Elementary Principles 281 - 8.3 (cont'd) b. C p = Cv + R C...

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8- 2 8.3 (cont’d) b. CCR pv =+ CT T TT p [] ( . . .) . .. . kJ / (mol C) o ⋅= + × × + ×− × −− 00252 1547 10 3012 10 0008314 0 0335 1547 10 3012 10 59 2 2 QH n C d T d T P T T == =⋅ + × × = × z z Δ 1 2 0 245 0 0335 1547 10 3012 10 9 65 2 25 1000 mol [kJ / (mol C)] 10 J Piston moves upward (gas expands). o3 (. ) [. . . ] . c. The difference is the work done on the piston by the gas in the constant pressure process. 8.4 a. ( ) () ( ) 66 o5 CH 40 C 0.1265 23.4 10 40 0.1360 [kJ/(mol K)] p l C × = b. C p v di b g bg o C [kJ / (mol C) 40 0 07406 32 95 10 40 2520 10 40 77 57 10 40 0 08684 58 2 12 3 °= + × × + × . . .] c. C p s b g b g C 313 K 0.009615 kJ / (mol K)] × × = 0 01118 1095 10 313 4 891 10 313 52 2 . [ d. Δ ± . ... . HT T v 3 kJ mol × × + × O Q P P = −−− 007406 32 95 10 2 2520 10 77 57 10 4 3171 5 2 8 3 12 4 40 300 e. Δ ± . . H TTT Cs × O Q P P = 0 01118 1095 10 2 4 891 10 3459 5 22 1 313 573 kJ / mol 8.5 H O (v, 100 C, 1 atm) H O (v, 350 C, 100 bar) 2 o 2 o a. ± H =−= 2926 2676 250 kJ kg kJ kg kJ kg b. ± . . . T T d T × + × × =⇒ z 0 03346 0 6886 10 0 7604 10 3593 10 8845 2 1 2 3 100 350 kJ mol 491.4 kJ kg Difference results from assumption in (b) that ± H is independent of P . The numerical difference is Δ ± H for H O v, 350 C, 1 atm H O v, 350 C, 100 bar
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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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