One mole of a gas at 300 K and 5.0 x 10^6 Pa occupies 4.0 x 10^-4 m^3. Taking R = 8.314 J K^-1 mol^-1, the compressibility factor (two decimals) is

JEE Main levelNumerical answerDifficulty 2 of 3rg114

One mole of a gas at $300 K$ and $5.0 \times 1 0^{6} Pa$ occupies $4.0 \times 1 0^{- 4} m^{3}$ . Taking $R = 8.314 J K^{- 1} mo l^{- 1}$ , the compressibility factor (two decimals) is

Numerical answer type, tolerance 0.01.

Z = \frac{P V}{n RT} = \frac{5.0 \times 1 0 ^{6} \times 4.0 \times 1 0 ^{- 4}}{1 \times 8.314 \times 300} = \frac{2000}{2494.2} = 0.8019

, i.e.

0.8

. The gas is more compressible than ideal.