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PROFESSOR: Last time, under the
tutelage of Professor Blendi,

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you started to see some of the
consequences, the more

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immediate consequences, of some
of the properties we've been

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discussing over the last few
lectures of binary

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liquid mixtures.
 

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And in particular he started
showing you some of the

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colligative properties by which
you see changes in the vapor

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pressure and the boiling point
and so forth, that arise

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because you've got liquids in
mixtures and this affects

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their chemical potentials.
 

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And today I want to start by
going through the fourth

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colligative property that's
commonly seen, what

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called osmotic pressure.
 

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And I mentioned this before.
 

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Today we'll go through a
quantitative calculation

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of its effects.
 

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And then we'll start in on a
different unit of the course.

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We'll start in on
statistical mechanics.

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So let's continue with the
colligative properties.

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And osmotic pressure plays a
super-important role, certainly

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in all of biology, because
of course inside biological

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cells you have solutions.
 

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Water is in there, along
with a whole bunch

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of other components.
 

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And so the chemical potential
of the water is strongly

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affected by all the
other constituents.

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And this has profound effects
on what's involved in

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maintaining an equilibrium
between what goes on inside

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and outside of the cell.
 

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And this is the case
in all organisms.

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And it's also exploited
of course in all sorts

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of other contexts.
 

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So what's happening with
osmotic pressure is, you've got

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some situation where there's a
semi-permeable membrane.

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It allows one of the
constituents but not

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others to pass through.
 

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And typically that's water.
 

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Although, of course,
it could be anything.

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So, in the lab the way
this typically looks is

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something like this.
 

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Here's your
semi-permeable membrane.

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Outside, we can imagine that
there's pure liquid A.

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Could be water, could
be something else.

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Inside, you've got A mixed
with B, one or more

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other constituents.
 

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And then there's some level
of the liquid outside

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the container.
 

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Of course, A can pass between,
pass through, the membrane.

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And so what ends up happening
is, if you just look at the

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situation, what you immediately
notice is that the liquid in

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here goes up higher than
the liquid around it.

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It's at a higher level.
 

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It seems counterintuitive
at first.

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And in fact it can be
enormously higher.

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But that's what happens.
 

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So there's extra height, h, of
liquid inside the container

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relative to what's outside it.
 

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And what that means is
the pressure inside here

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must be higher, right?
 

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Of course, the weight of
this liquid is pushing

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down on what's down here.
 

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So the pressure inside
here is higher than

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the pressure outside.
 

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But of course, from what you
know already about liquid

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mixtures, this should
in fact make sense.

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Because what's happening is
that if these things are at the

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same pressure, you know that
the chemical potential of A is

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going to be lower
in the mixture.

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It's always lower in the
mixture than it is in pure A.

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So there's nothing to
stop A from rushing in.

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So A does rush in.
 

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If the liquid level goes up, it
raises the pressure down here.

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But as the pressure rises,
now the chemical potential

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changes and goes up.
 

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So eventually, equilibrium
is reestablished.

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Because these aren't
at the same pressure.

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The additional pressure means
the chemical potential

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of A goes up.
 

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Of course, it goes down by
virtue of being in the mixture.

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And those two things will
eventually reach an equilibrium

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where the chemical potential of
A is the same everywhere,

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as it has to be.
 

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So what we'll calculate
now is, OK, how much

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does the pressure go up.
 

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When that happens.
 

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So let's take a look.
 

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And so of course, there are
all kinds of examples.

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In a typical one,
A would be water.

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And B would be one
or more sugars.

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All sorts of other potential
constituents also.

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OK, so let's just consider
the pressure at two places

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that are the same level.
 

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The same height.
 

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We'll label this
one point alpha.

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Let's put some color in here.
 

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We'll have points
alpha and beta.

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And we'll make this distance l.
 

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So now let's just first
calculate what the pressure

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is at both points.
 

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It's going to be higher
here, because of this

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additional liquid on top.
 

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So, if we look at point
alpha, well, let's start

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at point beta, actually.
 

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That's the simpler one
since it doesn't have

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the additional liquid.
 

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And first of all, there's
just atmospheric pressure.

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So there's an
external pressure.

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Certainly, typically, it would
just be atmosphere pressure.

