Elementary Principles 153

# Elementary Principles 153 - 5.60 O 2 TC = 154.4 K PC = 49.7...

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5- 41 5.60 O: 2 T K C = 154 4 . ; P a t m C = 49 7 . ; ω = 0 021 .; T K 6 5 C 208 2 . bg ; P atm = 83 ± m k g h = 250 ; R L a t m m o l K = 0 08206 . SRK constants: a L a t m m o l 22 =⋅ 138 b L m o l = 00221 m = 0517 α = 0840 . SRK equation : fV RT Vb a VV b P = 0=====>V = 2.01 L / mol E-Z Solve ² ²² ² ² di = + ⇒= = ± V 250 kg kmol 10 mol 2.01 L h 32.00 kg 1 kmol mol Lh 3 15,700 5.61 F P A - W = 0 where W = mg = 5500 kg 9.81 N yC O m s 2 2 = e j 53900 a. P W A N m atm 1.013 10 N / m atm CO piston 4 52 2 == × = 53900 015 1 301 2 π . . b. SRK equation of state: P = RT V-b a VV+b ² did i For CO : T , P atm 2c c = = 304 2 72 9 .. , =0.225 a = m atm / kmol , b = m / kmol, m C) 62 3 o 3654 0 02967 08263 25 1016 . , ( . ⋅= = 298 2 1016 3654 3 . . ² ² atm = 0.08206 K V - 0.02967 V V + 0.02967 =====> V = 0.675 m / kmol ma tm kmol K m kmol kmol m kmol
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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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