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I Why does external pressure on a liquid increase vapor pressure

  1. Jan 23, 2017 #1
    I was reading an online chemistry textbook that said "when a liquid is subjected to hydrostatic pressure (for example, by an inert, non-dissolving gas that occupies the vapor space above the liquid surface), the vapor pressure of the liquid is slightly raised." (link: http://www.chem1.com/acad/webtext/solut/solut-3.html, Section: Effects of Pressure on Entropy: Osmotic Pressure)

    Why does the pressure applied to the liquid increase vapor pressure? If you increase external pressure on the liquid, wouldn't you be making it more difficult for the liquid to transition into a gas? That's why a higher temperature is required to increase the vapor pressure. However, the online source makes the case that external hydrostatic pressure increases solvent molecules' tendency to escape from the liquid phase into the vapor phase. I'm confused as to why more gases would escape when there is an external pressure on the liquid....
     
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  3. Jan 23, 2017 #2

    Bystander

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    First approximation? No effect. Zero change in the partial pressure, which is all you are considering as a first approximation.
    Think about the "activity," or the "Poynting correction."
     
  4. Jan 23, 2017 #3
    Have you had a course in Thermodynamics yet, and, if so, are you familiar with the Gibbs free energy function?
     
  5. Jan 23, 2017 #4
    Hi Bystander, okay so you're saying that rather external pressure applied to a liquid has approximately no effect on the liquid's transitioning. Ah, I'm not familiar with the Poynting correction...
     
  6. Jan 23, 2017 #5
    Hi Chestermiller, I actually have not taken a course in thermodynamics, but I am familiar with Gibbs free energy to some extent as it's been taught in general chemistry courses. I do know that when we're talking about escaping tendency we are actually referring to Gibb's free energy (the phase that is most stable at a particular temperature and pressure has the most negative free energy). I also know that Gibb's free energy includes entropy and enthalpy terms. Entropy considerations are especially important when discussing colligative properties of solutions versus pure solvents (i.e. we see boiling point elevation b/c adding solute adds new energy microstates to the liquid phase but doesn't affect the vapor phase, causing the temperature required to make equal numbers of microstates in both phases).

    That being said, since my understanding of Gibbs free energy is still elementary, I am trying to figure out why more pressure would help gases escape? I was studying liquid - gas transitions, and osmotic pressure in particular, through the lens of pressure's effect(s) on entropy, but I am just completely confused right now.
     
  7. Jan 23, 2017 #6
    Are you familiar with the equation for the differential change in molar Gibbs free energy as a result of a small change in pressure at constant temperature:
    $$dG=VdP$$where V is the molar volume? Are you aware that, for equilibrium in this kind of system, the difference between molar Gibbs free energy of the species in the liquid and the partial molar Gibbs free energy of the same species in the gas phase is equal to zero?
     
  8. Jan 23, 2017 #7
    Thank you for your response! No I am not familiar with this. Do you mind explaining this relationship conceptually instead?

    If the difference btwn molar Gibbs free energy of the species in liquid and the partial molar Gibbs free energy of the same species in gas phase is equal to zero, then does a change in pressure (at constant temp) change this fact and create a nonzero difference in Gibbs free energy between the liquid and the gas phases of the same species?
     
  9. Jan 23, 2017 #8
    No. If we add as insoluable gas species B to the gas phase, that increases the total pressure. That causes the free energy of the pure liquid A to increase, via the equation I wrote, where V is the molar volume of liquid A (which is much smaller than the molar volume of A vapor). This increase of the free energy of species A in the liquid has to be matched by a corresponding increase in the free energy of A in the gas phase, in order to maintain phase equilibrium. So its equilibrium partial pressure has to change.
     
  10. Jan 23, 2017 #9
    Sorry if this seems like a dumb question: but why does the increase in total pressure cause the free energy of the pure liquid A to increase? (I know you referenced the equation but I'm trying to understand conceptually why increase in pressure leads to increase in free energy of the liquid?)

    Secondly: so due to the increase of free energy of species A in the liquid, molecules of species A will escape into the gas phase to maintain phase equilibrium? And this phenomenon happens while temperature is held constant?
     
  11. Jan 23, 2017 #10
    For the pure liquid, $$G=H-TS$$
    $$dG=dH-SdT-TdS$$
    $$dH=TdS+VdP$$ where P is the total pressure acting on the pure liquid (caused by the combination of the condensible and non-condensible gases in the gas phase) and V is the molar volume of the liquid. So, for the pure liquid,
    $$dG=-SdT+VdP$$
    At constant temperature, this gives$$dG=VdP$$
    Since a liquid is nearly incompressible, we can write: $$\Delta G_L=V\Delta P$$
    This is the change in free energy of the pure liquid as a result of adding a non-condensible gas to the gas phase to increase the total pressure. This must be matched by the change in free energy of the condensible gas. This all leads to the Poynting correction mentioned by Bystander.
     
