Naked Science Forum
Non Life Sciences => Physics, Astronomy & Cosmology => Topic started by: mcjhn on 05/06/2010 17:33:23
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removing nitrogen, oxygen etc. from a volume of air at room temperature/pressure so that only water vapour remains, what will be the temperature? If it isn't above 100C why doesn't the water vapour condense?
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removing nitrogen, oxygen etc. from a volume of air so that only water vapour remains, what will be the temperature? If it isn't above 100C why doesn't the water vapour condense?
Water, and for every other liquid and solid is the same, vaporizes at every temperature; otherwise how could clothes dry spreading them outside? There is an equilibrium between water vaporized and water condensed, which depends on temperature and pressure: at a given temperature the pressure is the water pressure at that temperature.
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but I thought it was the most energetic particles that left a liquid?
so if you took a volume of air at room temperature and pressure, somehow disappeared all other particles apart from water particles would you be left with a volume of pure water vapour at 23C (no equilibrium)?
imagine you could somehow do it instantaneously so the particles did not experience a pressure drop
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The temperature would remain the same (that's the equipartition principle) and the water wouldn't condense because there's not enough of it to exceed the saturated vapour pressure of water at that temperature.
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To explain take a triple point cell, whic is a simple device to provide a temperature reference point that is dependent only on the physical properties of water. It is a cell where water can exist simultaneously as a solid, a liquid and a gas, but only at around -0.1C Outside this precise point the water is either all liquid and gas or solid and gas, with it becoming all gas at well above 100C.
At the triple point all 3 phases exist, at the same temperature.
The probability that a molecule is vapour or liquid is dependent on the energy of the molecule, whilst the temperature is a measure of the average energy of the particles. Thus as the energy is spread over a large range, some have enough energy to be a vapour.
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but I thought it was the most energetic particles that left a liquid?
so if you took a volume of air at room temperature and pressure, somehow disappeared all other particles apart from water particles would you be left with a volume of pure water vapour at 23C (no equilibrium)?
imagine you could somehow do it instantaneously so the particles did not experience a pressure drop
The temperature would stay the same because of what Bored Chemist wrote: temperature is the average kinetic energy of a single molecule, so it wouldn't vary. You don't make work on the system and you don't heat it so you don't vary the system's energy; furthermore, for an ideal gas the kinetic energy of molecules don't vary with the average distance among them (which increases if you remove molecules at fixed volume). All you do is to vary the number of moles n. But PV = nRT so, if V and T stay the same, P is proportional to n and pressure DOES vary in your case.
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To explain take a triple point cell, whic is a simple device to provide a temperature reference point that is dependent only on the physical properties of water. It is a cell where water can exist simultaneously as a solid, a liquid and a gas, but only at around -0.1C Outside this precise point the water is either all liquid and gas or solid and gas, with it becoming all gas at well above 100C.
At the triple point all 3 phases exist, at the same temperature.
The probability that a molecule is vapour or liquid is dependent on the energy of the molecule, whilst the temperature is a measure of the average energy of the particles. Thus as the energy is spread over a large range, some have enough energy to be a vapour.
The triple point of water is +0.01C
Also, notwithstanding the fact that everyone knows that there are 3 states of matter, water forms a supercritical fluid under the right conditions.
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mcjhn Your thought experiment verges on magic but could just about be performed and we need to think it through properly. Let us assume we have some air saturated with water vapour at room temperature in a closed vessel and then removed almost all the oxygen and nitrogen by exposing some non volatile solid or liquid chemical absorbing agent to the air (this would leave a few residual gasses but that is not relevant to the experiment. the pressure in the vessel would drop considerably causing the remaining gas and water vapour to expand to fill the volume and it would cool considerably by adiabatic expansion. This temperature drop could cause some of the water vapour to condense out and there would be a mix of liquid water and water vapour in equilibrium in the vessel. I don't see why you thought it unlikely that water vapour would condense in the first place?
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ok, sorry, add to the room the a glass of water, the water vapour in the air and the liquid water are in equilibrium
the glass, water and air are at room temperature/pressure.
is it not only the more energetic water molecules that evaporate? (this is why wet skin gets chilly quickly in a breeze)
Graph of liquid water molecules
(https://www.thenakedscientists.com/forum/proxy.php?request=http%3A%2F%2Flh4.ggpht.com%2F_EI1LEVtoFmI%2FTAwm_fhy5cI%2FAAAAAAAACl4%2FkL7QOMbtQe4%2Fboltzmann.jpg&hash=d5ac812dbc46d0c4b96753e231b809c4)
therefore if you were to magically remove everything from the room apart from the water vapour, if you measured the instantaneous temperature it would be above room temperature as only the water molecules of higher than average energy evaporate??
