A gas has $Z = 0.9880$ at a molar volume of $5.00 L mo l^{- 1}$ . Neglecting the third and higher virial terms, its second virial coefficient $B$ in $mL mo l^{- 1}$ (nearest integer) is

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

Answer: -60 mL/mol

Z = 1 + B / V_{m}

, so

B = (Z - 1) V_{m} = (0.9880 - 1) (5.00) = - 0.0600 L mo l^{- 1} = - 60 mL mo l^{- 1}