CHEMICAL MATERIAL BALANCE - EXAMPLE 2.4 : A mixture consists of benzene (B), tol

CHEMICAL MATERIAL BALANCE - EXAMPLE 2.4 : A mixture consists of benzene (B), toluene (T) and xylene (X). At a temperature of 353 K, the data of vapor pressures : B : 754.12, T : 289.71, X : 91.19. Unit is mm Hg. The pressure P is 0.5 atm. The value of k for each substance is k = (vapor pressure) / P. (a) Calculate k for B, T and X. Let L / V = 0.65. (b) By using the equation V = F / [ (L / V) + 1 ], find the value of V when F = 100, then what is the value of L?

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CHEMICAL MATERIAL BALANCE - EXAMPLE 2.4 : A mixture consists of benzene (B), toluene (T) and xylene ..

Answer / kangchuentat

CHEMICAL MATERIAL BALANCE - ANSWER 2.4 : (a) 0.5 atm is 1 / 2 x 760 = 380 mm Hg. Values of k could be obtained : B : 754.12 / 380 = 1.985; T : 289.71 / 380 = 0.762; X : 91.19 / 380 = 0.240. (b) V = F / [ (L / V) + 1 ] = 100 / (0.65 + 1) = 100 / 1.65 = 60.61. L = 0.65 V = 39.39. 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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