(a) According to Raoult\'s law P A = P o A x X A P A = partial pressure of compo
ID: 700178 • Letter: #
Question
(a) According to Raoult's law
PA = PoA x XA
PA = partial pressure of component A
PoA = vapour pressure of pure A
XA = mole fraction A in liquid
say benzene mole fraction in liquid = XB
PoB = 155.7 kPa
PB = 155.7 kPa x XB -----------(1)
similarly for toluene
PT = 63.3 kPa x XT -----------------(2)
P = total pressure = PT + PB = 101.3 kPa given
add 1 and 2
PB + PT =101.3 kPa = 63.3 kPa x XT + 155.7 kPa x XB -----(3)
also we know that sum of all mole fractions = 1
XB + XT = 1
XT = 1 - XB
replace in 3
101.3 kPa = 63.3 kPa x (1 - XB) + 155.7 kPa x XB
XB = 0.411
Ans (a) = 0.411
(b)
In gas phase PB = P x YB
PB = partial pressure of benzene
P = total pressure
YB = mole fraction in vapour phase
from equation (1) PB = 155.7 kPa x 0.411 = 64 kPa
64 kPa = 101.3 kPa x YB
YB = 0.63
(c) we had total 50 moles benzene and 50 moles toluene
assuming that it behaves as an ideal gas
XB = nB/(nB + nT)
nB/XB = (nB + nT) ---(a)
nT/XT = (nB + nT) ----(b)
nB/nT = XB/XT = 0.411 / (1 - 0.411) = 0.698
nB = 0.698nT
similarly n'B/n'T = YB/YT = 0.63 / (1-0.63) = 1.703
n'B = 1.703n'T
also nT + n'T = 50 mol
nB + n'B = 50 mol
1.703n'T + 0.698nT = 50 mol
replace n'T = 50 - nT
1.703(50 - nT) + 0.698nT = 50 mol
85.13 - 1.703 nT + 0.698nT = 50 mol
35.13 = 1.005nT
nT = 35 mol
nT' = 15 mol
n'B/n'T = 1.703
n'B = 25.545 mol
nB = 50 - 25.545 mol = 24.455 mol
Ans = 24.455 mol
Explanation / Answer
The vapor pressures of benzene and toluene at 95C are, respectively, 155.7 kPa and 63.3 kPa. A gaseous mixture consisting of 50 moles of benzene and 50 moles of toluene was cooled to 95C. As a result of the cooling, some of the benzene and toluene condensed. If the total pressure above the condensate was 101.3 kPa, calculate:
(a) the mole fraction of benzene in the condensate (liquid).
(b) the mole fraction of benzene in the gas phase after cooling. (c) the number of moles of benzene in the condensate.
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