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It's pushing down here.
 

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It's also there.
 

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So that has to factor in.
 

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So there's our
external pressure.

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And then there is the product
of this height times the

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density of the solution.
 

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Times the gravitational
force constant.

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And I'll work through that
momentarily so you see that

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that indeed is the pressure.
 

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That's exerted by the
amount of liquid that's

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up above this level.
 

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But before I do that, let's put
the pressure at point alpha.

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So p alpha.
 

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And that's p external
plus l rho g.

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But now, there's this
additional height

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on top of that.
 

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So there's also h rho g.
 

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There's an extra pressure.
 

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We'll label this pi.
 

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So we can write p alpha is
equals to p beta plus h rho g.

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Which we'll say is equal
to p beta plus pi.

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Now, just to make sure we're
all on the same page about

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these quantities being the
pressure that's exerted

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by the liquid, let's
just look at that.

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So l times rho times g.
 

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It should be a pressure which
is a force per unit area.

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And in particular,
this is going to be a

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weight per unit area.
 

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Remember the weight is the
force that a mass exerts.

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And it exerts it because of
the gravitational constant.

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So what happens?
 

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Well, this is just a length.
 

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Rho is a density, right?
 

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So it's mass per unit volume.
 

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And of course, g is just our
gravitational force constant.

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So it'll be in units of
meters per second squared.

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So what's going to happen?
 

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If we multiply l times
rho together, this

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is units of length.
 

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This is units of volume.
 

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What we're going to get then
is mass per unit area.

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And that's what it
is, of course.

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It's distributing across
the area of that surface

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where point alpha is.
 

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So these two together,
l times rho, give us

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mass per unit area.
 

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So it could be in
units of kilograms.

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Per unit area, per
meters squared.

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And now we're going to multiply
that by the gravitational

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force constant.
 

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So now we're going to have
all together kilograms

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per meter second squared.
 

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In other words, it's
force per unit area.

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And so that's what we have.
 

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And now, of course, we have
the additional component.

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Due to the height here.
 

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Now, what do we know about
the chemical potential

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in both parts?
 

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The chemical potential has
to be equal outside and

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inside the container.
 

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So mu A at point alpha is just
mu A of the liquid at pressure

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p plus pi and temperature T.
 

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And that has to be equal
to mu A at point beta.

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But point beta is just
in the pure liquid.

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So that's just mu A star
of the liquid at pressure

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p and temperature T.
 

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So here's the chemical
potential at point beta.

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Here it is at point alpha.
 

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Those two have to be
equal to each other.

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And now we're just going
to use Raoult's law.

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And what does that tell us?
 

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We have RT log of xA plus mu A
star of the liquid at pressure

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p plus pi and T So there is
our chemical potential

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inside the container.
 

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And that's equal to mu
A star of the pure

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liquid at point beta.
 

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And so we can just rewrite
this as RT log xA.

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Plus mu A star liquid at
pressure p plus pi of T

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minus mu A star at pressure
p is equal to zero.

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OK., now we need the
pressure dependence of

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the chemical potential.
 

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But we certainly have an
expression for that.

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So we know that dG
is S dT plus V dp.

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The temperature is the
same on both sides.

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But we need to worry about what
happens at different pressures.

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So at constant T, dG is V dp.
 

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So now if we look at the
chemical potential, which is

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just the Gibbs free energy
per mole, then d mu A

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star is VA star molar dp.
 

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In other words, the difference
in the chemical potential

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is, this changes as a
function of pressure.

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Is going to be given by the
molar volume of a under

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these conditions in
the pure liquid dp.

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It's just the potential, the
pressure dependence of G.

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It's not the kind of topic
I would have thought would

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drive away a lot of people.
 

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But, you never know.
 

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OK.
 

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For the rest of you who
are willing to bear with

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me, let's continue.
 

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00:15:06 --> 00:15:08
So now let's just integrate.
 

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We need to know the change in
mu A at a finite jump in

220
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pressure from inside to
outside the container.

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So we're just going
to integrate.

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00:15:19 --> 00:15:34
So if I want mu A star at
pressure p plus pi minus mu A

223
00:15:34 --> 00:15:42
star at pressure p, then I just
have to integrate from p to

224
00:15:42 --> 00:15:50
p plus pi VA bar star dp.
 

