So there are n factorial different ways then, to make the same contribution to
the ensemble partition funcion. And if we choose to take q as the product
of the molecular partition functions, that is capital q the ensemble as that
product. Each being allowed to take on the various
values will over count that term. Will make it again, and again, and again
with different permutations. How many permutations?
N factorial of them. So we should divide by n factorial in
order to remove that that non-allowed distinguishability.
And come up with an ensemble partition function capital Q.
And we actually employed this. I told you I'd derive it, and now we
have. in, and it is appropriate then, for
Fermions and bosons, when they are indistinguishable particles.
So let me return to this question of, is it possible for two particles to be in
the same state, and the same energy level.
And the rule that one can use is, if this expression here, number of particles
divided by volume, times Planck's constant squared, divided by eight times
the mass times Boltzmann constant, times the temperature, all raised to the three
halves power. If that expression is considerably less
than one then Boltzmann statistics are valid the division by n factorial to
compute the ensemble partition function is valid the particles in the system will
obey Boltzmann statistics. Let's just explore how often that might
happen that this inequality is satisfied. And let's just take a simple system.
Molecular hydrogen as a gas, at one bar pressure.
And 300 Kelvin. So, under those conditions.
We can replace N over V using the ideal gas equation of state as P over kT.
And that allows us, then, to derive. The pressure in SI units would be 10 to
the 5th Pascal. The Boltzmann's constant is expressed
here, and the temperature. And that all adds up to 2.414 times 10 to
the 25th per cubit meter. Meanwhile, if we evaluate the quantity in
parentheses, we have h squared over 8 m k t.
Here's Planck's constant squared, 8, the mass of molecular hydrogen is 3.35 times
10 to the minus 27th kilograms. Here is Boltzmann's constant, here is 300
kelvin again. And the net when raised to the three, has
power 2.486 times 10 to the minus 31st cubic meters.
Multiplying these two quantities together, we end up with the full
expression is equal to 6 times 10 to the minus 6th.
Right, so six one-millionths, which certainly is a great deal less than one.
Then hence, we would expect Boltzmann's statistics to be satisfied for hydrogen
gas at this pressure. And, indeed, we see, in order to get up
to a value closer to one, given that pressure appears here linearly.
And this is about hm, it's about a, let's round it and say it's one one hundredth
thousands. That would be ten to the minus fifth.
You would need to go to a hundred thousand times higher pressure.
That'd be one hundred thousand bar. So that's a whole lot of pressure, and
obviously everyday systems are not under that kind of pressure.
Alright we've worked with the ensemble partition function.
That's great and now have a better feel for that en-factorial in the denominator
and its dependence on the molecular partition function.
So next we're going to look at the molecular partition function and talk
about its construction.