Unformatted text preview: Solutions to Chemical and Engineering Thermodynamics, 3e 5
5.1 (also available as a Mathcad worksheet) $ $ (a) G = H  TS at P = 25 MPa and T = 22399 C = 49714 K . . . equal with $ $ $ G V = H V  TS V = 28031  497.14 6.2575 = 307 .8 J g . the accuracy $ $ $ G L = H L  TS L = 962.11  497 .14 2.5547 =  307.9 J g of tables (b) T ( C) T ( K ) $V $V $L H  TS G U  V  W 225 250 300 350 498.15 28063  498.15 6.2639 = . 52315 28801  523.15 6.4085 = . . 57315 30088  573.15 6.6438 = . . 62315 31263  . . 62315 6.8403 = . 314 .1 J g 472 .5 799 .1 11362 . 400 67315 32393  673.15 7 .0148 =  14827 . . . (Note: All Gibbs free energies are relative to the internal energy and entropy of the liquid phase $ $ $ $ $ $ being zero at the triple point. Since H L ~ U L , and G L = H L  TS L , we have that G L = 0 at the triple point.) $ $ $ (c) T ( C) T ( K) HL  TS L GV 160 170 180 190 200 210 43315 675.55  . 45315 763.22 . 43315 19427 = . . 165.9 J g  185.7  206.3 227 .9 250.4 44315 719 .21  44315 2.0419 = . .  453.15 2.1396 = 46315 807.62  46315 2.2359 = . . 47315 852 .45  47315 2.3309 = . . 48315 897.76  48315 2.4248 = 273.8 . . RESULTS (d) T ( C) 150 160 180 200 220 $ m3 kg V 0.001091 0.001102 0.001127 0.001157 0.001190 c h 224 0.001197 to 0.07998 T ( C) $ m3 kg V c h 225 250 300 350 400 0.08027 0.08700 0.09890 010976 012010 . . (e) Will compute CP from CP ~ FG H$ IJ H T K =
P $ $ H (T + T )  H ( T ) T Solutions to Chemical and Engineering Thermodynamics, 3e T ( C) CP kJ kg K 150 170 180 190 200 210 a a f f 224 4.6225 to 3.200 4.328 4.392 4.430 4.472 4.518 4 .572 250 300 350 400 T ( C) CP kJ kg K 2.952 2 .574 2.350 2.260 These results are plotted below. 5.2 dU dV & = Q P dt dt & dS Q & Closed system entropy balance: = + Sgen dt T (a) System at constant volume and constant entropy Closed system energy balance: dV dS = 0 and =0 dt dt & dU Q & & & & = Q and 0 = + S gen Q = TSgen dt T Solutions to Chemical and Engineering Thermodynamics, 3e and dU 0 or U = minimum at equilibrium at constant V and S. dt dU & & =  TSgen ; T > 0 ; Sgen 0 dt & & (b) System at constant entropy and pressure again Q =  TS gen . Now dP dV d = 0 P = ( PV ) . Thus dt dt dt dU dV d & & = Q P =  TS gen  ( PV ) dt dt dt and dU d d dH & + ( PV ) = (U + PV ) = =  TS gen 0 dt dt dt dt Therefore, enthalpy is a minimum at equilibrium at constant S and P. ...
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 Spring '07
 Rethwisch
 Thermodynamics, Chemical and Engineering Thermodynamics, DT DT DT

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