225
00:15:50 --> 00:15:52
And now I'm going to assume,
and this is certainly a

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safe assumption, that this
quantity this, molar volume,

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00:15:56 --> 00:15:57
isn't going to change.
 

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00:15:57 --> 00:16:00
In other words, the molar
volume of liquid A isn't going

229
00:16:00 --> 00:16:04
to change significantly, going
from the pressure out here

230
00:16:04 --> 00:16:05
to the pressure in here.
 

231
00:16:05 --> 00:16:08
On the grand scale of things
it's a small pressure change.

232
00:16:08 --> 00:16:11
We can assume that the liquid
is incompressible over that

233
00:16:11 --> 00:16:13
small pressure change.
 

234
00:16:13 --> 00:16:15
Which means that this is
constant, since all we

235
00:16:15 --> 00:16:27
have then is simply
VA bar star times pi.

236
00:16:27 --> 00:16:32
So then, substituting back
here, so now we've calculated

237
00:16:32 --> 00:16:35
this difference, it's
a simple result.

238
00:16:35 --> 00:16:42
So then we simply have that
RT log xA plus VA bar

239
00:16:42 --> 00:16:50
star pi is equal to zero.
 

240
00:16:50 --> 00:16:55
Now, you saw last time, and
I'll work through this quickly

241
00:16:55 --> 00:17:04
again, that the log of xA can
be written approximately as

242
00:17:04 --> 00:17:11
minus nB over nA if you
remember, this came from

243
00:17:11 --> 00:17:17
writing that this is equal to
the log of one minus xB.

244
00:17:17 --> 00:17:21
But we're assuming that B is
the minor constituent here.

245
00:17:21 --> 00:17:25
So xB is a small number.
 

246
00:17:25 --> 00:17:30
So that this is roughly
equal to minus xB.

247
00:17:30 --> 00:17:38
And that's just minus
nB over nA plus nB.

248
00:17:38 --> 00:17:41
And since nB is much smaller
than nA, this is approximately

249
00:17:41 --> 00:17:48
equal to minus nB over nA.
 

250
00:17:48 --> 00:17:51
So we're going to make
this substitution.

251
00:17:51 --> 00:17:58
Also, this quantity, this molar
volume of A over the pure

252
00:17:58 --> 00:18:05
liquid, since the concentration
of B is low, we can assume

253
00:18:05 --> 00:18:09
that this is just the molar
volume of A in general.

254
00:18:09 --> 00:18:16
So, in other words, we
can write VA bar star.

255
00:18:16 --> 00:18:21
We can just write it as VA bar.
 

256
00:18:21 --> 00:18:24
That is, we're not going to
worry about changes in the

257
00:18:24 --> 00:18:28
molar volume, either as a
function of pressure or a

258
00:18:28 --> 00:18:30
function of concentration at
the low concentrations

259
00:18:30 --> 00:18:32
that we're working.
 

260
00:18:32 --> 00:18:39
And then, note that nA
times VA per mole is

261
00:18:39 --> 00:18:41
just the total volume.
 

262
00:18:41 --> 00:18:43
Oops, not bar.
 

263
00:18:43 --> 00:18:44
Not the molar quantity.
 

264
00:18:44 --> 00:18:46
This is the molar quantity
multiplied by the

265
00:18:46 --> 00:18:49
number of moles.
 

266
00:18:49 --> 00:18:54
So it just gives us VA, the
total volume, occupied by A.

267
00:18:54 --> 00:19:05
And using these two results,
then we have that RT times

268
00:19:05 --> 00:19:11
negative nB over nA,
that's this result.

269
00:19:11 --> 00:19:21
Plus VA over nA times pi is
equal to zero. nA will cancel.

270
00:19:21 --> 00:19:27
And finally, since almost all
the volume is due to A, again

271
00:19:27 --> 00:19:31
because B is the minor
constituent, we can approximate

272
00:19:31 --> 00:19:39
further that VA is
approximately equal to V.

273
00:19:39 --> 00:19:44
So finally getting rid of the
nA on the denominators, we're

274
00:19:44 --> 00:19:58
left with a simple expression.
pi times V is RT nB.

