A van der Waals gas has T c = 300 K and P c = 50 atm . The compressibility factor of 1 mol of this gas at 600 K and a molar volume of 0.40 L (up to two decimal places) is ( R = 0.0821 L atm K − 1

Question

A van der Waals gas has T c ​ = 300 K and P c ​ = 50 atm . The compressibility factor of 1 mol of this gas at 600 K and a molar volume of 0.40 L (up to two decimal places) is ( R = 0.0821 L atm K − 1 mol − 1 )

TestHub · Accepted Answer

b = R T c ​ / ( 8 P c ​ ) = 0.0821 × 300/400 = 0.0616 L / mol a = 27 R 2 T c 2 ​ / ( 64 P c ​ ) = 27 × 0.00674 × 90000/3200 = 5.12 atm L 2 / mol 2 . P = RT / ( V − b ) − a / V 2 = 49.26/0.3384 − 5.12/0.16 = 145.6 − 32.0 = 113.6 atm . Z = P V / RT = 113.6 × 0.40/49.26 = 0.92 . Answer: 0.92.

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