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Raoult's law and pressure changes

  1. Jul 11, 2013 #1
    When I have an ideal mixture of 2 liquds, A and B. so the partial pressure of A is PA=PA°xA while for B, it's PB=PB°xB. So if I were to halve the volume, the PA doubles. When this happens would the pressure try to decrease itself? I would think it would decrease as in a pure substance it would decrease itself. So how would the composition of the gases change? Would they remain the same such that both gaseous A and B return into solution such that their mole fraction remains the same?

    Then when the solution is heated, how would the composition change? PA° increases and so does PA. But I'm not too sure how xA changes. I'm thinking it would remain the same as both liquids A and B would turn into a vapour so their mole fractions remain the same. Is this correct?

    Thanks :)
     
  2. jcsd
  3. Jul 11, 2013 #2
    I don't understand what you mean about halving the volume. You have an ideal liquid mixture of two components in equilibrium with an ideal gas mixture of the same two components. The system is at a total pressure of PA+PB. Are you proposing to then halve the total volume, the volume of gas, or the volume of liquid.

    When you heat the solution, what is the constraint? Do you hold the total pressure constant?

    If you are going to start doing problems like this where you begin to investigate the effects of changes, you need to first establish the initial state of the system: number of moles of each species in combined liquid and gas, and fraction of system which is liquid.

    Chet
     
  4. Jul 12, 2013 #3
    Hi for the first question, I'm having the volume of the gas such that the pressure should double. But I was thinking - for a pure liquid the partial pressure remains the same despite the changes in volume. So, I assumed that the partial pressure of the components A and B would remain the same. And since PA=PA°xA and since PA° is only affected by temperature, it remains constant and xA also has to remain constant to allow PA to remain constant. But same goes for B too, so xB also remains the same. So both gaseous A and B return into the liquid state and maintains the mole fraction xA and xB to be constant?

    In the second case, I'm not too sure if the total pressure would remain the same actually. Isolating one of the gases A, PA=PA°xA and PA° increases as temperature increases. But I'm not too sure if PA should increase or remain constant when this happens. What would actually happen here?

    Thanks :)
     
  5. Jul 13, 2013 #4
    As I said previously, if you are going to start doing problems like this where you begin to investigate the effects of changes, you need to first establish a base case for the system, and solve the appropriate equations for the base case. Then you can start looking at the effects of changes in the parameters. If you are willing to start working on a base case, I can help you. Let the gas and liquid be contained in a cylinder, and let the number of moles of each species within the cylinder be fixed. Let T0 be the temperature of the system in the base case, and let xA be the mole fraction of species A in the liquid. In terms of xA, PA0(T0) and PB0(T0), what is the total pressure of the system? In terms of xA, PA0(T0) and PB0(T0), what is the mole fraction of species A (yA) in the gas phase?

    After you have reported back with these result, I will continue specifying the base case.

    Chet
     
  6. Jul 14, 2013 #5
    Ohh ok let me make a case. PA° at 25 degrees is 10Pa and PB° at 25 degrees is 15Pa and the number of moles of A in the liquid phase is 10 moles while for B it's 20 moles. So the partial pressure of A, PA=10/30x10=3.33Pa and for B PB=20/30x15=10Pa. So when I halve the volume I would think that the partial pressures of A and B should remain the same? So the total number of moles of A and B would increase but the mole fraction of A and B would remain the same.

    So we let A be moles of gaseous A going back into liquid phase and B be moles of gaseous B going back into liquid phase. So (10+A)/(30+A+B)=1/3 and (20+B)/(30+A+B)=2/3 then with these 2 equations we can solve for the number of moles going back into the liquid phase?
     
  7. Jul 14, 2013 #6
    Whoa. This isn't exactly what I had in mind. We need to slow down a little at take things in smaller bite sized chunks. I wanted to do everything symbolically, without specifying any particular numbers yet.

    As a base case, I want to specify the temperature T, the mole fraction of species A in the liquid xA, and the total number of moles of each species nA and nB in the vessel, and then I want to determine the total pressure, the number of moles of liquid phase, the number of moles of vapor phase, the mole fractions of the two species in the vapor, and the volume of the vessel.

