0:07
So we've just seen that there is a relationship between an ensemble
partition function. And an equation of state by having worked
with the trial partition function shown here.
Which involves a molecular or atomic partition function combined with a Nth
power divided by N factorial assembly. Of the ensemble partition function, where
these taken to the nth power and dividing by N factorial are associated with
non-interacting indistinguishable particles.
And we showed that when we took that ensemble partition function and solved
for the pressure, we recovered the ideal gas equation of state.
So in today's video I'd like to look at a different partition function.
0:59
And that's the one shown here. And so it continues to be a function of
number of particles, volume, beta. There's a 1 over N factorial.
There is a term 3N over 2 power that continues to include within it things
like the mass of the gas, here's beta, here's Planck's constant.
V minus N times r all to the Nth power and the exponential of s beta N squared
over V. And these two new parameters if you will
new constants r and s are positive constants.
And so that defines an ensemble partition function.
And you see the dependence on N and on V and on beta and I'll give you a moment
here to think a little bit. And work with that partition function and
see some limiting behavior. And then we'll come back and work some
more with it. All right.
So you've seen how this partition function reduces under certain limiting
behavior. And one thing I might point out here.
It's tempting to notice that here you have a term that involves something
raised to the Nth power, the 3N over 2 but there's an N if you like, in the
exponent. Here's something raised to the Nth power.
And hiding in this exponential is an Nth power.
It's N squared but I could take exponential of the N, take all that to
the N. Remember that a power of a power is like
something to the product of those two powers.
so it looks in a sense as though you could write this ensemble partition
function as some primitive partition function all raised to the Nth power.
Not unlike the ideal gas case. That's incorrect, though.
You shouldn't view it that way. There is not a molecular partition
function here and that's because it would not depend only on volume and beta.
There would still be a dependence on N in this term.
There would still be a dependence on N in this term and of course you can't have a
molecular partition function that depends on how many particles there are on in an
ensemble. That, that doesn't make logical sense.
So this particular ensemble partition function doesn't decompose into a
molecular component. It's just good for the ensemble.
But let's continue to work with it in particular.
Let's explore the expectation value of the pressure, which will be derived by
taking the partial derivative of the log of the partition function.
With respective volume and multiplying by kT.
So first, let's expand the log of this ensemble partition function.
And so if I take the log and I separate out all the various terms, the products,
the powers, they multiply the log, I'll get 3 halves as I pull this 3 half,
sorry, 3N over 2, 3 halves N, multiplying all the material inside that logarithm.
So 2 pi m in the numerator minus h squared and beta in the denomenator plus
again a power will come down N log this term.
Plus the log of an exponential is just the argument of the exponential and then
in ally here is the minus log N factorial here in this term.
So let's just move that to the next slide to continue working with it.
And if we differentiate that now with respect to volume.
Well, happily nothing in this term depends on volume, that's going to go
away. Here are the only two terms that depend
on volume and those are pretty simple differentiations to do.
So we'll end up with partial log Q, partial V is N over V minus N times r
minus S beta and squared, all divided by V squared.
4:42
That is we're going to multiply times kT. That derivative and we end up then with
pressure is equal to NkT over V minus Nr minus sN squared over V squared.
So let me rearrange that equation a bit and in particular, I'll take the minus sN
squared over V squared, I'll bring it over on the other side and add it to
pressure. And then I'll multiply both sides, times
V minus NR. So here's the multiplier over here and
that removes the term and the denominator over here.
And finally, let's take N, the number of particles equal to Avogadros number so
that we'll be working with molar quantities.
And we'll express a and b in molar units in which case, I will get P plus a over V
bar squared times V bar minus b. Equals what is Avogadro's number times
Boltzmann's constant, that's the universal gas constant, RT.
And so you see that we have recovered the Van de Waals Equation of State.
So that partition function is the ensemble partition that is consistent
with the Van de Waals Equation of State. So with that in hand, seeing that
derivation of course we can work with the partition function in other ways as well.
So I think I'll pause here for a moment and let you consider the internal energy
of a Van de Waals gas. Good we've had the opportunity now to see
the relationship between an ensemble partition function and an associated
equation of state for two cases. The ideal gas and the Van der Wall's gas.
It gives you some indication of how powerful an entity, the ensemble
partition function can be. So let's continue to work with it more in
the next video. [SOUND]
Dr. Christopher J. CramerDistinguished McKnight and University Teaching Professor of Chemistry and Chemical Physics
