Is the vapor pressure partial

Water and vapor pressure: vapor pressure table for water and ice, relationship with boiling and boiling point, formula for calculation by interpolation

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1. General

Water can appear in the 3 physical states solid, liquid and gaseous. In the solid state, the water molecules have gathered together to form a crystal lattice in which each water molecule has its fixed place. It can still shiver in its place, but not leave it easily. Since the molecules cannot move past each other, a piece of ice retains its shape. In the liquid state, on the other hand, the molecules are still connected to one another by their forces of attraction to form a coherent mass, but have largely dissolved their crystal lattice. They cannot move away from each other, but they can easily move past each other, so that liquid water does not have its own shape: Instead, it adapts to the vessel in which it is currently located. In the gaseous state, the water molecules have overcome their mutual forces of attraction and, in principle, whiz around freely in the area. In doing so, they constantly collide with other gas molecules. In pure water vapor these can of course only be additional water molecules, otherwise other existing gas molecules as well. The colliding molecules bounce off one another, so that each molecule in a gas phase moves around in a wild zigzag course.

The energy of the molecules in the water (how violently or quickly they move) is very different. Molecules that exceed a certain minimum energy and are currently on the surface can, thanks to their "momentum", overcome the forces of attraction to their neighboring molecules, "tear themselves away" from them and leave the liquid water. The higher the temperature, the more water molecules have this minimum energy and the more leave the liquid water every second. Conversely, water molecules that are currently in the gas space can also dip back into the water surface. The more of them are in the gas space, the more immerse into the liquid water per second.

Equilibrium is reached when every second as many water molecules submerge as they come out of the water.

An increase in temperature results in:
More water molecules per second leave the liquid water. This increases the amount of water molecules in the gas space and thus the number of those that submerge again per second until equilibrium is reached again. In equilibrium there are now more water molecules in one cm³ gas space than in equilibrium at the lower temperature.

The pressure in a gas depends on the number of gas molecules per cm³ and on their energy. Therefore, the pressure exerted by the water molecules is greater at a higher temperature than at a lower temperature.

The vapor pressure of a substance can be measured directly if it is placed in a gas-tight container and the air is then removed with a vacuum pump. The pressure that is established in the gas space in the bottle at the respective temperature is the vapor pressure. If it is not too low, it can be measured with a manometer.

In principle, every solid and every liquid substance has a vapor pressure. For some substances, however, it is very low at normal ambient temperatures and can therefore only be measured with great effort. The following table shows the vapor pressures of water and ice for different temperatures:

Vapor pressures of ice and water in mbar

Temp.
With increasing temperature, the properties of liquid and gaseous water converge more and more. At 374.12 ° C ("critical temperature"), the properties have perfectly matched each other, so that there is no longer any difference between liquid and gaseous water. Consequently, neither boiling nor condensation can take place above this temperature.

The vapor pressure data shows the minimum pressure required to liquefy (condense) water vapor at a given temperature or, conversely, at what temperature water boils at a given pressure.





2. Vapor pressure effects in everyday life

If warm water is put into a bottle, it is tightly closed and then shaken, the vapor pressure equilibrium is quickly established. The vapor pressure of the water is added to the pressure of the air that was already in the bottle, so that the pressure in the bottle increases by the amount of the vapor pressure. The higher pressure manifests itself in the fact that when the bottle is opened, some of the air-water vapor mixture hisses out. When it is closed and shaken again, it hardly hisses the second time. Because now there is less air in the bottle and the partial pressures of air and water vapor, when added together, give the ambient pressure.

A similar observation is made on a freshly filled petrol can. When closed and shaken, the (relatively high) vapor pressure of the gasoline inflates it a little and when you open it, a mixture of gasoline and air can be heard escaping. (The whole thing happens without shaking, it just takes longer because the equilibrium is achieved much more slowly.)





3. Vapor pressure and boiling

If the pressure of the steam trying to expand a steam bubble is greater than the ambient pressure (air pressure) trying to compress the steam bubble, then steam bubbles can form in the first place and then the water begins to boil. At an air pressure of 1013 mbar, water boils at 100 ° C because its vapor pressure at this temperature is 1013 mbar. If the air pressure is lower, then water will boil at a lower temperature because the vapor pressure required for bubble formation is then also lower.





4. Vapor pressure formula:

The frequently cited Clausius-Clapeyron equation is a theoretically derived equation that does not cover all influences and does not lead to really useful results in practice.

Since the temperature-vapor pressure curve is very non-linear, interpolations and especially extrapolations should be carried out with the following equation, which is based on the Clausius-Clapeyron equation:

Log (p) = K1 + K2 * 1 / T

p: vapor pressure
Log (p): logarithm of the vapor pressure
T: absolute temperature
K1, K2: constants

T (in K) = temperature (in ° C) + 273.15
Temperature (in ° C) = T (in K) - 273.15

How to do it:

The 2 known value pairs (p1; T1) and (p2; T2) are inserted into the equation. You get 2 equations with K1 and K2 as the 2 unknowns and you have to calculate them from them.

With the calculated K1 and K2 inserted into the equation, the p-values ​​for all T-values ​​between T1 and T2 can now be calculated (interpolation) and also a little beyond that (extrapolation).

If you have problems solving 2 equations with 2 unknowns, you should learn that first and then take care of vapor pressure formulas. As a control for those who at least try, here are the calculated constants for the value pairs (20 ° C; 23.4 mbar) and (40 ° C; 73.7 mbar):

K1 = 9.17059
K2 = -2286.97

and a comparison of the calculated vapor pressure values ​​with the actual values:

Temp. (° C) Vapor pressure (mbar) calculated measured =================================== 20 23.40 23.4 25 31.63 31.7 30 42.32 42.4 35 56.10 56.21 40 73.70 73.7 50 124 123 60 202 199 70 321 311
(You can see from the 70 ° C values ​​at the latest that you shouldn't overdo it with extrapolating.)





5. Credits:

D'Ans-Lax: Pocket book for chemists and physicists, Volume 1: Macroscopic chem.-phys. Properties, Springer-Verlag, 3rd edition, 1967

Weast, Handbook of Chemistry and Physics, 64th Edition, 1983-1984, CRC Press Inc., Boca Raton, Florida

Meyer / Schiffner, Technical Thermodynamics, VEB Fachbuchverlag, Leipzig






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