    Starting out, let:
    xA=x
    xB=1-x
    yA=y
    yB=1-y

    From Raolt's law, the total pressure P is given by:
    [tex]P=xP_{A0}(T)+(1-x)P_{B0}(T)[/tex]
    From this, the mole fraction of species A in the vapor is given by:
    [tex]y=\frac{xP_{A0}(T)}{P}=\frac{xP_{A0}(T)}{xP_{A0}(T)+(1-x)P_{B0}(T)}[/tex]
    From this, the mole fraction of species B in the vapor is given by:
    [tex]1-y=\frac{(1-x)P_{B0}(T)}{P}=\frac{(1-x)P_{B0}(T)}{xP_{A0}(T)+(1-x)P_{B0}(T)}[/tex]
    Next, let's get the number of moles of liquid ml and the number of moles of vapor mv in the vessel:
    [tex]m_lx+m_vy=n_A[/tex]
    [tex]m_l(1-x)+m_v(1-y)=n_B[/tex]

    Solve this pair of linear algebraic equations for ml and mv in terms of x, y, nA, and nB. After you do that and report back, I will tell you how to get the total volume of material in the vessel (i.e., the total volume of the vessel). After that, we can start exploring what happens if you start making changes.

    Chet
     
  8. Jul 14, 2013 #7
    Oops sorry about that I thought you meant creating a scenario Haha.

    Ok I solved for mv which is (nBx-nA)/(x-y) and mi is (nA-(nBxy+nAy)/(x-y))/x

    Thanks for the help!
     
  9. Jul 14, 2013 #8
    That is not what I got. I got:

    [tex]m_l=\frac{n_By-n_A(1-y)}{y-x}[/tex]
    [tex]m_v=\frac{-n_Bx+n_A(1-x)}{y-x}[/tex]
    As a check, if I add these two equations together, I get:
    [tex]m_v+m_l=n_A+n_B[/tex]
    The latter, of course, is a requirement.

    I'm getting the feeling that you don't want to pursue this any further. The next step is to get the volume of liquid and the volume of vapor, which add up to the total volume of the vessel. Getting the volume of vapor is very easy since you know the pressure, the temperature, and the number of moles of vapor. To get the volume of liquid, you need to work with the specific volumes of the two pure species at the temperature and pressure of the system. If you don't want to pursue this further, I understand.

    chet
     
  10. Jul 14, 2013 #9
    Hi Sorry about the short reply. It was tough to type out everything so I took a picture of it. I'm not too sure why we got different answer though. Here is the link http://i.imgur.com/DEv5E3f.jpg could you help me check where I went wrong? I'm still very interested in this topic thought. Sorry if I came across as being uninterested.

    Thanks so much for the help
     
  11. Jul 14, 2013 #10
    I can see on the link where you made your first algebraic error (there may be more). Please go back and check your algebra. The error was involved in getting mv.

    The next thing I'd like you to do is to use the ideal gas law to determine the volume of vapor in the vessel in terms of mv, P, and T.

    chet
     
  12. Jul 17, 2013 #11
    So from here we would use PV=nRT to get our total vapour volume. So V=nRT/P=(nA+nB)8.314T/P?

    Sorry for the late reply. Had to complete some assignments.
     
  13. Jul 17, 2013 #12
    Try again. When using the ideal gas law to calculate the volume of vapor in the container, do you use the total number of moles of liquid and vapor in the container, or just the number of moles of vapor?

    Chet
     
  14. Jul 17, 2013 #13
    Oops I should just use the moles of the vapour so n is equals to mv?

    Thanks again!
     
  15. Jul 17, 2013 #14
    Yes. Now lets turn attention to calculating the volume of liquid in the container. For an ideal solution, the volume of liquid does not change when the pure constituents are mixed. So all we have to do to get the volume of the liquid phase is add up the volumes of the pure liquid constituents at the same temperature and total pressure as the solution. Assume that the pure liquid species are incompressible and, for our purposes, do not exhibit significant thermal expansion. Let ρA represent the molar density of pure species A, and let ρB represent the molar density of pure species B. Knowing the number of moles of liquid ml and the mole fractions of species A and B in the liquid (x and 1-x), you should be able to write down an equation for the volume of liquid in the container. Please write down this equation, and report back. Also, from these results, what is your equation for the total volume of the container?

    Chet
     
  16. Jul 17, 2013 #15
    Ok thanks!

    For liquid A, the number of moles is mx so we can use mx/VA=ρA so VA=mx/ρA and for liquid B, the number of moles is m(1-x) so m(1-x)/VB=ρB so again we can use VB=m(1-x)/ρB. And so the total volume would be mx/ρA+m(1-x)/ρB?

    Thanks so much Chet!
     
  17. Jul 17, 2013 #16
    Good. So the total volume of the container is

    [tex]V=m_l\left(\frac{x}{ρ_A}+\frac{(1-x)}{ρ_B}\right)+\frac{m_vRT}{P}[/tex]

    This completes the specification and analysis of the base case. Now, if you still have the energy and desire, we can start looking at the effects of changes. But, rather than looking at large changes like halving the volume or doubling the pressure, we should first begin by infinitecimal changes in some of the parameters. Why? Because if we can't do that, we certainly won't be able to handle large changes. Pick a parameter you want to vary first so we can start to analyze the situations involving change. Possibilities include:

    Temperature, while holding the total pressure constant
    Mole fraction A in the liquid
    Number of moles of A
    Volume, holding the temperature constant
    Volume, under adiabatic conditions
    Other?

