Vapor Pressure

Well, I think you need to start here:

Vapor pressure - Wikipedia, the free encyclopedia
Your question is nonsensical and exhibits a lack of understanding of the subject.

Maybe you could re-phrase it?

Have you got access to a set of Steam Tables, Guv?

My basic explanation of vapour pressure is that the surface of a liquid is like a box full of ping pong balls just one ball too full. The pressure from the balls pushing from the side and below forces that one ball to pop out.

This is a visualization of the molecular situation, the molecules on the surface are being pushed from the sides and below causing the ones on the surface to tend to pop out until the vapour pressure from above is equal to the pressure on the sides and below. Pressure is pressure, no matter if it comes from the liquid or the vapour. Once the vapour pressure is high enough the liquid stops evaporating.

Drew K

Just to expand a bit...

In my model of ping pong balls, after a ball is pushed out of the liquid and enters the vapour phase it floats around in the gas mixture. This is a simplistic model of what actually is happening. In solid state the molecules are rigid and unmoving (much) In a liquid phase the molecules are separated by a distance and move around (hence the nature of a liquid to conform to the shape of it's container). In a gas phase the molecules are farther apart and also conform to the shape of their container.

It is the pressure that has a large control over the liquid and gas phases. Increase pressure and the ping pong balls are pushed harder by the gas than the liquid and they get closer together until they condense and fall from the gas. Lower the pressure and the liquid pushes more molecules into the gas.

Temperature also has a factor in liquid gas and solid phases. Higher temp usually results in more movement of molecules and higher pressures.

Steam tables and quality (percentage of gas vs vapour) tables can give you values for how much gas and liquid there is at a specific pressure and temperature.

Drew K

Well, to start off, unless your water is boiling, the VP could not be even one bar. Let me rephrase; at standard atmospheric pressure, the vapor pressure of water at 100ºC (yes, the boiling point!) is 1 atmosphere (1.01 bara). The boiling point of a compound is reached when its vapor pressure equals the ambient pressure. The Antoine equation, used to calculate vapor pressures, parameters have been determined for a great many compounds.

PV=nRT is not for liquids. (maybe that's why it is called the Ideal Gas Law). There are equations of state that encompass liquids.

Well, I've given you some basic info and some links. Go read up!

I have found this site very useful - lots of data:

http://webbook.nist.gov/chemistry/

Good find.

The vapor pressure of a liquid depends only on its' temperature. In a closed vessel containing air at 2 atmospheres pressure, water at 20 deg C will have a vapor pressure of 17.5 mm of Hg (1 atmosphere = 760 mm of Hg). The pressure it exerts is referred to as the "partial pressure." The partial pressure of the air would be (760 X 2) - 17.5 = 1502.5 mm Hg. This 17.5 mm of Hg will be maintained regardless of the overall pressure in the vessel. If you increased the overall pressure to 3 atmospheres, but kept the temperature at 20 deg C, the water still only exerts 17.5 mm of Hg. If you evacuated the vessel below 17.5 mm Hg, the water would boil off in order to maintain that 17.5 mm Hg. pressure, until its all gone.

Hope this helps and not clouded the issue.

The ideal gas law PV = nRT, applies to normal pressures, but breaks down at very high pressures, due to interaction between the molecules, which is not significant at the lower pressures.

its' its?.........(it is?).................it's?

Wat wood I no, my gramma is terible most the time enyway.

It may be a simple oversight, but you have provided an example with

"In a closed vessel containing air at 2 atmospheres pressure", and

" The partial pressure of the air would be (760 X 2) - 17.5 = 1502.5 mm".

Don't you think you might want to change something? You may have something valuable to say, but this is not it.

Not really an oversight. If the question poser had stated "2.0000 atmospheres pressure," rather than just "2 atmospheres," then I would have been more pedantic.

If it suits you better, I can change that to "air at 1.976973684210526 atmospheres pressure." Needeless to say, the total pressure in the system is:

pp(H2O) + pp(N2) + pp(O2) + pp(Ar) + pp(CO2) + etc. (the rest being negligible)

Actually, there is small but measurable effect on vapor pressure from system mechanical pressure. I believe this acts to increase the vapor pressure of the liquid under mechanical pressure, just slightly, depending on how much overall pressure is applied to the vessel in question.

