TUTOR: Here I'm going to describe
the simplest cubic equation of state
which is the Van der Waals equation.
And then do a couple calculations,
to demonstrate how a real gas
can really differ significantly from an ideal gas.
So first, let's write down the Van der Waals equation.
This relates pressure
in terms of ideal gas constant,
absolute temperature, the volume
and this is specific volume,
volume per mole minus a constant.
This constant is positive.
And then a second constant of again,
specific volume squared.
So this, the Vander der Waals equation.
Lemme pause and label each term.
So I've labeled,
the terms are of course being the gas constant.
So this constant B relates to the fact that
molecules actually occupy space.
So ideal gas law, we assume molecules occupy no space,
we can compress a gas down as far as we'd like.
But for a real gas,
as we compress and get molecules close to each other,
we can't get that volume close to zero.
And so, this term B is accounting for that fact.
So as we try and compress in this term V minus B,
as V gets close to B that term of course becomes small
which means the pressure will end up being
much higher than ideal gas pressure.
The second term is related to the fact
that there are attractive interactions between molecules.
And so because A is positive, then the second term
with the negative sign in front of it
means the pressure due to attractive forces,
the pressures may lower
than what we expect from ideal gas law.
These constants can be related
to critical pressures and critical temperatures.
So A can be written as ideal gas constant squared.
The critical temperature,
of course, absolute degrees like Kelvin,
and the critical pressure and temperature
you need for each molecule.
So the A terms may be different for a different molecule.
And the B term, again the gas constant,
critical temperature over eight times the critical pressure.
So let's look at some simple calculations.
And we'll do these for carbon dioxide,
critical pressure 73.9 bar
and the critical temperature 304.2 Kelvin.
So I can calculate A and B,
and lemme just write those down.
So now let's do a couple calculations,
and we're gonna pick the temperature of 275 Kelvin.
And let's suppose the volume is 0.20.
And I wanna determine the pressure.
So the pressure if it were an ideal gas,
and I'm gonna report this three significant figures.
So this is ideal gas.
And now what's the pressure
for the Vander Waals equation state?
Well, it's 54.1 bar, dramatically different.
And the pressure is lower
because of attractive interactions
and because this negative term here
is what's most significant.
Correction, if you like the ideal gas law,
and so this is, let me pause and just write that down.
So let's look, again, we'll keep the temperature 275.
But let's look at a case where the volume is smaller.
Now, notice that this volume
is starting to get close to this B value.
So we're trying to push the molecules very close together.
We probably don't really consider this a gas,
but we can still apply the nine ideal
equation of state, this cubic equation,
the Van der Waals equation.
Again, the pressure if it were an ideal gas
for 57 bar, again, this is three significant figures
and the pressure Vander Waals equation, 1707 bar.
On now those, we have the opposite situation,
the real gas has a higher pressure than the ideal gas.
This because now,
this first term
is dominating where we have a specific volume
close to the B value.
So this denominator becomes very small,
takes very high pressure
to push these molecules close together,
probably not something we really consider a guess.
But the main point here is there are two contributions,
attractive forces and repulsive forces
in which one dominates and how much different we are
from ideal gas depends on what pressure
and temperature conditions we're at.