[curiouscat]

I am trying to verify my understanding of something:

Suppose I've an equimolar mix of two liquids in an isothermal, const. volume container whose Vap. Pressures at that T are P1 and P2.

I  think this ought to hold, but want to verify my understanding (been a long time since I took that class!):

(1) Pressure exerted if they are miscible P = 0.5 P1 + 0.5 P2

(2)  Pressure exerted if they are immiscible P = P1 + P2

Am I making a blunder?

[Corribus]

If we're assuming an ideal mixture, then (1) is basically Raoult's law.  However for (2)... if they are immiscible liquids, one is going to be sitting on top of the other because the densities will be different.  So how do you have a vapor pressure for the more dense liquid when it essentially has a sealing layer?  Unless you mix it up...

Ah, here you go:

http://www.chemguide.co.uk/physical/phaseeqia/immiscible.html

Helps?

[curiouscat]

If we're assuming an ideal mixture, then (1) is basically Raoult's law. 

Yes, implicitly I was assuming that. Sorry, should have said: Assuming fugacity coefficients and activity coefficients are unity.

However for (2)... if they are immiscible liquids, one is going to be sitting on top of the other because the densities will be different.  So how do you have a vapor pressure for the more dense liquid when it essentially has a sealing layer?  Unless you mix it up...

Ah, yes. Another missing assumption. Stirred it is. (modelling a stirred biphasic reactor essentially)

Helps?

Yep. Needed a confirming voice.

[curiouscat]

As a follow up to that: Very often, the two immiscible phases are themselves having both components and not pure liquids. In that case, would P exerted be the sum of two pressures as calculated in my (1) applied to each phase? (Again assuming ideal mixtures and agitation)

e.g. Say you took an equimolar mix of Dodecane (vap. pr. P2) and Ethanol (vap pr. P1) at 0°C. Per phase diagram below, it splits into:

(Phase A) 30% EtOH 70% Dodecane
(Phase B) 85% EtOH  15% Dodecane

Would total pressure then be ( 0.3 P1 + 0.7 P2 ) + (0.85 P1 + 0.15 P2)


Never saw this worked out like this before so am a bit suspicious of my logic.

Especially, can mixture ideality and immiscibility be invoked simultaneously (need it to derive my expression)? Or is that simplifying assumption too bizarre.

[Corribus]

Your guess would be my guess.  However in such a case I find it difficult to imagine either phase would be an ideal mixture.  If the two liquids are mostly immiscible, I would think their intermolecular interactions would be on the strong side.

[curiouscat]

Your guess would be my guess.  However in such a case I find it difficult to imagine either phase would be an ideal mixture.  If the two liquids are mostly immiscible, I would think their intermolecular interactions would be on the strong side.

Yes. Agreed.

Do you have any intuition what value these γ might be? Roughly.

It's an interesting problem.

P = ( 0.3 γ1_low  P1 + 0.7 γ2_high  P2 ) + (0.85 γ1_high P1 + 0.15 γ2_low  P2)

It is especially interesting, if these γ values are mostly close to or higher than one. I see an interesting possibility. Which is why the values are interesting.

[Corribus]

No idea.  But I also think that for two immiscible solvents A and B, the amoung of B found in A will be far less than 30%.  I could be wrong, of course.

[curiouscat]

No idea.  But I also think that for two immiscible solvents A and B, the amoung of B found in A will be far less than 30%.  I could be wrong, of course.

Don't know generally what's typical (you might be right about 30% being too high), but for this specific example isn't it 30% correct based on this figure? That's why I used 30%.

I hope I am reading that figure correctly. To me it looks like at 0°C the dodecane rich phase has about 30% ethanol in it.

[Corribus]

I forgot about the figure.  Yes, I agree with you.