Themes > Science > Chemistry > Miscellenous > Help file Index > Liquid and
Solid Properties > Phase Diagrams
A phase diagram is a diagram that shows the state of a substance at different
temperatures and pressures. The sketch below shows a phase diagram for a
hypothetical substance.
The line from the red A to the red B is the line along the liquid-vapor equilibrium line.
At pressures and temperatures along this line, the substance is in equilibrium
between the liquid and vapor phases. The line ends at the critical temperature and
pressure of the substance, beyond which the there is no difference between the
liquid and gaseous phases. Substances past this point are known as supercritical
fluids.
In the same vein, the A-C line is the line where solid and liquid co-exist, and A-D is
the area where solid and vapor coexist. The three lines meet at the triple point of the
material.
The slope of the A-C line depends on the relative densities of the solid and the
liquid. A positive slope means that the solid is more dense than the liquid, a negative
slope means the opposite. In the diagram above, as the pressure increases the solid
phase becomes stable at higher and higher temperatures: this is because it is the
denser phase and under higher pressures it requires more energy to overcome the
exterior pressure and change to the less dense liquid phase. Only a few materials
have a liquid phase that is denser than the solid: water is the most important
example.
Example: If we begin at the point marked A on the phase diagram below and
increase the temperature, what will happen to the substance?
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Solution: The point marked A on the phase diagram is in the solid region, so the
substance begins as a solid. As we increase the temperature at constant pressure,
we move right along the red line in the diagram below. The line passes first through
the liquid area, then the vapor, so the substance moves from solid->liquid->vapor.
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RAOULT'S LAW AND NON-VOLATILE
SOLUTES
This page deals with Raoult's Law and how it applies to
solutions in which the solute is non-volatile - for example, a
solution of salt in water. A non-volatile solute (the salt, for
example) hasn't got any tendency to form a vapour at the
temperature of the solution.
It goes on to explain how the resulting lowering of vapour
pressure affects the boiling point and freezing point of the
solution.
Important: If you haven't already read the page about saturated
vapour pressure, you should follow this link before you go on.
Use the BACK button on your browser to return to this page when
you are ready.
Raoult's Law
There are several ways of stating Raoult's Law, and you tend to
use slightly different versions depending on the situation you are
talking about. You can use the simplified definition in the box
below in the case of a single volatile liquid (the solvent) and a
non-volatile solute.
The vapour pressure of a solution of a non-volatile solute
is equal to the vapour pressure of the pure solvent at that
temperature multiplied by its mole fraction.
In equation form, this reads:
In this equation, Po is the vapour pressure of the pure solvent at
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a particular temperature.
xsolv is the mole fraction of the solvent. That is exactly what it
says it is - the fraction of the total number of moles present
which is solvent.
You calculate this using:
Suppose you had a solution containing 10 moles of water and
0.1 moles of sugar. The total number of moles is therefore 10.1
The mole fraction of the water is:
A simple explanation of why Raoult's Law works
There are two ways of explaining why Raoult's Law works - a
simple visual way, and a more sophisticated way based on
entropy. Because of the level I am aiming at, I'm just going to
look at the simple way.
Remember that saturated vapour pressure is what you get when
a liquid is in a sealed container. An equilibrium is set up where
the number of particles breaking away from the surface is
exactly the same as the number sticking on to the surface again.
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Now suppose you added enough solute so that the solvent
molecules only occupied 50% of the surface of the solution.
A certain fraction of the solvent molecules will have enough
energy to escape from the surface (say, 1 in 1000 or 1 in a
million, or whatever). If you reduce the number of solvent
molecules on the surface, you are going to reduce the number
which can escape in any given time.
But it won't make any difference to the ability of molecules in the
vapour to stick to the surface again. If a solvent molecule in the
vapour hits a bit of surface occupied by the solute particles, it
may well stick. There are obviously attractions between solvent
and solute otherwise you wouldn't have a solution in the first
place.
The net effect of this is that when equilibrium is established,
there will be fewer solvent molecules in the vapour phase - it is
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less likely that they are going to break away, but there isn't any
problem about them returning.
If there are fewer particles in the vapour at equilibrium, the
saturated vapour pressure is lower.
Limitations on Raoult's Law
Raoult's Law only works for ideal solutions. An ideal solution is
defined as one which obeys Raoult's Law.
Features of an ideal solution
In practice, there's no such thing! However, very dilute solutions
obey Raoult's Law to a reasonable approximation. 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 1000 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
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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 2000, or 1 in 5000 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.
The nature of the solute
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 0.1 moles of sodium chloride, there would
actually be 0.2 moles of particles in the solution - and that's the
figure you would have to use in the mole fraction calculation.
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 A2 in solution).
If it does either of these things, you have to treat Raoult's law
with great care.
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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.
Raoult's Law and melting and boiling points
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.
Use the BACK button on your browser to return to this page when
you are ready.
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The line separating the liquid and vapour regions is the set of
conditions where liquid and vapour are in equilibrium.
It can be thought of as the effect of pressure on the boiling point
of the water, but it is also the curve showing the effect of
temperature on the saturated vapour pressure of the water.
These two ways of looking at the same line are discussed briefly
in a note about half-way down the page about phase diagrams
(follow the last link above).
If you draw the saturated vapour pressure curve for a solution of
a non-volatile solute in water, it will always be lower than the
curve for the pure water.
Note: The curves for the pure water and for the solution are often
drawn parallel to each other. That has got to be wrong!
Suppose you have a solution where the mole fraction of the water is
0.99 and the vapour pressure of the pure water at that temperature
is 100 kPa. The vapour pressure of the solution will be 99 kPa - a
fall of 1 kPa. At a lower temperature, where the vapour pressure of
the pure water is 10 kPa, the fall will only be 0.1 kPa. For the curves
to be parallel the falls would have to be the same over the whole
temperature range. They aren't!
If you look closely at the last diagram, you will see that the point
at which the liquid-vapour equilibrium curve meets the solidvapour curve has moved. That point is the triple point of the
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system - a unique set of temperature and pressure conditions at
which it is possible to get solid, liquid and vapour all in
equilibrium with each other at the same time.
Since the triple point has solid-liquid equilibrium present
(amongst other equilibria), it is also a melting point of the system
- although not the normal melting point because the pressure
isn't 1 atmosphere.
That must mean that the phase diagram needs a new melting
point line (a solid-liquid equilibrium line) passing through the
new triple point. That is shown in the next diagram.
Now we are finally in a position to see what effect a non-volatile
solute has on the melting and freezing points of the solution.
Look at what happens when you draw in the 1 atmosphere
pressure line which lets you measure the melting and boiling
points. The diagram also includes the melting and boiling points
of the pure water from the original phase diagram for pure water
(black lines).
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Because of the changes to the phase diagram, you can see that:


the boiling point of the solvent in a solution is higher than
that of the pure solvent;
the freezing point (melting point) of the solvent in a
solution is lower than that of the pure solvent.
We have looked at this with water as the solvent, but using a
different solvent would make no difference to the argument or
the conclusions.
The only difference is in the slope of the solid-liquid equilibrium
lines. For most solvents, these slope forwards whereas the
water line slopes backwards. You could prove to yourself that
that doesn't affect what we have been looking at by re-drawing
all these diagrams with the slope of that particular line changed.
You will find it makes no difference whatsoever.
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