One mole of a gas occupies 1.0 l at 27°C. The compressibility factor of gas at this temperature is 0.85. If b = 0.05 l mol⁻¹, what is the value of van der Waal's constant a for the gas?

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Using the truncated virial form Z = 1 + (b - a/RT)/(Vₘ): 0.85 = 1 + (0.05 - a/24.63)/1.0 (with RT = 0.0821 × 300 = 24.63). So 0.05 - a/24.63 = -0.15, giving a/24.63 = 0.20 and a = 4.926 ≈ 4.93 atm l² mol⁻². Ans: (b)

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