1chcek this..
benzene and toluene form nearly ideal solutions.at 300K p°toluene=32.06mm and p°benzene=103.01mm.
a liquid mixture is composed of 3 mole of toluene and 2 mole of benzene.if the presurre over mixture is reduced,at what pressure does last trace of liquid disappear.
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1this is not the way... there is a very nice concept in this question...
which i am able to 'feel' but cudn workout..
see, let the vap pr of soln initially was Po.
now, if we reduce the pressure by ΔP to P' ... then to maintain the initial vap pr some A and some B vaporizes to become vapours such dat the mole fractions of A and B in the vapour is retained.
now, mole fraction of A and B in vapour is obviously ≠their mole fractions in the solution.
1Answer is 44.4mm(approx.)
Here's the solution:
For last drop the mole fraction is 3/5 and 2/5 approximately...but these are the mole fraction in Vapour Phase and not in liquid phase.
A is toluene...and B is benzene
according to formula:
\frac{1}{Ps}=\frac{Ya}{Pa}+\frac{Yb}{Pb}Put values:
Ya = 3/5
Pa = 32.06 mm
Yb = 2/5
Pb = 103.01 mm
Value of Ps is 44.4(approx.)
Thank You
I Hope This will Help You
1YES! I GOT IT NOW!
the concept is dat only which i had explained above...
we need to fing that pressure for which the mole fractions of A and B in vapour becomes 3/5 and 2/5 ...
why--->> that is the mole fraction in liq state... so, if whole liq becomes vapour, then the mole fractions of A and B in the vapour becomes same as that in liq initially..
now, if P is the vapour pressure at that instant ;
mole fraction of A =xA
...................... B =xB
then, 1/P = xA/P0A + xB/P0B ... [the same formula used by mrnobody1]
=> 1/P = 3/5/32.06 + 2/5 /103.01
=> P= 44.183 mm ≈ 44.2 mm
1but mrnobody why to substitue mole fractions 2/5 and 3/5.according to what concept u substituted it.please explain.
1derive the expression for total pressuer using mole fraction in the vapour phase(it can be done using daltons law... everybody please try it)
(1/p)=Xa/pa0 + Xb/pb0
substituting for mole fractions 2/5 and 3/5 and p0 values u get
P=44.26mm
1everybody nice going....well try.i hope soon we will get the concept as well as correct answer behind the question.
1@kalyan, dhakkan lagake pani ubalna... saare pani becomes air or wat ?! .. waise... dekhke ati hun :P
kalyan i saw it jus now :P :P :P
water doesnt disappear[3]
1@kalyan, to turn the whole liquid out, the vap pressure need not be equal to atm pressure [boiling point]
coz... the whole story is going on in some closed container i think .. orelse .. all these wont occur...
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11YA=p1Xa/p1Xa + p2Xb
and the expression 4 YB is similar
so
after solving we get
YA= 0.308
and
YB= 0.692
1Partial pressure of solution is
P=paxa+pbxb
wherein pais partial pressure of toulene
& xa is mole fraction of toulene.
pb is partial pressure of benzene
& xb is mole fraction of benzene
This gives vapour pressure as 60.44 mm
It completely turns as liquid when vapour pressure = atmospheric pressure
But in order to find this out we either need the volume of container or the information as to pressure is measured in mm of which substance (eg water, mercury etc.)[2][2]
11YA= mole fraction in the vapour phase
and p1= the po of toluene
and similarly the other 2