Introductory Chemical Engineering Thermodynamics

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Solutions to Chemical and Engineering Thermodynamics, 3e 5.18 (a) Assume the vapor phase is ideal, and that H vap is approximately constant (or an average H vap can be used). ln P P H R T T 2 1 2 1 1 1 = - - F H G I K J vap (1) F H I K = - + - + F H I K = × = × ln . . . . . . . . 2 026 1013 1 222 0 27315 1 1780 27315 352 10 2 93 10 3 4 H R H R H vap vap vap K J mol (b) H T H T H T vap sat. vap, sat. liq, ( 29 = ( 29- ( 29 = ( 29- ( 29 - ( 29- ( 29 ( 29 = - F H G I K J - - F H G I K J L N M M O Q P P H T H T H T H T H T T H H T H H T C C T T sat. vap, sat. liq. , IG IG vap IG sat. vap, IG sat. liq., (c) T T T r C = = + + = 200 27315 2831 27315 0851 . . . . H H T C T r IG sat. vap J mol K - F H G I K J = = 0 851 506 . . and H H T C T r IG sat. liq. J mol K - F H G I K J = = 0 851 4469 . . H T vap K 44.69 J mol ( 29 = - = × 556 45 506 2205 10 4 . . . (d) The reason for the discrepancy is probably not the inaccuracy of corresponding states (since Z C = 0272 . which is close to 0.27) but rather the assumption of an ideal vapor phase in the Clausius-Clapeyron equation. We correct for gas-phase nonideality below. at T = ° 178 C , T r = 0811 . , Z = 082 . T = ° 222 C , T r = 0890 . , Z = 071 . The average value of the compressibility is Z = + ( 29 = 1 2 082 071 0765 . . . We now replace eqn. 1 with ln . . . P P H ZR T T H 2 1 2 1 4 4 1 1 0 765 2 93 10 2 24 10 = - - F H G I K J = × × = × vap vap J mol J mol c h which is in much better agreement with the result of part (c). A better way to proceed would be to compute the compressibility as a function of temperature, i.e., find Z Z T P = ( 29 , and then integrate dP dT H P Z T P RT = ( 29 vap , 2
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Solutions to Chemical and Engineering Thermodynamics, 3e 5.21 (also available as a Mathcad worksheet) 5.22 (also available as a Mathcad worksheet) 5.21 From Eq. 5.7-4 T= P* V*T/ H bar 10 5 Pa .
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