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Solution_F12_EMS_160_HW6

Solution_F12_EMS_160_HW6 - You are responsible for the...

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Unformatted text preview: You are responsible for the purchase of oxygen a cylindrical vessel of diameter 0.2 meters and hei pressure of 200 atm at 300K in Waals gas? The prefer that the gas behaved ideally or as a van der meters. Would you “1 and b = 0.0318 l/mole. .atmmole van der Waals constants for oxygen are a = 1.3612 77': Va/wnt 0/ If? fir} A? 77/14 5' fi/fl’xfi/i 2 = d. #23? ’73 I I . r o’z-Ml WI, // #2 jar x/xa’m, fl-r 17M cab/aw “’1 =' fl]: 2’“ ’2'” =- 5// M/zx ‘7 Minx/7: 5W 1' f /1-/ f” /.’r 4 Va» #6!" WfiA/ffg/ 3 (f 4- fi”)[V/- m5) :' fiflf ,2 n 1‘ #30” {2.36; ’ WHkfij/l/(f 4e :-" )(iwy -- may”) , «Mad/2.40m if! ma/r/t If /'.r #441 /a// Wfihfl}? Jb f7” 7%: by? /[ #70 3M bCMV/b/ 1.? Mad. ' ‘fg 8.5 For sulfur dioxide, Tcr = 430.7K and PCI = 77.8 atm. Calculate £1. 7 :32... grL,,:14_,-_r_fi 273)? 27} ’94, K ,5 e.- 4m7~miui : “517%,?” 774%! 1 I 2 4 :- 2751/9. : 27:54:77; 72/ .-.- {.77 4.415» rip/0" the critical van der Waals constants for the gas the critical volume of van der Waals $02, the pressure exerted by 1 mole of SO; occupying a volume of 500 cm3 at 500K. Compare this with the pressure which would be exerted by an ideal gas occupying the same molar volume at the same temperature. ‘ V5,. 2 3} .-: MW 47mm Wm. mme - 15 -—.€.. ,—. HIM/Mm _ £33 V"?! V" fif—Aiio/7 aiz : Maw/pm) ' —' é: : W122? L” .- rem/9» 50 9.10 The activity coefficient of Zn in liquid Zn-Cd alloys at 435°C can be represented as In m '3 0.875 Xéd - 0.30 Xéd Derive the corresponding expression for the dependence of In 75d on composition and calculate the activity of cadmium in the alloy of X Cd = 0.5 at 435°C. “a”, a; :- Mn- x2; - an”; M fag, :- /r7:’oXa¢Xa "lié/‘I’W- a ~%34 fa, ,. . /,75'an,,ng v“ bexa/J’W .— Hag, =- /,75X},¢Xg,, - 3'7Xbfl'xb)¢xaa ‘ p'iixa'rdyh + nix;an % 72/ =- diizix; + 4.5 x; i Al’ 75/..- 0.5 , X24: A!!! “’45 44¢ 5 55-77 9.11 The molar excess Gibbs free energy of formation of solid solutions in the system Au-Ni can be represented by T st = XNiXAu(24140XAu + 38280XN1 - 14230XAuXNi)[1‘ 2660] J Calculate the activities of Au and Ni in the alloy of X M = 0.5 at 1100 K. ‘ 14' 2:4” g); 6”: xfl-z)[H/éfiz + away,” _, ,fiuthxflZy _;2:’: : [333235924 - lilsmz-a- 42/Mx’- #52?fo J“. 57/ -—- (/2 ' a = [312/9 -— lfliddx + M44.” 2:"— 142/2121 + 42%? 2: ’7 ’«f1/ :61» fl” )2» ’5'!) 4' J/OIJ' = fi3/fixf/lfiAJ/2K '. 1:4“ : A37 M 6m =- Mx/ar .— p.17; 1 - 3*" = ' m W 6 (“I A?“ ’ ‘3’) '- 4’5/ :l3/fir/wfifiifi. +45% x37] . 5w;- /,7 a»; 51%;: /,‘7x.¢.;.=- elf ...
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