Why vapour pressure does not depend on surface area?

If we prepare a solution i.e., we dissolve small amount of solute in excess of solvent. We see that on the surface of the liquid of solution, number of solvent molecules became less bacause some molecules of solute are also available on the surface of the liquid. Hence vapour pressure will be decreased i.e., vapour pressure of the solvent in the solution will be less than that of the pure solvent.

Pure solvent Solution
Vapour pressure= P₀ Vapour pressure= P_s

If P₀ be the vapour pressure of solvent in pure state and P_s be the vapour pressure of the solvent in impure state i.e., vapour pressure of the solution, then it will always exist that,
P₀ > P_s
At constant temperature the vapour pressure of pure liquid will always be constant. When the liquid is in equilibrium with vapours. If Kp is the equilibrium constant for this equilibrium
Kp = P₀
Kp only depends upon temperature. Therefore vapour pressure of the liquid is also constant and depends only on temperature but it does not depend upon the surface area or nature of the vessel.

Hello student
Vapour pressure does not depend upon the surface area bacause vapours produced from the surface increases on increasing the surface area. Therefore the ratio of numbers of vapours produced to surface area remain the same and vapour pressure being independent is not affected.
We also know that vapour pressure is an intensive property and is independent of size and shape of vessel which is another resaon.