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In a textbook I am referring to, there is a section on the effect of external pressure affecting the vapour pressure of a liquid. enter image description here Source: A Textbook Of Physical Chemistry Volume 3 - KL Kapoor - Applications of Thermodynamics (Pg 26, 27).

(There is a small typing error in Eq 1.3.1, it is meant to be $\left(\frac{\partial \mu_{B(l)}}{\partial p} \right)_T = V_{m, B(l)}$)

They have provided a qualitative analysis as to why the vapour pressure changes with external pressure, however according to the coexistence curve, there should be a fixed vapour pressure at a fixed temperature and it should not vary with external pressure.

Also, at equilibrium, shouldn't the vapour pressure of the gas be equal to the external pressure?

What does the author imply here? From what I have learnt about vapour pressure, it should be true that vapour pressure depends only and only on temperature.

Clarification

I read about the Poynting effect @Chet Miller suggested. With reference to this answer, it seems like vapour pressure and saturated vapour pressure are different terms.

However I could not find the difference between the two. I think as the name implies, saturated vapour pressure is the one where pure fluid is only present in the system, however vapour pressure is a more general term, which is used (in this case) when there are other non-condensable gases in the system.

Further, would it make a difference if instead of non-condensable gases to create external pressure, we used, say, a piston. What would be the distinction between saturated vapour pressure and vapour pressure in that case?

Any help is greatly appreciated!

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  • $\begingroup$ Can you provide correct reference of the book? $\endgroup$ May 19, 2023 at 19:03
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    $\begingroup$ @MathewMahindaratne books.google.co.in/books/about/… The photo I've given is from this book, pages 26, 27 cover section 1.7 I talked about $\endgroup$
    – omega
    May 19, 2023 at 19:37
  • $\begingroup$ Note that this effect is used for medicinal compressed mixture N2O + O2 1:1, where pure N2O at pressure equal to its partial pressure in the mixture would be liquid, but the mixture is gaseous. The effect is significant due closiness of critical T near 35 Deg C. $\endgroup$
    – Poutnik
    May 20, 2023 at 5:14
  • $\begingroup$ Vapor pressure is sometimes used confusingly in 2 different meanings: 1/ Vapor saturated pressure (total or partial) 2/ Vapor partial pressure (saturated or not). The usual case is the first meaning, but sometimes is better to be explicit. $\endgroup$
    – Poutnik
    May 20, 2023 at 18:45
  • $\begingroup$ when there are other incompressible gases in the system - Being compressible is nature of all gasses. $\endgroup$
    – Poutnik
    May 20, 2023 at 18:49

2 Answers 2

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They have provided a qualitative analysis as to why the vapour pressure changes with external pressure, however according to the coexistence curve, there should be a fixed vapour pressure at a fixed temperature and it should not vary with external pressure.

This is correct, if you vary the pressure for a pure fluid in the coexistence curve, you end up at the liquid or gas phase.

Also, at equilibrium, shouldn't the vapour pressure of the gas be equal to the external pressure?

This is somewhat ok. Maybe a better formulation is that the gas and liquid phases are in mechanical equilibrium, and we call this pressure the vapor pressure or saturation pressure. However, the author likes the concept of "vapor pressure of a liquid" that may bring confusion.

What does the author imply here? From what I have learnt about vapour pressure, it should be true that vapour pressure depends only and only on temperature.

All you have said is fine, but the author is showing another thing, which I will try to explain. To help us we have the following figure:

enter image description here

Note that both phases are not in mechanical equilibrium. If you have seen the mathematical treatment of osmotic pressure, it is another case where the pressure of both phases are not equal.

Considering equal chemical potentials for a constant temperature in both phases yields \begin{align} \require{cancel} \mu_\alpha(p,T) &= \mu_\beta(P,T) \\ V_\alpha(p,T) dp - \cancel{S_\alpha(p,T) dT} &= V_\beta(P,T) dP - \cancel{S_\beta(P,T) dT} \\ V_\alpha(p,T) dp &= V_\beta(P,T) dP \\ \left(\frac{dp}{dP}\right)_T &= \frac{V_\beta(P,T)}{V_\alpha(p,T)} \tag{1} \end{align}

Eq. (1) establishes the restriction in order to not have exchange of matter between phases. Both molar volumes are evaluated at different pressures. Moreover, for an evolution in both phases:

  1. Since $\alpha$ is saturated liquid, we may consider its molar volume constant without considerable error.
  2. If we approximate the behaviour of phase $\beta$ to that of an ideal gas. Eq. (1) takes the form \begin{align} \left(\frac{dp}{dP}\right)_T &= \left[\frac{RT}{V_\alpha(p,T)}\right] \frac{1}{P} \\ \int_{p_1^\pu{sat}}^{p_2^\pu{sat}} \; dp &= \left[\frac{RT}{V_\alpha(p,T)}\right] \int_{P_1}^{P_2} \frac{dP}{P} \\ p_2^\pu{sat} &= p_1^\pu{sat} + \left[\frac{RT}{V_\alpha(p,T)}\right] \ln\left(\frac{P_2}{P_1}\right) \tag{2} \end{align}

If the pressure of phase $\beta$ increases, then the pressure of phase $\alpha$ must increase, or vice versa. Eq. (2) may be "useful" in order to obtain the final saturation pressure $p_2^\pu{sat}$ for an initial saturation pressure $p_1^\pu{sat}$, when the phase $\beta$ is compressed from $P_1$ to $P_2$.

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  • $\begingroup$ Thanks for the reply! A few questions: When we are considering phase $\beta$ as an ideal gas, shouldn't we write $V_{\beta} (P, T) = \frac{RT}{P}$? The implications of the final equation also seem to be confusing... Does it mean on changing the external pressure, we can change the saturation pressure? (Without change in temperature) $\endgroup$
    – omega
    May 20, 2023 at 3:57
  • $\begingroup$ Shouldn't it be $V_\alpha(p,T) dp - \cancel{S_\alpha(p,T) dT} = V_\beta(P,T) dP - \cancel{S_\beta(P,T) dT}$? $\endgroup$
    – ananta
    May 20, 2023 at 5:52
  • $\begingroup$ @ananta Yes, wow. Thanks. $\endgroup$ May 20, 2023 at 9:26
  • $\begingroup$ @omega Yes, I've made the changes in the final equation so it is more clear. Yes, you cannot leave unchaged both pressures or you would have transfer of matter from one to another. $\endgroup$ May 20, 2023 at 10:00
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They are talking about a pure liquid and and a gas composed of two or more chemical components, one of which the same vapor as the pure liquid and the rest of which are non-condensible. So the total pressure is > the equilibrium vapor pressure of the pure liquid. The corruption they are obtaining is the so-called Poynting correction.

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