One mole of $CO X_{2}$ ( $a = 3.59 L^{2} atm mo l^{- 2}$ , $b = 0.0427 L mo l^{- 1}$ ) occupies $0.200 L$ at $300 K$ . Using the van der Waals equation and $R = 0.0821 L atm K^{- 1} mo l^{- 1}$ , the compressibility factor (two decimals) is

Numerical answer type, tolerance 0.02.

Answer: 0.54

P = \frac{RT}{V _{m} - b} - \frac{a}{V _{m}^{2}} = \frac{24.63}{0.1573} - \frac{3.59}{0.0400} = 156.58 - 89.75 = 66.83 atm

Z = \frac{P V _{m}}{RT} = \frac{66.83 \times 0.200}{24.63} = 0.5427

, i.e.

0.54

. Strong attraction in

CO X_{2}

gives

Z

well below 1.