$0.500 mol$ of a gas with compressibility factor $0.92$ is held in $1.20 L$ at $320 K$ ( $R = 0.0821 L atm K^{- 1} mo l^{- 1}$ ). The pressure in atm (two decimals) is

Numerical answer type, answer in atm, tolerance 0.02.

Answer: 10.07 atm

P = \frac{Z n RT}{V} = \frac{0.92 \times 0.500 \times 0.0821 \times 320}{1.20} = \frac{12.0851}{1.20} = 10.0709 atm

, i.e.

10.07 atm

. The ideal value would be

10.95 atm