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So, the pressure exerted by the vapour particles of A at any particular temperature is called the vapour pressure of A at that temperature. Vapour pressure is exhibited by all solids and liquids and depends only on the sort of liquid and temperature.
Now, if another liquid B is added to this container, the B particles will occupy the space between A particles on the surface of the solution. For any given liquid there are a fraction of molecules on the surface having sufficient energy to escape to the vapour phase.
Since we have a fewer number of A particles on the surface at this point, the number of vapour particles of A in the vapour phase will be lesser.
This is going to result in lower vapour pressure of A. Now if B is also volatile, we will have a fewer number of B particles in the vapour phase as compared to pure liquid B. However, they are not easily found and are rare. The solution in the last diagram wouldn't actually obey Raoult's Law - it is far too concentrated. I had to draw it that concentrated to make the point more clearly.
In an ideal solution, it takes exactly the same amount of energy for a solvent molecule to break away from the surface of the solution as it did in the pure solvent. The forces of attraction between solvent and solute are exactly the same as between the original solvent molecules - not a very likely event! Suppose that in the pure solvent, 1 in molecules had enough energy to overcome the intermolecular forces and break away from the surface in any given time. In an ideal solution, that would still be exactly the same proportion.
Fewer would, of course, break away because there are now fewer solvent molecules on the surface - but of those that are on the surface, the same proportion still break away. If there were strong solvent-solute attractions, this proportion may be reduced to 1 in , or 1 in or whatever. In any real solution of, say, a salt in water, there are strong attractions between the water molecules and the ions.
That would tend to slow down the loss of water molecules from the surface. However, if the solution is sufficiently dilute, there will be good-sized regions on the surface where you still have water molecules on their own. The solution will then approach ideal behaviour.
There is another thing that you have to be careful of if you are going to do any calculations on Raoult's Law beyond the scope of this site. You may have noticed in the little calculation about mole fraction further up the page, that I used sugar as a solute rather than salt. There was a good reason for that!
What matters isn't actually the number of moles of substance that you put into the solution, but the number of moles of particles formed. For each mole of sodium chloride dissolved, you get 1 mole of sodium ions and 1 mole of chloride ions - in other words, you get twice the number of moles of particles as of original salt.
So, if you added 0. Unless you think carefully about it, Raoult's Law only works for solutes which don't change their nature when they dissolve. For example, they mustn't ionise or associate in other words, if you put in substance A, it mustn't form A 2 in solution. Note: This isn't a problem you are likely to have to worry about if you are a UK A level student.
Just be aware that the problem exists. The effect of Raoult's Law is that the saturated vapour pressure of a solution is going to be lower than that of the pure solvent at any particular temperature.
That has important effects on the phase diagram of the solvent. The next diagram shows the phase diagram for pure water in the region around its normal melting and boiling points. The 1 atmosphere line shows the conditions for measuring the normal melting and boiling points. Note: In common with most phase diagrams, this is drawn highly distorted in order to show more clearly what is going on.
If you haven't already read my page about phase diagrams for pure substances, you should follow this link before you go on to make proper sense of what comes next. Buy Now! Managed Services By: www. To maintain dry conditions in a room of air space m3 containing 2. The cooled air has a. Online GK. Flights in a rotary dryer are provided to. Pick out the correct statement. In the constant rate period of the rate of drying curve for batch drying,.
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