For a gas $B = - 0.050 L mo l^{- 1}$ and $C = 0.0016 L^{2} mo l^{- 2}$ at a certain temperature. Its compressibility factor at a molar volume of $0.400 L mo l^{- 1}$ , keeping terms up to $C$ (three decimals), is

Numerical answer type, tolerance 0.002.

Answer: 0.885

Z = 1 + \frac{B}{V _{m}} + \frac{C}{V _{m}^{2}} = 1 + \frac{- 0.050}{0.400} + \frac{0.0016}{0.160} = 1 - 0.125 + 0.010 = 0.885