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# Introductory Chemical Engineering Thermodynamics

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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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