Elementary - 6.55(cont'd xnp = 0.500 mol n C5 H 12(l mol xip = 0.500 mol i-C5 H12(l/mol Total pressure P =xnp pnp xip pip = 0.50(6717 0.50(7883 =

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6- 38 6.55 (cont’d) xn np = 0500 . mol - C H (l) / mol 51 2 2 0.500 mol -C H (l)/mol ip xi = ** Total pressure: = + 0.50(6717) 0.50(7883) 7300 mm Hg np np ip ip Px p x p ⋅⋅ = + = * 2 0.500(6717) 0.46 mol -C H (v)/mol 7300 np np np xp yn P == = 2 1 1 0.46 0.54 mol -C H (v)/mol ip np yy i =− = When the last drop of liquid evaporates, np = . mol -C H (v) / mol 2 2 0.500 mol -C H (v)/mol ip yi = xx yP p p PP P np ip np np ip ip + = + =+= = (( .. 120 120 6725 7960 1 7291 oo C) C) mm Hg 2 0.5*7250 mm Hg 0.54 mol n-C H (l)/mol 6717 mm Hg np x 2 1 1 0.54 0.46 mol -C H (l)/mol ip np i = b. When the first drop of liquid forms, y np = . m o l n - C H ( v ) / m o l 2 ip = . mol -C H (v) / mol 12 12 P = (1200 + 760) = 1960 mm Hg * * 6.84471 1060.793/( 231.541) 6.73457 992.019/( 231.541) o 0.500 0.500 980 980 1 () 10 10 63.1 C dp dp np ip TT np dp ip dp dp pT T −+ + += + = + = ⇒= ( ) 6.84471 1060.793 63.1 231.541 10 1758 mm Hg np p ( ) 6.73457 992.019 63.1 231.541 10 2215 mm Hg ip p 2 *o 0.5*1960 mm Hg
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This note was uploaded on 02/27/2012 for the course CHEMICAL E 312 taught by Professor Cheung during the Fall '11 term at The University of Akron.

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