1
$\begingroup$

Experiment: We have a cylinder with a piston that is essentially weightless, frictionless, and exposed to atmospheric pressure. The cylinder has a volume of 2V, but half of it is filled with water and the other half is just filled with air that has the same density as the atmosphere. The water starts at barely above its freezing point (so its vapor pressure is 0). At this point, the piston is just floating above the water because the air pressure on the inside of the cylinder equals the air pressure on the outside.

Now, let's say we heated the water to 70 Celsius**. The piston would begin to expand because the water vapor pressure will go up. Eventually the rate of water vapor evaporating and water vapor condensing will reach an equilibrium and the piston would stop expanding. How do we know what the vapor pressure of the water is? The volume is now larger so the air is now less dense (so its partial pressure is less than 1 atm) which means the rest of the pressure is accounted for by the water vapor.

This is my guess for how scientists would calculate vapor pressure. But it honestly just seems a lot easier if we created a new term called "vapor height." Vapor height would be defined as the $V/A$ generated by a solution at a certain temperature.

**We are assuming the air molecules don't increase in temperature.

$\endgroup$
  • $\begingroup$ Why do you say that the vapor pressure of a solid is zero? This is not the case. $\endgroup$ – Chemomechanics Apr 3 '17 at 23:34
  • $\begingroup$ If your question is strictly about measurement, then vapor pressure can be measured by measuring capacitance of air, and such sensors are available. $\endgroup$ – Deep Apr 4 '17 at 5:30
  • $\begingroup$ Nova, you need to research this, instead of proposing a method that will not work and is impractical. $\endgroup$ – David White Apr 7 '18 at 17:17
1
$\begingroup$

The piston would actually go up forever, in your case, until real-world thermodynamics gets in your way. At some point the heat transferring into or out of your device is going to cause the entire device to be at uneven temperatures.

The measurement you want is actually far easier to capture than you're thinking it is. You put a quantity of liquid in a container and pull a vacuum on it to get all of the other stuff (like air molecules) out. This can be done via a vacuum pump, or it could be even simpler: you could simply have an accordion like container, squish all of the air out of it until there's only liquid remaining (like squeezing a toothpaste tube), and seal it. Then, if you pull the accordion open, expanding the chamber, you'll create a region filled with nothing but water vapor over an area with water in it (because we didn't let any air in).

However, you did it, you now have a rigid container containing nothing but liquid water and water vapor. Nothing else. This is important because it means the partial pressure of the water vapor is exactly equal to the pressure inside the container. Now, all you have to do is let it equalize thermally. If you want to know the vapor pressure of water at 40C, heat the entire device up to 40C and let it equalize.

Now, all you have to do is read the pressure in the chamber with a pressure gauge. The pressure you read is the vapor pressure of the water at that temperature. As long as there is at least some liquid and some gas in the chamber, this will be the correct pressure.

$\endgroup$
  • $\begingroup$ Why would the piston, in the situation I proposed, expand forever? The number of water molecules in the gas phase would eventually reach an equilibrium with the liquid phase because at all the temperatures below the BP, rate of evaporation = rate of condensation. $\endgroup$ – Nova Apr 4 '17 at 21:09
  • $\begingroup$ Also, you're suggesting that if I simply had a piston with water in it (no air), it would also expand until all the water molecules would eventually turn into gas. But that wouldn't happen because the chemical potential of them condensing would be much too high for that to happen. $\endgroup$ – Nova Apr 4 '17 at 21:13
  • $\begingroup$ @Nova The partial pressure is a fixed pressure at a given temperature. Your piston has no feedback which causes it to exert more pressure on the inside of the container as it moves outwards. It will move outwards, increasing the volume of the container, which causes more water molecules to become a gas, raising the pressure back to the vapor pressure, which pushes the piston out further. This will be true as long as the vapor pressure is above atmospheric pressure. If the vapor pressure is at or below atmospheric pressure, the piston wont move at all. $\endgroup$ – Cort Ammon Apr 4 '17 at 22:07
  • $\begingroup$ In the real world, you would see the contents of the chamber (gas and liquid) would decrease in temperature due to the heat of vaporization as molecules evaporate, which would decrease the vapor pressure unless you added heat to maintain 70C. Alternatively, in the real world, you might simply run out of liquid in the container. $\endgroup$ – Cort Ammon Apr 4 '17 at 22:09
  • $\begingroup$ The pressure gague in my answer is very similar to your piston, but they are designed to increase in volume very little as more pressure is applied. The piston does not have that property. It moves as long as the pressure on the inside is greater than the pressure on the outside. $\endgroup$ – Cort Ammon Apr 4 '17 at 22:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.