Many times study of fact that with pressure of a substance only depends on the temperature and nature of the substance recently I study about dew point and bubble point.

Then I came up with this question to which I didn't find any on Google :-

Suppose we have a system as shown in figure: enter image description here In the above system the piston is touching the liquid surface that is there is no vapour currently.

Now we slowly move the piston up, keeping the temperature same and the vapour starts forming in the container. enter image description here

Now the question is how will this pressure applied by the vapours of the liquid so form (i.e. the vapour pressure at equilibrium) will vary with the change in external pressure at the same temperature ?

Will the vapour pressure be equal to the saturated vapour pressure of that substance at that temperature ? or if it changes will it not contradict the statement used in almost all articles over the net teaching vapour pressure , stating that vapour pressure of a substance only two points on temperature and nature of that substance?


As the approximation, the saturated vapour pressure does not depend on the total pressure, only on the substance and temperature. (*)

If there is nothing but the vapor in the gaseous phase under the piston then the amount of vapor and its pressure accomodate to external pressure. (**)

  • If the external pressure is kept greater than saturated vapor pressure for given temperature, all vapour will eventually condense and released heat will dissipate, assuming constant temperature is maintained.
  • If the external pressure is kept lower than saturated vapor pressure, all liquid will eventually evaporate and needed evaporation heat will be extracted from surrounding.
  • If the external pressure is by chance equal to the saturated vapor pressure, liquid and vapor will maintain the equilibrium with the same vapour pressure as the external one.

The piston stays in rest if and only if the net force on it is zero, therefore if actual vapour pressure ( saturated in equilibrium ) is equal to the total external pressure caused by external, non-vapor based forces acting on the piston. Otherwise it starts moving toward the lower pressure.

There is assumed the lifting of the piston is slow enough for liquid to have time to establish saturated vapour pressure at any time. Fast piston lifting can end with higher volume and less than saturated vapor pressure, which may not be possible to be topped again due lack of remaining liquid.

(*) - Finer analysis tells us that for gas+vapor mixtures, liquid saturated vapor pressure increases little with the total pressure. This is proportional to liquid molar volume and such an increase is usually negligible. Unless temperature and pressure are near the liquid critical point where it can make the difference.

(**) - Atmospheric pressure is just one of possible sources or components of external pressure on the piston. Atmospheric pressure can be

  • the only part of external pressure, like in idealized scenario with a free moving, massless, frictionless piston
  • missing, e.g. if the system is placed in vacuum
  • a part of total external pressure, like if the weight caused by the piston mass cannot be neglected and/or if explicit mechanical external force is applied.
  • $\begingroup$ What is the relation between vapour pressure and the external pressure ? Are they equal ? Can we apply Boyle's law here ? $\endgroup$
    – user112102
    Jul 17 '21 at 0:20
  • $\begingroup$ Does the piston move if they are not equal? Boyle's law is for ideal gases, and as approximation for real "permanent gases. It is not applicable for vapor, if increased pressure above saturated vapor pressure causes condensation. Even before condensation starts, there is already big deviation from this law. $\endgroup$
    – Poutnik
    Jul 17 '21 at 9:35
  • $\begingroup$ What if the question states to consider vapor as ideal gas will Boyle's law be valid then ? I don't think that it is necessary for piston to move when the pressure is unequal and the reason I'm saying that is because of a particular case in which we move the piston all the way up until all the liquid become vapour at that point the sum of external pressure and vapour pressure should be equal to atmospheric pressure ( maybe ) ? $\endgroup$
    – user112102
    Jul 18 '21 at 4:21
  • $\begingroup$ Such assumption does not make sense, ideal gas never condenses, what is the direct opposite to being a vapor. $\endgroup$
    – Poutnik
    Jul 18 '21 at 14:08
  • $\begingroup$ No, external pressure being sum of atmospheric pressure and eventually additional pressure would be equal to saturated vapour pressure, assuming lifting the piston is slow enough. (Fast moving would end with higher volume and less than saturated vapor pressure.) $\endgroup$
    – Poutnik
    Jul 18 '21 at 14:20

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