For nitrogen ( $a = 1.39 L^{2} atm mo l^{- 2}$ , $b = 0.0391 L mo l^{- 1}$ ) the second virial coefficient $B = b - a / RT$ at $300 K$ , in $mL mo l^{- 1}$ (one decimal, $R = 0.0821 L atm K^{- 1} mo l^{- 1}$ ), is

Numerical answer type, answer in mL/mol, tolerance 0.3.

Answer: -17.3 mL/mol

a / RT = 1.39/ (0.0821 \times 300) = 1.39/24.63 = 0.05644 L mo l^{- 1}

B = 0.0391 - 0.05644 = - 0.01734 L mo l^{- 1} = - 17.34 mL mo l^{- 1}

, i.e.

- 17.3 mL mo l^{- 1}

. The negative sign shows attraction dominates at

300 K