Why Is Aqueous Tension Treated Separately in Equilibrium Calculations?

[phoenixXL]

Question
In a container of capacity 1-Litre, air and some liquid water is present in equilibrium at total pressure of 200 mm of Hg. This container is connected to another 1-Litre evacuated container. Find total pressure inside the container when equilibrium is again stabilized. Aqueous Tension at this temp. is 96 mm of Hg.

Solution( Given in the book )
Total Pressure = 200mm
Pressure of Gas = 200 - Aq. Tension
==> P₁ = 200-96 = 104mm Hg

Now the volume gets doubled ( Temp and moles of the gas is conserved, Hence Boyle's law holds good over here )
P₁V₁ = P₂V₂
==> P₂ = 52mm Hg

After equilibrium is established
P_total = 52 + Aq.Tension = 148mm Hg

Problem
In the solution we separated the pressure of the gas and the water vapour, but WHY?
Won't the vapour of water(Aq. Tension) also follow the ideal gas equation, so that we can put the total pressure of the container(200mm Hg) in the equation.

Aqueous Tension at this temp. is 96 mm of Hg.

Moreover, the above statement implies that the aqueous tension depends merely on temp. and not on the volume of the container. But being a gas (water vapour) its pressure should reduce upon increase in the volume.I have searched the internet to clear my concepts about aqueous tension, vapour pressure and ideal gas equation but to no profit.
Please help me out. Thanks for your time.

 

[Borek]

Never heard the term "aqueous tension", but it looks like it is just an equivalent of "partial pressure due to the presence of the saturated water vapor".

As long as there is a liquid water present, pressure of saturated vapor is constant (if you increase the volume of the container, some of the liquid evaporates, till the pressure is that of a saturated vapor).

 

[phoenixXL]

I just wonder how I couldn't think of it.

As long as there is a liquid water present, pressure of saturated vapor is constant (if you increase the volume of the container, some of the liquid evaporates, till the pressure is that of a saturated vapor).

So, if I increase the volume of the container then the instant I increased the volume I disturbed the vapor-liquid equilibrium. Thereafter, the pressure of the gas is less than aqueous tension, so some more of it evaporates until equilibrium. When this equilibrium is established, then again the pressure exerted by the water vapor over water surface is same as that of the aqueous tension.

I'm getting the feel of the answer but still an inch behind.

Can you please answer,
What is that when we say that the Water Vapor is in equilibrium with Water (obviously in a closed container). What is the necessary and sufficient condition to have this?

Thanks a lot once again :)

 

[Borek]

If the liquid exists (after a long enough period of time), system is in equilibrium.

Minimum amount of the substance required (be it water or anything else) can be calculated if you know the volume and the pressure of the saturated vapor (tabulated for most common substances). But in any case if you wait long enough to be sure system is at equilibrium and the liquid is still persent, you can be sure the vapor pressure is that of a saturated vapor.

 

[phoenixXL]

Thanks a lot, I highly appreciate your to the point answers. Thank You.

 

[phoenixXL]

One side thought I had,

The saturated vapor pressure is constant for any given temperature.
Is this an observation or any theory is working behind it ?

 

[Borek]

Is this an observation or any theory is working behind it ?

Both.

Sorry for not going into details, but it is not something for a short post and I have no time for a longer one ATM.

 

[phoenixXL]

Sorry for not going into details, but it is not something for a short post and I have no time for a longer one ATM.

I'm very much interested in the concept(theory) behind it, I will be waiting for your reply. :)

 

[Borek]

Many ways to skin that cat.

But it is mainly about intermolecular forces and the average kinetic energy of molecules (3/2kT). To simplify things let's assume in the gas phase molecules are far apart enough so that their interactions are negligible (that is, we have an ideal gas above the liquid). In the liquid molecules attract each other. Molecule on the liquid surface is attracted from one side only, so if its energy is high enough it can leave the liquid and evaporate. The higher the temperature, the higher the average molecule energy, the higher the fraction of the molecules with the energy high enough to leave the solution. But it can also go other way around - if its energy is low enough and it is close to the liquid surface, it can get captured. These two processes define the pressure of the saturated vapor. As for a given temperature both intermolecular forces and average kinetic energy of molecules is constant, pressure of the saturated vapor is constant as well.

 

[phoenixXL]

Thank you, Sir
your answer cleared the theory part.