For a van der Waals gas the virial coefficients are $B = b - a / RT$ and $C = b^{2}$ . A gas has $b = 0.0400 L mo l^{- 1}$ and $a = 1.60 L^{2} atm mo l^{- 2}$ . At $487 K$ and molar volume $0.400 L mo l^{- 1}$ ( $R = 0.0821 L atm K^{- 1} mo l^{- 1}$ ), $Z$ to three decimals is approximately

Answer: C

a / RT = 1.60/ (0.0821 \times 487) = 1.60/39.983 = 0.04002

, so

B = 0.0400 - 0.04002 = - 0.00002

, essentially zero (this is the Boyle temperature). Then

Z = 1 + B / V_{m} + b^{2} / V_{m}^{2} = 1 + - 0.00004 + \frac{0.0016}{0.160} = 1.0100 \approx 1.010

. The third virial term alone gives the deviation at

T_{B}