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Lecture19 - ChE 210 Meeting 19 Outline(F&R Section...

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Unformatted text preview: ChE 210: Meeting 19 February 24, 2011 Outline (F&R Section 6.3-6.4) Homework Due Friday 03/04/2011 - Raoult’s Law and Henry’s Law 6.8, 6.18, 6.23, 6.35, 6.37 - Vapor-Liquid Equilibrium Definitions - Txy and ny Diagrams - Example Problem Process Eguigment i i . .. , 0181 i Had-ac, \ L‘C fox. em mg” m (:k tic L.) " .DJYI Ca \ CG"~.‘¢U¢' 1‘31 \4 «fifiw {If}: :gfwwwwwgww ‘ & 1““ fit 3:66.93 ‘ ‘ V 1' Sif‘um r315 it» gfi§3m ('Mi‘k‘fi kLi L. WMW$ .. fifsz Te 333%; ”k "i wwwww fwfwf ' 3%,ng ww wee; 1/5 ChE 210: Meeting 19 February 24, 2011 Raoult’s Law and Henm’s Law \ i . * ::' {u \‘VLK Ix“:- ;_:.,’,,\,\z‘L ’\ \ Raoult s Law. pA a yAP = xApA(T) . n . l . (- ‘i’ ‘ ELTLlT‘Y-“K’ / :fo A 9 L i “(I i A, .I._ " \l \J ‘ i l ) ‘ , . I _ z \ \ Tkh “t l 9. rd” C 1.4 A“ “j A: 3 n 14’ 5“ £92,? L" a“. \ “1‘ ,, = 1‘ k)» v keshxgfilyiw) ( \l {I \ [V L\ ‘1 x,“ \UQ.‘) .. K ’k Jigflj. I )r-‘ 1 l ) Henry’s Law: 10A 5 MP = XAHA(T> 5k "f . , m ’0 ’9 1 1“ \, '“‘<:i\\\/:lt L ‘g\x_.-.;T=. ‘t H _; \;Q\ :\ ' ' (a , W »\ Mi 12 gin/L1 L i‘xlf-Lg‘dli;k‘tfgx ‘r'xfi L, J Vapor-Liguid Eguilibrium Definitions Saturated vapor — the vapor phase contains the maximum amount of vapor possible at the specified temperature and pressure Superheated vapor — the vapor (or gas) phase contains less than the maximum amount of vapor possible at the specified temperature and pressure Dew point temperature— temperature at which a superheated vapor becomes a saturated vapor in a gas containing a single superheated component Bubble point temperature — temperature at which the first vapor bubble forms as a liquid is heated Dew point pressure — pressure at which a drop of liquid is formed from a vapor as the pressure is increased Bubble point pressure — pressure at which a bubble of vapor is formed from a liquid as the pressure is decreased Relative saturation: sr = pi x100 * Pi (T) 2/5 ChE 210: Meeting 19 February 24, 2011 Molal saturation: sm = —- Absolute saturation: 581 = pi_Mi (P - Pi )de *mflmdifl \r‘c‘afléci 05; :ji‘m‘cmofl (filter?) 59 €0,735 (“c-1H "to butter Txy and ny Diagrams 115 110 ' 34m; :05 1 1309 m £3 mew % 12:38 “g €935 E 13.90 5% 9c: 5% was M V m 3, 85 5 g 900 g as 99999 g 353:} « £1. ’35 3913 ,., “3’0 ~ 69f} 55 i’ _-_> m » - 50g} ..... i {3 {332 0.4 0.6 958 if.) £3 05233 8.5} {3.5 093 LG was frantian benzene Male fizafliaa {53:12am PM ‘1 atm Tm 16693 {a} “firs; diagram {22) Pk}? diagram Xx 3:" F‘ i. _. 9; / _ it) i ; : A kid“ 1.); [MM \d 1‘” {'(f.”{‘-_{'.j hfibfldbtk hy- (-1177? r’i/iil\;”\ rK/‘r‘ifli’i/f‘fb. 3/5 ChE 210: Meeting 19 February 24, 2011 Problem 6.35 mv = ' 5 ;<(H,V= “f 526 “c. P ‘2 53 0 (Tim Condenser ms,E = 1kg mas = 1 kg mH,L = mm; = 0.78 kg mH,s = 0.05 kg H: hexane (3 " BORIS Rn: 161610.1’3 weave“ of hexame A 30 need m H‘L ( Deqcee; a? 3%66d’mx 13‘fo L1 @1”\\(,|'"G~Ufif> ( “1 m . ‘ (“N2 m0 ‘5th gram) 92‘ m“ bod (N2. H ) 3" I) I?) mum averqu‘wfrt’ — 0 Free spec \ {)th Con ($3th " 5:42.54) ‘ f?h\1$ \CLLJC; (5&3 0,370 ,: “S? : $1; 15"“) w/ ("H LT) mm .flmmm) Q D ck“? 4/5 ChE 210: Meeting 19 February 24, 2011 Condemer : . 7 Unknbw‘b K025 Wm VH2.» rm 7“»: Wm H’xH‘F ) ‘ a may boJ LN; H) — 0 {30C spec; " 0’1 pms Cm W“,D+‘h«_,bfil and my WNW =\) ‘ 94 (DNS law (3m? : 107: i381) .m.m\5e)(gacoyyo) ‘—~* *beccww DQWd 0? D e? ¥ Overau : 5 x ’ a ‘ . b LiflanN£mNz mv VHN 7N“ fink“) ‘01 mew boJ (N2 H) “0 Pivc spec - \ @3th Com (VH0: + \INZN 1‘) - ' ‘ ., >5 ' PMS law (YHN P 'PH (015°C) 9mm Pm’xum’) \DO9F 5/5 ...
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