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JEE Main levelSingle correct optionDifficulty 2 of 3rg039

A gas has $T_{c} = 150 K$ and $P_{c} = 48 atm$ . Using the van der Waals relations and $R = 0.0821 L atm K^{- 1} mo l^{- 1}$ , its constant $b$ is closest to

A $0.016 L mo l^{- 1}$
B $0.032 L mo l^{- 1}$
C $0.064 L mo l^{- 1}$
D $0.128 L mo l^{- 1}$

Answer: B

b = \frac{R T _{c}}{8 P _{c}} = \frac{0.0821 \times 150}{8 \times 48} = \frac{12.315}{384} = 0.0321 L mo l^{- 1}

, i.e. about

0.032 L mo l^{- 1}

(argon-like).

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