Elementary Principles 115 - 5.4 (cont'd) e. Basis: 1 kg...

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5- 3 5.4 (cont’d) e. Basis 1 kg slurry x kg crystals V m crystals x kg crystals kg / m cc 3 c c 3 :, ⇒= bg di ρ 1- x kg liquid V m liquid 1-x kg liquid kg / m kg VVm x x cl 3 c l 3 sl 3 c c c l b g b g , = = + = + ρρ 11 1 5.5 Assume P atm atm = 1 PV RT V = 0.08206 m atm kmol K K 4.0 atm kmol 10 mol mm o l 3 3 3 ±± . . =⇒ = 3132 1 00064 == 1 0 0064 10 45 3 mol 29.0 g 1 kg m a i r mo l g kg m 3 3 . . 5.6 a. V= nRT P mol L atm mol K 373.2 K 10 atm L = = 100 0 08206 306 .. . b. % . . error = 3.06L - 2.8L L 28 100% 9 3% ×= 5.7 Assume P b a r atm = 1013 . a. PV nRT n bar m kmol K 25+ 273.2 K .08314 m bar kg N kmol kg N 3 2 2 = +⋅ = 10 1013 20 0 0 28 02 249 3 . b. PV PV nV T T P P n V ss s s s s s s = n m 273K
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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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