Red box around the energies of water vapour molecules
(https://www.thenakedscientists.com/forum/proxy.php?request=http%3A%2F%2Flh4.ggpht.com%2F_EI1LEVtoFmI%2FTAwpnpmhziI%2FAAAAAAAACmA%2FLZALCFBcE8k%2Fboltzmann1.jpg&hash=b605ae80cf5c92390c9abd17c4ff1179)
if the water vapour molecules did not have these higher than average energies why don't they condense back into a liquid?
yeah your right i don't think its possible but as a thought experiment..
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You are not thinking clearly. That graph describes the energy a molecule at the surface of liquid water needs to evaporate. Once a molecule has evaporated it collides with other gas molecules almost instantly and exchanges energy it therefore does not have that energy it has a different energy and the energy is spread out among the other molecules in the gas according to the statistics molecules in gases and liquids only have specific energies for very brief periods known as the mean free time between collisions.
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mcjhn Your thought experiment verges on magic but could just about be performed and we need to think it through properly. Let us assume we have some air saturated with water vapour at room temperature in a closed vessel and then removed almost all the oxygen and nitrogen by exposing some non volatile solid or liquid chemical absorbing agent to the air (this would leave a few residual gasses but that is not relevant to the experiment. the pressure in the vessel would drop considerably causing the remaining gas and water vapour to expand to fill the volume
The pressure drops because you remove moles of gas, and you are left with the partial pressure of water only, but it doesn't expand at all, the volume stays the same.
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ok, sorry, add to the room the a glass of water, the water vapour in the air and the liquid water are in equilibrium
the glass, water and air are at room temperature/pressure.
is it not only the more energetic water molecules that evaporate? (this is why wet skin gets chilly quickly in a breeze)
Graph of liquid water molecules
(https://www.thenakedscientists.com/forum/proxy.php?request=http%3A%2F%2Flh4.ggpht.com%2F_EI1LEVtoFmI%2FTAwm_fhy5cI%2FAAAAAAAACl4%2FkL7QOMbtQe4%2Fboltzmann.jpg&hash=d5ac812dbc46d0c4b96753e231b809c4)
therefore if you were to magically remove everything from the room apart from the water vapour, if you measured the instantaneous temperature it would be above room temperature as only the water molecules of higher than average energy evaporate??
Red box around the energies of water vapour molecules
(https://www.thenakedscientists.com/forum/proxy.php?request=http%3A%2F%2Flh4.ggpht.com%2F_EI1LEVtoFmI%2FTAwpnpmhziI%2FAAAAAAAACmA%2FLZALCFBcE8k%2Fboltzmann1.jpg&hash=b605ae80cf5c92390c9abd17c4ff1179)
if the water vapour molecules did not have these higher than average energies why don't they condense back into a liquid?
yeah your right i don't think its possible but as a thought experiment..
This is a pefect example of confusing didactics. I would never explain things in that way. Yes, it is true that only the more energetic molecules leave the liquid surface, but that distribution is not static: it rearranges immediately forming again molecules with those energies; the difference is that the new *average* energy of every molecule is slightly smaller, because of the latent heat of vaporization lost. The same with those molecules which went in the air: their energies rearrange immediately (immediately means after a few collisions with other air's molecules, which takes fractions of milliseconds). Temperature is related to the *average* energy of the molecule, you don't even imagine which paradoxes could arise if you don't consider this.
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The killer point is, I think, that in order to leave the liquid, the molecules have to lose some energy to it. A molecule at the surface is attracted back by dipole-dipole interactions with the water below it.
They have to overcome that force and, in doing so they lose (on average) their "excess" energy.
You also need to think about the other half of the equilibrium. Molecules hitting the water might stick. If they do so they carry energy to the water.
The water is at equilibrium when these two processes are happening at the same rate.
That means the average energy of a water molecule is exactly the same whether it's in the liquid or vapour state.
If it were not then molecules would evaporate or condense until the temperature was the same and the vapour was at equilibrium with the liquid.
Incidentally, this works in exactly the same way if you just have some water in a closed system with no air present.
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The killer point is, I think, that in order to leave the liquid, the molecules have to lose some energy to it. A molecule at the surface is attracted back by dipole-dipole interactions with the water below it.
They have to overcome that force and, in doing so they lose (on average) their "excess" energy.
This is confusing too. Actually the liquid gives energy to the leaving molecules, it's for this reason that the liquid becoms colder. But energy is kinetic and potential; the kinetic energy of a leaving molecule decreases, while its potential energy increases. Probably it would be better to avoid this subject...