Talk:Vapor pressure

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I think the last mentioned temperature is not the melting point but the triple point. Is that correct?

Actually, no. When the vapor pressures of liquid and solid are equal only if the two phases (liquid and solid) are in equilibrium. If these pressures are also equal to the total pressure, then the gas phase is in equilibuim, too ( three phase -> triple point) -- Olof

And another question: to which substances does this discussion apply? Wood for instance doesn't have a vapor pressure, does it?

Sure it does. This is why I put wood in the garage for a year before putting it in the fireplace: some of its components ( water ) sublimate. However, wood is a multicomponent material, so it is better to think of vapor pressures of each of its components. -- Olof

And yet another one: what is the vapor pressure of solid water at 50 degrees Celcius? AxelBoldt

Higher than the vapor pressure of liquid water at 50 degrees Celsius. The fact that we know that it is higher doesn't imply that we are able to actually measure the value, although it could be estimated by extrapolating data from the temperature range where water is stable.

I've got to admit, though, that this page is quite confusing, and could use a better explanation of all the concepts. -- Olof

1 comparing and contrasting the dependence of the temperature.......
2 Technical accessibility
3 Collection of Gas over water
4 Vapor Pressure & Temperature
5 'notes'?

[edit] comparing and contrasting the dependence of the temperature.......

In a simple distillation, am I correct in assuming that the temperature of the liquid and vapor are equal? I am trying to compare and contrast the dependence of temperature on the distillate volume of cyclohexane and toluene. I find it easy to compare the temperature with the mixture, but would like to find some contrasting factors.

Email is charlespr01@hotmail.com Chuck

[edit] Technical accessibility

This article needs more context around or a better explanation of technical details to make it more accessible to general readers and technical readers outside the specialty, without removing technical details. See below for more information.

This article could use some work to make it more accessible to people who are not already familiar with the principles of thermodynamics and kinetic theory. I'm sure that schoolchildren and even adults who haven't had basic science education would find the explication of fundamental concepts here to be quite helpful. I've already drafted similar improvements for Phase (matter) (see my 10 July 2005 edits there), which are illustrative of what I have in mind. -- Beland 13:29, 10 July 2005 (UTC)

[edit] Collection of Gas over water

Perhaps a mention should be made about how Vapor Pressure affects the Collection of a Gas over water? In High School Chemistry this is the main use we have for Vapor pressure. Interpretivechaos 02:00, 30 January 2006 (UTC)

Hey you posted on my birthday. But yeah, I agree. I'll add it eventually. --M1ss1ontomars2k4 02:14, 14 May 2006 (UTC)

[edit] Vapor Pressure & Temperature

If we know the Vapor Pressure of a liquid at a given temperature, is there a way to recalculate the vapor pressure at a new temperature.

As an example, I believe toluene has a VP of around 22 at 68degrees. If it was say 80 degrees or 40 degrees can we calculate what the new vapor pressure is?

HAZMAT

Vapour pressure relates to temperature in a non-linear way, but you can still predict the vapour pressure using the Clausius-Clapeyron_relation or the Antoine equation.

Clausius -

$\ln \frac{P_2}{P_1} = \frac{-\Delta H_{vap}}{R} * \left (\frac{1}{T_2} - \frac{1}{T_1} \right)$

where P is in torr and T is in Kelvin and R is the gas constant (~8.3145 J mol^-1 ^K-1) and ΔH_vap is 38.06 kJ/mol.

I am apparently really bad at using TeX and preview is randomly breaking on it. ln ( P₂ / P₁) = -ΔH_vap / (RT) * (1/T₂ - 1/T₁)

Plugging stuff in (assuming your VP listed is in.. i have no idea really atm? bar? I'm going to go with bar, and that you mean celsius)

P₂ = e^([-38060 J/mol / 8.3145 J/(molK)] * [1/293.15K - 1/277.594K]) * 22 bar

or $P_2 = e^{\frac{-38060 \frac{J}{mol}}{8.3145 \frac{J}{molK}} * \left ( \frac{1}{293.15K} - \frac{1}{277.594K} \right ) } * 22 bar$

P₂ = 52.777

There are other ways of using this relation to do the same thing using calculus aswell.

Antoine - You just look up the numbers and plug them in really. Just make sure you have the correct units/correct formula. Sometimes it is ln, sometimes it is log, sometimes it is kelvin, sometimes it is celsius and various other variations —kotepho 2006-03-19 21:03Z

[edit] 'notes'?

Yes, we need a better column title than 'notes', but 'Boiling Point' is simply misleading. Using 'Boiling point' implies that the temperature listed in the column is the temperature at which liquid and gas coexist, as temperature is changed at constant pressure. That's not what these temperatures in this column are. Instead, this column is used to note the temperature at which the given vapor pressure is measured.