The Boyle temperature of a gas is $716 K$ and its van der Waals constant $a = 3.59 L^{2} atm mo l^{- 2}$ . Taking $R = 0.0821 L atm K^{- 1} mo l^{- 1}$ , the critical volume $V_{c}$ of the gas in $L mo l^{- 1}$ (three decimals) is

Numerical answer type, answer in L/mol, tolerance 0.002.

Answer: 0.183 L/mol

From

T_{B} = a / R b

b = \frac{a}{R T _{B}} = \frac{3.59}{0.0821 \times 716} = \frac{3.59}{58.784} = 0.06107 L mo l^{- 1}

. For a van der Waals gas

V_{c} = 3 b = 0.1832

, i.e.

0.183 L mo l^{- 1}