ratio $= \frac{- ( b P + R T )}{a} = - \frac{18.0 + 24.6}{6.0} = - \frac{42.6}{6.0} = - 7.10$ mol dm $^{- 3}$ .

$∴$ Ratio $= - 7.10$ .

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JEE Advanced 2025 Paper 1, Chemistry Section 3 Q3: Deviation From Ideal Gas Behaviour

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