At $300 K$ and $12.5 atm$ a gas has $Z = 0.9625$ . Assuming $Z = 1 + BP / RT$ and $R = 0.0821 L atm K^{- 1} mo l^{- 1}$ , the second virial coefficient $B$ in $mL mo l^{- 1}$ (nearest integer) is

Numerical answer type, answer in mL/mol, tolerance 1.

Answer: -74 mL/mol

B = \frac{( Z - 1 ) RT}{P} = \frac{( - 0.0375 ) ( 24.63 )}{12.5} = \frac{- 0.9236}{12.5} = - 0.07389 L mo l^{- 1} = - 73.9 mL mo l^{- 1}

, i.e.

- 74 mL mo l^{- 1}