Hey TIA! Are you going to come back with some feedback? Since you failed to register on this site, you will not be automatically notified of replies through email, so we will likely not hear from you again.

hi Mikerho and all,

i am reading and try to understand. i will comeback with feedback after i have some idea on what all of you trying to figure out

~ i'm a very bad engineering graduates! i get confused all the time ~

right now i'm still studying all of your comments..keep on 'teaching' me because i'm reading, just that i'm silent reader until i can come out with some comments

TIA

Basically, if you have a liquid, such as water, at a given temperature, some of the molecules will have enough energy to escape and enter the gaseous phase. In the gas, some water molecules will reenter the liquid. Eventually, there is an equilibrium point at which the rate of molecules returning to the liquid is the same as the number leaving. The pressure of the water vapor at that point is the "vapor pressure" for that temperature. Each substance has its own pressure (partial pressure) and the total pressure is the sum of the partial pressures of the components.

I have heard of a pump sold to pump air pressure into a partially filled soda bottle to prevent the beverage from losing the dissolved carbon dioxide and going flat. Obviously, this would not work because the partial pressure of carbon dioxide is totally independent of the partial pressure of nitrogen, oxygen, argon, etc. in the air. (The carbon dioxide percentage in air is very low, .04%)

You might have noticed that rubber helium party balloons deflate partly over time. It is the same principle. The partial pressure of helium inside is much greater inside than outside, so the helium escapes. Likewise, the partial pressure of air on the inside is less inside than outside, so the helium diffuses out and the air diffuses in until equilibrium is reached.

Vapor pressure is just that, the pressure exerted by vapor from the liquid in the container, and by the way this is independent of the system being a closed or open system. For example, water boils in an open pan, or a closed one when the temperature reaches the point (boiling point) where vapor pressure is equal to system pressure. The "normal" boiling point definition is for an open system at sea level, where system pressure is exactly one atmosphere.

The ideal gas law is somewhat irrelevant in this instance, although it might reveal something about the volume of the vapor if compressed or expanded. In your closed system container, with the "vapor" pressure being at 2 atm, and the liquid in your "experiment" being water, the temperature will be 121 degree C, which is hot water, and at 3 atm, the temperature will be 130 degree C. These are the "saturation" temperatures for water that is boiling in closed systems where the entire system pressure is composed of water vapor. If you cool the system off without venting it, there will be a vacuum (less than ambient pressure) at some point, and the vessel might stand a chance of collapsing.

"The vapor pressure inside is 2 bar, do the water also has 2 bar pressure if the container is at 2 bar?"

Dear OP, your all the queries were answered in previous comments except the above. For this answer is a big YES but only on top surface of water in contact with vapor at 2 bar. At other places of the container, the pressure depend on hight of the water column from the top surface of water.

For example, if hight of the container is 1m, in half full condition top surface of water is 0.5 m from the base of container. Water pressure at the bottom will be approximately 2+0.5/10 = 2.05 bar, at 0.1 m hight 2+0.4/10 = 2.04 bar and so on..

Do you have any further doubt regarding your post?

Hi All teachers!

tq for your kind support

i've read all the comments and take a look at your given link (not reading much yet )

i think i understand VP especially with refering to steam tables.

the real situation that make my colleague come out with the situation is:

we have a pipeline connect to pump, based on process simulation

shows the liquid has vapor pressure.so how to eliminate the vapor?

preliminarily i answered, decrease the system temp. and pressure.

how to make this possible? thats the question

(i'm sorry i not come out with this question straight away because

only recently i know the real situation)

TIA

If you have vapour in a pipeline, you have sections where there is low pressure creating vapour lock. Increase the pressure and you will not have any evaporation from liquid to gas.

If the vapour condition is between the source and your pump you will have to either install a booster pump or use a larger pipe.

How large of a pipeline is this (diameter and flow conditions (velocity and pressure))? What are the elevation changes?

Drew K