A gas has $B = - 0.0400 L mo l^{- 1}$ and $C = 0.0016 L^{2} mo l^{- 2}$ at a certain temperature. Keeping terms up to $C$ in $Z = 1 + B / V_{m} + C / V_{m}^{2}$ , the molar volume in $L mo l^{- 1}$ (three decimals) at which $Z$ returns to exactly 1 is

Numerical answer type, answer in L/mol, tolerance 0.002.

Answer: 0.04 L/mol

Z = 1

requires

B / V_{m} + C / V_{m}^{2} = 0

, i.e.

V_{m} = - C / B = - 0.0016/ (- 0.0400) = 0.0400 L mo l^{- 1}

, i.e.

0.04

. At this molar volume the attraction term and the excluded volume term cancel; at larger

V_{m}

Z

is below 1 and at smaller

V_{m}

it is above 1.