One mole of oxygen ( $a = 1.36 L^{2} atm mo l^{- 2}$ , $b = 0.0318 L mo l^{- 1}$ ) is confined to $0.500 L$ at $350 K$ . By what percentage is the van der Waals pressure lower than the ideal gas pressure? Give the answer to one decimal place. ( $R = 0.0821 L atm K^{- 1} mo l^{- 1}$ )

Numerical answer type, answer in %, tolerance 0.2.

Answer: 2.7 %

Ideal:

P = RT / V_{m} = 57.470 atm

. van der Waals:

P = \frac{RT}{V _{m} - b} - \frac{a}{V _{m}^{2}} = \frac{28.7350}{0.4682} - \frac{1.36}{0.25} = 61.373 - 5.44 = 55.933 atm

. Percentage lowering

= \frac{57.470 - 55.933}{57.470} \times 100 = 2.67

, i.e.

2.7

percent.