  12. Jan 23, 2017 #11
    Ah... I'm sorry I still am confused. While I can follow along with the math, I need to be able to rationalize the math in my head in layman's terms. So at constant temperature, the equation tells us that there will be a change in free energy of the pure liquid due to an increase in total pressure - this is difficult for me to grasp because I would think that an increase in total pressure acting on the liquid would cause liquid molecules to come closer together (resulting in stronger intermolecular forces) and therefore be less likely to enter the gas phase (in other words, require more energy to do so). Now I get that liquids are nearly incompressible, but why is the increase in total pressure raising the free energy of the liquid, conceptually speaking?
     
  13. Jan 23, 2017 #12
    Sorry. As a continuum mechanics guy, I don't feel qualified to speculate about how this all plays out at the molecular level.
     
  14. Jan 23, 2017 #13
    That's fair. Just to clarify though - in cases when a non-condensible gas is added such that it increases the total pressure exerted on the liquid, what happens when temperature is not constant? Would temperature rise?

    I'm just trying to reconcile the following two ideas:

    1) If we increase the external pressure on the liquid, a higher temperature would be required so that the vapor pressure equals the external pressure and hence the boiling point of the liquid increases. (increase in temp = increase in vapor pressure)

    2) If we increase the external pressure on the liquid while temperature is held constant, a higher total pressure on the liquid increases the vapor pressure (from molecules escaping the pure liquid into the gas phase).
     
  15. Jan 23, 2017 #14
    OK. I'll tell you what I'm going to do. I'm going to give you what you need to quantify these types of questions on your own. Then you can report back.

    Let's do it for water, with air as the non-condensible gas.

    Let:

    p*(T) = equilibrium vapor pressure of pure water vapor in equilibrium with pure liquid water at temperature T (no non-condensible air present)

    ##p_e(T,P)## = partial pressure of water vapor in equilibrium with liquid water at temperature T and total pressure P (equal to partial pressure of air plus ##p_e(T,P)##)

    ##p_a## = partial pressure of (non-condensible) air

    For equilibrium between the liquid water and the water vapor in the gas phase, we must have that

    $$V(P-p^*(T))=RT\ln(p_e/p^*(T)=V(p_a+p_e-p^*(T))\tag{1}$$where V is the molar volume of liquid water (0.018 liters/mole) and R is the universal gas constant (0.082 (l atm)/(K mole)

    Note from Eqn. 1 that, if the partial pressure of the air in the gas phase ##p_a## is equal to zero, ##p_e## is equal to the equilibrium vapor pressure of pure water p*(T)

    Why don't you try a few calculations with this at a selected temperature (say 50 C) and with various values of the non-condensible air partial pressure ##p_a## (say, up to 1 atm) just to get your feet wet? See how ##p_e## varies as a function of the air partial pressure.
     
  16. Jan 24, 2017 #15

    DrDu

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    Let me try to put what happens in this effect into words: This is an effect already present when gasses are assumed to behave ideally. Hence in the gas phase the vapour and the other gas don't interfere in any way. When some volume of the liquid vapourises, the volume available for the inert gas increases slightly. With this volume increase, also the entropy of the gas increases, as the particles have more places where to be, statistically speaking. Furthermore, this increase in entropy is also proportional to particle number, which is proportional to the pressure at constant temperature. Hence the higher the pressure the more liquid will evaporate.
     
  17. Jan 24, 2017 #16
    Hi Dr. Du. Thanks for your input. But, I'm not sure it fully addresses the question asked by the OP. My understanding is that he was asking: why, at the molecular scale, increasing the pressure on a (virtually incompressible) pure liquid increases the tendency of the liquid to want to evaporate (i.e., increases its chemical potential). We know mathematically that its chemical potential (i.e., Gibbs free energy) increases by ##V\Delta P## (even without a gas phase present, e.g. with a piston), but how does this relate to the behavior of the liquid at the molecular scale? Additional thoughts??
     
  18. Jan 24, 2017 #17

    DrDu

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    Good point. I would say that the pressure compresses the bonds whence they become weaker.
     
  19. Jan 24, 2017 #18

    PeterDonis

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    From the article you linked to--the sentence after the one you quoted:

    "The pressure acts to compress the liquid very slightly, effectively narrowing the potential energy well in which the individual molecules reside and thus increasing their tendency to escape from the liquid phase."

    Does this make sense?
     
  20. Jan 24, 2017 #19

    DrDu

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    I would rather say that the potential between two atoms making up the compound is approximately parabolic with constant width (you may also assume a van der Waals or 6-12 potential) and compression leads to a deviance of the interatomic distance from the least energy minimum. To formulate it differently, at larger distances than the equilibrium distance the atoms attract, but at lower distances they start to repell each other (Pauli repulsion).
     
  21. Jan 24, 2017 #20

    PeterDonis

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    But such a configuration shouldn't be stable. If we view the effect of increasing pressure as narrowing the potential well, then the new configuration is stable--it's the least energy configuration given the new narrower potential.
     
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