Answer / kang chuen tat (malaysia - pen

Answer 42 - Let x(1) = x(2) = 0.5 for 50 : 50 mole fraction mixture, then P = x(1) p(1) g(1) + x(2) p(2) g(2) or 1.08 = 0.5 (0.82) exp [ (0.5) A (0.5) ] + 0.5 (1.93) exp [ (0.5) A (0.5) ]. Then exp [ (0.5) A (0.5) ] = 1.08 / [ 0.5 (0.82 + 1.93) ] = 0.785, A = - 0.97. P for 30 : 70 mixture is P = x(1) p(1) g(1) + x(2) p(2) g(2) = (0.3) (0.82) exp [ (-0.97) (0.7) (0.7) ] + (0.7) (1.93) exp [ (-0.97) (0.3) (0.3) ] = 1.39 bar (proven) where g(1) = exp [ A x(2) x(2) ], g(2) = exp [ A x(1) x(1) ]. The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.

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