A gas obeys $P V_{m} = RT (1 + \frac{B}{V _{m}})$ with $B = - 0.0700 L mo l^{- 1}$ at $300 K$ . Its molar volume at $10.0 atm$ , in $L mo l^{- 1}$ (three decimals, $R = 0.0821 L atm K^{- 1} mo l^{- 1}$ ), is

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

Answer: 2.391 L/mol

Multiplying by

V_{m}

P V_{m}^{2} - RT V_{m} - RTB = 0

, i.e.

10.0 V_{m}^{2} - 24.63 V_{m} + 1.7241 = 0

V_{m} = \frac{24.63 + 24.6 3 ^{2} - 4 \times 10.0 \times 1.7241}{20.0} = \frac{24.63 + 537.673}{20.0} = \frac{24.63 + 23.1878}{20.0} = 2.3909

, i.e.

2.391 L mo l^{- 1}

. The ideal value would be

2.463 L mo l^{- 1}