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Topic: Distillation lab calculations  (Read 9219 times)

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Offline mandy

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Distillation lab calculations
« on: January 27, 2007, 09:30:41 PM »
1)  Compound A has a vapour pressure of 400 mm Hg at 95 degree celcius whereas compound B has a vapour pressure of 250 mm Hg at the same temperature.  If A and B are miscible, what is the vapour pressure of a 3:1 mixture of A and B?

I'm thinking that this question has to do with Raoult's Law. 
P(total) = P(A) x N(A) + P(B) x N(B)
            = (400 mm Hg)(0.75) + (250 mm Hg)(0.25)
            = 300 + 62.5
            = 362.5 mm Hg
Have I done this right?  I'm not sure about the mole fraction parts, or if the temperature given in the question is important somehow.

2)  The boiling point of benzene is 80 degree celcius.  What is the vapour pressure of benzene at this temperature? 

I have no clue how to do this question.  Any hints to get me started or helpful websites I could use?

Any help is appreciated.

Offline lavoisier

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Re: Distillation lab calculations
« Reply #1 on: January 28, 2007, 11:03:57 AM »
This looks more like physical chemistry to me, but I will be bold and try to answer.

1) if you assume that A and B form an ideal mixture, and that the 3:1 ratio is in moles, then I think your answer is correct. The temperature is always the same, therefore you don't need to consider it.

2) I think this shows what's wrong about how physical chemistry is taught: they make you study so many formulae and theorems with no apparent relation with reality, that at the end of the day you miss the sense of what you're talking about.
What is the boiling point of a substance? And why do substances boil at different temperatures under different pressures? If you can't answer these two questions, then your teacher is not very good, because he makes you use words (boiling point, temperature, vapour pressure) that mean nothing to you, let alone the formulae to calculate them.

Offline mandy

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Re: Distillation lab calculations
« Reply #2 on: January 28, 2007, 12:43:15 PM »
  I know that the boiling point of a substance is the temperature at which the vapour pressure of the liquid equals the external pressure exerted on the liquid.

  But is there a formula to find the pressure if i am given the boiling point?  If there is, I haven't been taught it.

Offline lavoisier

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Re: Distillation lab calculations
« Reply #3 on: January 30, 2007, 01:56:48 PM »
At this point it's only a matter of logic.

If at the boiling point a substance has a vapour pressure equal to the pressure exerted on the liquid, then its vapour pressure at that temperature is the pressure exerted on the liquid!!!!!
I couldn't possibly say it clearer than that (I think this is called a tautology).

And that's why serious books give boiling points with the pressure (like bp760 benzene = 80.C).

Offline doc1965

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Re: Distillation lab calculations
« Reply #4 on: February 01, 2007, 03:16:37 PM »
I believe that the boiling point and vapor pressure are related in that what the boiling point is of a substance is when the vapor begins to form.  I agree with who ever said we are not being taught that well cuz there is nothing in our lab or class that helps us.  Mandy, are you taking organic Chem at Ottawa U? I figured it out but it took me a long time as well as reading over the lab manual about 300 times.  Good luck

Offline lavoisier

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Re: Distillation lab calculations
« Reply #5 on: February 01, 2007, 03:49:31 PM »
doc1965,
actually any liquid can reach equilibrium with its vapour at any temperature, if you are free to chose the pressure you want. You don't need to get to the boiling point to have vapour.
There is the famous Clausius-Clapeyron equation, which correlates the vapour pressure p with the temperature T by means of two parameters A and B:

ln p = A - B/T

A and B are often measured experimentally (or even better, by the Antoine equation, which has an additional parameter C), but for ideal systems they are actually thermodynamic functions of the substance being considered.

What happens at the boiling point is that the vapour pressure of the liquid is equal to the external pressure exerted, as it was said before.
As a consequence of the phase rule, when you have the two pure phases (liquid and vapour) and the external pressure is kept constant, the temperature can't change, so if you supply heat you get vaporisation, if you substract heat you get condensation, but the temperature remains the same (= the boiling point).

Offline Borek

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Re: Distillation lab calculations
« Reply #6 on: February 01, 2007, 04:01:14 PM »
actually any liquid can reach equilibrium with its vapour at any temperature, if you are free to chose the pressure you want.

Almost. You must be sure there is enough substance ;)
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