One mole of a real gas occupies $0.40 L$ at $50 atm$ and $300 K$ . Taking $R = 0.0821 L atm K^{- 1} mo l^{- 1}$ , the compressibility factor $Z$ of the gas (two decimals) is

Numerical answer type, tolerance 0.01.

Answer: 0.81

Z = \frac{P V}{n RT} = \frac{50 \times 0.40}{1 \times 0.0821 \times 300} = \frac{20}{24.63} = 0.8120

, which rounds to

0.81

. Since

Z

is less than 1, attractive forces dominate at this state.