A 2.0 L vessel contains 100 g of N e ( M = 20 g / mol ) at 300 K. Taking the attraction term as negligible and b = 0.017 L / mol , the compressibility factor of the gas is ( R = 0.0821 L atm K − 1 mol

Question

A 2.0 L vessel contains 100 g of N e ( M = 20 g / mol ) at 300 K. Taking the attraction term as negligible and b = 0.017 L / mol , the compressibility factor of the gas is ( R = 0.0821 L atm K − 1 mol − 1 )

TestHub · Accepted Answer

n = 100/20 = 5 mol . P ( V − nb ) = n RT gives P = n RT / ( V − nb ) = ( 5 × 0.0821 × 300 ) / ( 2.0 − 0.085 ) = 123.2/1.915 = 64.3 atm . Z = P V / n RT = 2.0/ ( 2.0 − 0.085 ) = 1.044 , about 1.04. Trap: 0.96 comes from the wrong sign ( 1 − nb / V = 0.9575 ) ; 1.00 is the ideal value. Answer: (C).

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