Two moles of a gas at $500 K$ and $80 atm$ have $Z = 1.15$ . Using $R = 0.0821 L atm K^{- 1} mo l^{- 1}$ , the volume of the gas in litres (two decimals) is

Numerical answer type, answer in L, tolerance 0.02.

Answer: 1.18 L

V = \frac{Z n RT}{P} = \frac{1.15 \times 2 \times 0.0821 \times 500}{80} = \frac{94.41}{80} = 1.1802 L

, i.e.

1.18 L

. The volume is 15 percent larger than the ideal value of

1.0263 L