Introductory Chemical Engineering Thermodynamics

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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. ...
View Full Document

Ask a homework question - tutors are online