One mole of $O X_{2}$ ( $a = 1.36 L^{2} atm mo l^{- 2}$ , $b = 0.0318 L mo l^{- 1}$ ) is held at $400 K$ in a volume of $0.300 L$ . Using the van der Waals equation, the compressibility factor is closest to ( $R = 0.0821 L atm K^{- 1} mo l^{- 1}$ )

Answer: B

P = \frac{RT}{V _{m} - b} - \frac{a}{V _{m}^{2}} = \frac{32.84}{0.2682} - \frac{1.36}{0.0900} = 122.45 - 15.11 = 107.33 atm

Z = \frac{P V _{m}}{RT} = \frac{107.33 \times 0.300}{32.84} = 0.981

, closest to 0.98. The two corrections nearly cancel here.