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Topic: Mixtures of volatile liquids.  (Read 5892 times)

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Offline Twickel

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Mixtures of volatile liquids.
« on: September 02, 2012, 08:46:59 AM »
 Hi
I have a few questions that I am uncertain about.
At 353K the vapour pressures of benzene and bromobenzene are 757 and 66 mmHg respectively. Assume the mixture obeys Raoults Law, for a mixture with xbenzene=0.3, calculate the pressure at which this mixture will begin  boil and then calculate the mole fraction of benzene in the vapour phase at this pressu

How do I go about calculating those? Which equations do I use?

Thanks

Offline Borek

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Re: Mixtures of volatile liquids.
« Reply #1 on: September 02, 2012, 10:00:30 AM »
Start with a boiling point definition.
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Offline curiouscat

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Re: Mixtures of volatile liquids.
« Reply #2 on: September 02, 2012, 10:19:55 AM »
yiP=xiPisat

y1P=x1P1sat
y2P=x2P2sat

y1+y2=1
x1+x2=1

x1=0.3

7 variables
3 known x1 P1sat P2sat
4 unknown
4 eqns

Solve for P,y1,y2

Offline Twickel

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Re: Mixtures of volatile liquids.
« Reply #3 on: September 02, 2012, 10:27:27 AM »
Start with a boiling point definition.

I think boiling point is when the vapour pressure of the liquid is greater then that of the surroundings. All I have is this equation xB(gas) = PB/P = xB(liq)PB*/P

From a pressure vs mole fraction diagram, do I find the pressure at which the we hit the bubble point?

Offline Twickel

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Re: Mixtures of volatile liquids.
« Reply #4 on: September 02, 2012, 10:34:44 AM »
yiP=xiPisat

y1P=x1P1sat
y2P=x2P2sat

y1+y2=1
x1+x2=1

x1=0.3

7 variables
3 known x1 P1sat P2sat
4 unknown
4 eqns

Solve for P,y1,y2

I do not know any of those equations, what do they y's mean, is this the only possible way of solving this problem ( the pressure at which we boil)?

Offline Stepan

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Re: Mixtures of volatile liquids.
« Reply #5 on: September 02, 2012, 11:00:32 AM »
yiP=xiPisat

y1P=x1P1sat
y2P=x2P2sat

y1+y2=1
x1+x2=1

x1=0.3

7 variables
3 known x1 P1sat P2sat
4 unknown
4 eqns

Solve for P,y1,y2

I do not know any of those equations, what do they y's mean, is this the only possible way of solving this problem ( the pressure at which we boil)?

Raoults Law  :'(

Offline Twickel

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Re: Mixtures of volatile liquids.
« Reply #6 on: September 02, 2012, 11:17:11 AM »
Ok, What I did was P benzene= 0.3 x 757= 227.1, P bromobenzene = 0.7x66 = 46.2

Total pressure is 273.3

I think thats the answer, but I don't know what this means, why is the pressure at which this boils the total pressure of the vapours?
« Last Edit: September 02, 2012, 11:53:40 AM by Twickel »

Offline Borek

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Re: Mixtures of volatile liquids.
« Reply #7 on: September 02, 2012, 12:04:05 PM »
I think boiling point is when the vapour pressure of the liquid is greater then that of the surroundings.

Equal, not greater.

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Offline Twickel

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Re: Mixtures of volatile liquids.
« Reply #8 on: September 02, 2012, 12:12:06 PM »
thank you, I still do not understand why the total pressure is when the system begins to boil.

Also is calculating the mole fraction of benzene in gas just equal to 0.3x757/273. i.e Pbenzne/total presure

Offline Borek

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Re: Mixtures of volatile liquids.
« Reply #9 on: September 02, 2012, 02:08:20 PM »
thank you, I still do not understand why the total pressure is when the system begins to boil.

You know the pressure of the vapor over liquid. You lower the external pressure till you come to a moment when the vapor pressure of the liquid equals external pressure - and that's the moment when the liquid starts to boil. Consult the definition.

Quote
Also is calculating the mole fraction of benzene in gas just equal to 0.3x757/273. i.e Pbenzne/total presure

Yes. Think about Avogadro's principle.
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Offline Twickel

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Re: Mixtures of volatile liquids.
« Reply #10 on: September 05, 2012, 06:28:44 AM »
Thank you, I now understand.

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