    Chet
     
  18. Jul 20, 2013 #17
    I was thinking just changing the volume keeping the volume constant. and leaving the system to change on its own.

    In this case, I would think the mole fraction of the liquids remain the same and so will the partial pressures of the different gases. Would this be true?
     
  19. Jul 20, 2013 #18
    Did you mean changing the volume while keeping the temperature constant?
     
  20. Jul 20, 2013 #19
    Yup. Would what I said be true? Thanks :)
     
  21. Jul 21, 2013 #20
    Actually, no. Before jumping to any conclusions, we should let the math tell us what it has to say.

    OK. We are going to change the volume by dV, holding T, nA, and nB constant. Now, if we increase the volume under these conditions, the total pressure P will have to decrease, and if we decrease the volume under these conditions, the total pressure P will have to increase. Notice that the volume V is determined in the very last equation in our sequence of relations, and the total pressure P is determined in the very first equation in our sequence. For what we are doing here, it will be much more convenient to specify a change in pressure dP, and then solve for the changes in all the other parameters. So lets do that. From the first equation in our sequence, if we increment the pressure by dP (at constant temperature), what does the mole fraction change of species A, dx, have to be?
     
  22. Jul 21, 2013 #21
    Hi thanks again for the post. I'm not sure how to differentiate actually. Because my math module is next semester. But looking at the formula, PA=PA°xA I know that my PA° is going to be fixed as the vapour pressure isn't dependent on the volume above it.

    So I thought if the total pressure were to double, that would mean that the partial pressure of each gas doubles right? If this were to happen, that would mean each of my mole fraction would have to double. So shouldn't this be impossible?

    Thanks so much :) and sorry my mathematics skills aren't good enough to follow what you required.
     
  23. Jul 21, 2013 #22
    If the pressure doubled, and the temperature were constant, the mole fractions of A and B would have to change. We know this from the equation:
    [tex]P=xP_{A0}(T)+(1-x)P_{B0}(T)[/tex]
    Suppose the pressure increased by ΔP, and the mole fraction of A changed by Δx. Then from this equation, we would have:

    [tex]Δ P=Δ xP_{A0}(T)-Δ xP_{B0}(T)[/tex]
    Therefore, the change in mole fraction x would be:
    [tex]Δ x=\frac{Δ P}{P_{A0}(T)-P_{B0}(T)}[/tex]
     
  24. Jul 26, 2013 #23
    Hi chestermill, sorry for the late rely. Been really busy with my school work.

    I thought about this and I still don't quite understand why the total pressure should change. For example, if I had a pure liquid and it has a certain vapour pressure over it when we decrease the volume of the gas the partial pressure wouldn't double. It would remain the same. So I thought I could apply that here too where I have 2 liquids together. Shouldn't their combined vapour pressure also stay the same despite their volume changes? If so wouldn't their mole fractions in the liquid remain the same?

    But I'm not too sure about that because, when I have an ideal solutions with a liquid and a solute. So if we were to halve the volume of the gas at a constant temperature, if the partial pressure remains constant, that would mean that the mole fraction of the liquid would have to stay constant. But that wouldn't be possible as the only way to reduce the vapour pressure is if the gas goes back into the liquid state but since only the liquid has a vapour over it, vapour would have to condense back to a liquid. So the mole fraction would have to increase.

    So I'm guessing that my assumption that the total vapour pressure of the 2 liquids would remain the same despite the volume changes of the vapour? But I don't really understand why this is so.. The pure liquid would cause its own vapour pressure to stay the same. But why when we have 2 liquids and mix them together, the total vapour pressure would change according to the ideal gas law PV=nRT?

    Thanks so much :)
     
  25. Jul 26, 2013 #24
    This only applies to a pure liquid.

    This is basically a correct assessment.
    I'm not sure I fully understand what you are asking here. It is true that the total gas pressure will satisfy the ideal gas law (if you are assuming ideal gas behavior), but the number of moles of each species in the gas can (and does) change, and, more importantly, the mole fractions of the two species in the gas can (and does) change.

    The only way that the system can behave like a pure liquid is if the vapor pressures of the two species are identical at all temperatures. We should be happy that this is the case. Differences in the vapor pressures of pure liquids is what enables us to purify solutions of liquids by the process of distillation.
     
  26. Jul 27, 2013 #25
    Hi chestermiller thanks for reply :)

    So only a pure liquid would have a constant vapour pressure. But why wouldn't the two liquids case have their vapour pressures to remain constant like in the pure liquid case?

    Thanks :)
     
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