Elementary Principles 150 - 5.56 b g T bC H g = 369.9 K, P...

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5- 38 5.56 PV RT B V B= RT P BB c c o1 ± ± =+ ⇒ + 1 ω bg From Table B.1 T CH OH K, P atm T C H atm From Table 5.3-1 CH OH = 0.559, C H = 0.152 B CH OH) T BCH ) T B CH OH) T c3 c 8 c 33 8 o3 r 8 r 13 r b g ej e j == =− = = 5132 7850 369 9 42 0 0083 0422 373.2K 513.2K 0619 0 422 0 083 373.2K 369.9K 0333 0139 0172 373.2K 11 4 .. (. . . . . . . . . . . . .6 .6 .6 .6 .2 ωω 513.2K BCH) T 369.9K 138 r e j 4 44 0516 373.2K 0 0270 .2 .2 .2 . . . . . = B(CH OH) = RT P La tm mol K 78.5 atm B(C H ) = RT P mol K 42.0 atm 3 c c L mol 38 c c L mol + = −− = + = = ch 0 08206 513.2K 0 619 0559 0516 0 4868 0 08206 369.9 K 0 333 0152 0 0270 02436 . . . . . . By y B B B B B L / mol = -0.3652 L / mol B L/mol mix i j ij j i ij ii jj ij mix =⇒ = + + +− 05 05 0 4868 030 030 04868 2 030 070 03652 070 070 03166 . . . . . . di b g b g b g PV RT V-B = 10.0 atm 373.2K mol K VV + 0 . 3 1 6 6 = 0 2 mix 2 ± ± . ±± F H G I K J 008206 Solve for V:V = 1 1- 4 0.326 mol / L .3166 L / mol 2 0.326 mol / L
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This note was uploaded on 02/27/2012 for the course CHEMICAL E 312 taught by Professor Cheung during the Fall '11 term at The University of Akron.

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