- [Instructor] So here we're gonna use
the first law in differential form
and definition of Gibbs free energy
to derive the Clapeyron and the Clausius-Clapeyron equations
which relate for vapor and liquid equilibrium
saturation pressure to temperature.
So let's start off with the first law,
writing in differential form.
And we'll write it for the case
where we have a reversible process,
and you'll see why we're doing this.
It's because we can now write the Q reversible
in terms of entropy change in temperature,
and we can write work in terms of pressure
and volume change.
And now what I'm gonna do is,
since we're interested in Gibbs free energy,
I'm going to write differential form definition,
if you like of enthalpian terms of U plus PV.
And now I'm gonna substitute
from this previous equation,
dU is TdS minus PdV.
And I'm going to expand this term
as PdV plus VdP.
And you'll notice that these two terms cancel.
So then I'll rewrite the equation.
Well, now I'll write the definition for Gibbs free energy
and also write the differential form of the equation.
So now what we're going to do is substitute this into here.
And let me pause and make that substitution.
And then for this last turn, I'm gonna do as we did before
and expand it out into two terms.
And again, we're gonna notice
that this term cancels with this term.
So I'll rewrite the equation.
So if we imagine we were to start
with a liquid and vapor in equilibrium,
and vaporize some more of the liquid for example,
this equation says that if I vaporize,
starting the liquid, pause for a second,
if I were to vaporize at constant pressure
or constant weight,
this takes place at constant temperature,
and so that says, going from liquid to vapor,
dG must be zero because there's no pressure change
and no temperature change,
that is the Gibbs free energy to liquid
equals the Gibbs free energy to vapor at a given temperature
where we have vapor-liquid equilibrium.
If I went to a different temperature,
that would indeed also be true.
And so if I were to make comparison,
slightly different temperature
and therefore slightly different pressure,
the Gibbs free energy of the liquid
would still be equal to Gibbs free energy
of the gas at the new condition,
which means any change in the Gibbs free energy
of the liquid must be identical to the change
in the Gibbs free energy of the vapor
as we go from say, temperature one to temperature two.
So this must be true.
And now since I know that dG is equal to this,
I can substitute it, we're looking at two temperatures
and two corresponding pressure.
So let me write that.
Series equation written.
Substitute in for dG in terms of dP and dT terms.
And notice now I have written pressure as Psat
since we're looking at saturation pressure.
I'm gonna rearrange this equation.
Again, I'm gonna pause and do that rearrangement
and group together the dT and the dP terms.
So in order to simplify this,
I'm gonna go back to our definition of Gibbs free energy,
and Gibbs free energy for a vapor, H.
Well, I'm gonna also write the same thing for liquid.
And of course I set Gibbs free energy of liquid
equal to Gibbs free energy of vapor
'cause we're looking at these two phases being equilibrium.
Now I'm gonna rearrange these two terms that are equal.
Well, delta, this H-vapor minus H-liquid,
is delta H of vaporization.
I'll bring this T from the other side of the equation.
And likewise, the right side is delta S of vaporization.
And so I'm going to go back now
and substitute in here for delta S of vaporization
in terms of delta H of vaporization and T.
So I made the substitution
and also put the dT term on the right side
and the volume of vapor and volume of liquid
on the left side.
And so this then is the Clapeyron equation.
And so just from a pretty simple starting point
of the first law in our definitions,
we show how saturation pressure is related
to temperature and heat of vaporization.
Now this is for vapor and liquid,
but equivalent equation is true
for solid and vapor equilibrium,
where the volume change obviously is the vapor and the solid
and we have the heat of sublimation.
And it's also true for solid-liquid equilibrium,
where then we have delta H for melting.
We can further simplify this if we substitute
for the vapor in terms of the ideal gas law.
And we assume the volume of the vapor
is much greater than the volume of the liquid,
which would be true if we're looking
at relatively low pressures.
So let's make that substitution, and I'll also simplify.
So let me rearrange and simplify this,
where you will note that this pressure
that was in denominator of the denominator
is now in denominator on the other side of the equation.
So let's simplify this equation a bit.
And what we're gonna do is look at differential of 1 over T,
which might be easier to visualize as this.
And so the differential then, -1, T to -2, differential T.
Or I can write this as,
dT over T square with a minus sign.
And likewise, we can look at the differential
of the natural log of Psat,
and that's equal to 1 over Psat,
times the differential of Psat.
So you can see we're gonna make the substitutions
into this equation.
We're going to substitute for this term,
and then of course the right side.
And we make those two substitutions,
we end up with the heat of vaporization
over R, minus sign.
And now differential 1 over T
equals differential of the log of Psat.
Well, this equation is known
as the Clausius-Clapeyron equation,
and it's often seen in an integrated form
where we first make the assumption
that delta H of vaporization is constant.
And so over, not too large to temperature rains,
that's a pretty good approximation.
And then you'll see the integrated form between two limits,
temperature T2 and temperature T1
corresponds to saturation pressures P2 and P1.
And so this also is, if you like another form
of a Clausius-Clapeyron equation
that relates the saturation pressure to temperature.
So if we know heated vaporization,
and we know the saturation pressure,
at one temperature,
we can calculate the saturation pressure,
another temperature.