At 150 K and 100 atm the density of A r ( M = 40 g / mol ) is 1.0 g / mL . Its compressibility factor Z is approximately 3.3 × 1 0 x . Find x. ( R = 0.0821 L atm K − 1 mol − 1 )

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Density = 1.0 g / mL = 1000 g / L , so V m ​ = 40/1000 = 0.040 L / mol . Z = P V m ​ / RT = 100 × 0.040/ ( 0.0821 × 150 ) = 4.0/12.3 = 0.325 , about 3.3 × 1 0 − 1 . x = − 1 . Answer: -1.

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