275
00:19:58 --> 00:20:01
So we have a very simple
expression which is called the

276
00:20:01 --> 00:20:06
van't Hoff expression, and look
at how it resembles

277
00:20:06 --> 00:20:08
the ideal gas law.
 

278
00:20:08 --> 00:20:11
This is the pressure times
the volume equals the

279
00:20:11 --> 00:20:13
number of moles times RT.
 

280
00:20:13 --> 00:20:17
Of course, really it's
a change in pressure.

281
00:20:17 --> 00:20:20
And this number of moles is the
number of moles of a guest

282
00:20:20 --> 00:20:24
constituent in a solution.
 

283
00:20:24 --> 00:20:29
But it has the same form
that you're familiar with.

284
00:20:29 --> 00:20:33
We also, sometimes it's
convenient to substitute,

285
00:20:33 --> 00:20:40
since nB over V, that's
just the concentration.

286
00:20:40 --> 00:20:50
That's the number of
moles over the volume.

287
00:20:50 --> 00:20:53
And in that case, the
expression is rewritten as pi

288
00:20:53 --> 00:20:58
is RT times c, where it's
understood that this is the

289
00:20:58 --> 00:21:06
concentration of the solute.
 

290
00:21:06 --> 00:21:14
OK, any questions?
 

291
00:21:14 --> 00:21:17
Let's work through a
numerical problem.

292
00:21:17 --> 00:21:21
These things can seem a little
bit, they can seem simple, but

293
00:21:21 --> 00:21:25
sometimes when confronted with
a problem, sometimes

294
00:21:25 --> 00:21:27
it may seem less than
obvious what to do.

295
00:21:27 --> 00:21:30
And also the results can seem
a little bit surprising.

296
00:21:30 --> 00:21:33
So I just want to work through
a simple numerical example

297
00:21:33 --> 00:21:34
to see how these play out.
 

298
00:21:34 --> 00:21:51
So let's try, let's imagine we
have 10 grams of unknown solid.

299
00:21:51 --> 00:22:09
Dissolved in 1000
grams of water.

300
00:22:09 --> 00:22:22
The vapor pressure is 25.195
torr at 27 degrees C.

301
00:22:22 --> 00:22:35
And the pure water vapor
pressure is 25.000.

302
00:22:35 --> 00:22:43
No, .200 torr at the
same temperature.

303
00:22:43 --> 00:22:53
So first, let's just calculate
the molecular weight

304
00:22:53 --> 00:22:54
of the solid.
 

305
00:22:54 --> 00:22:58
Now, this you I think,
saw from last time.

306
00:22:58 --> 00:22:59
So this is straightforward.
 

307
00:22:59 --> 00:23:02
It's going to come from the
vapor pressure lowering not

308
00:23:02 --> 00:23:05
from the osmotic pressure,
which we'll do next.

309
00:23:05 --> 00:23:19
So the delta p of H2O is minus
xB times p star of H2O.

310
00:23:19 --> 00:23:21
Alright, this is
from last time.

311
00:23:21 --> 00:23:24
The expression for the
vapor pressure reduction.

312
00:23:24 --> 00:23:31
Due to the concentration of
the mole fraction of B.

313
00:23:31 --> 00:23:42
And then xB is 0.005
torr over 25.2 torr.

314
00:23:42 --> 00:23:45
Right?
 

315
00:23:45 --> 00:23:49
And that's 1.98 times
10 to the minus 4.

316
00:23:49 --> 00:23:57
So that's our mole
fraction of B.

317
00:23:57 --> 00:24:06
And then xB is nB
over nA plus nB.

318
00:24:06 --> 00:24:13
So, now we have this, it's 10
grams divided by the molecular

319
00:24:13 --> 00:24:23
weight, which is in grams per
mole, over 1000 grams of water.

320
00:24:23 --> 00:24:35
Divided by 18 grams per mole
plus 10 grams divided by the

321
00:24:35 --> 00:24:40
molecular weight in
grams per mole.

322
00:24:40 --> 00:24:44
But since the solution is
overwhelmingly water, we can

323
00:24:44 --> 00:24:47
get rid of this term and
make the math simple.

324
00:24:47 --> 00:24:48
Can you see that?
 

325
00:24:48 --> 00:24:52
I guess barely.
 

326
00:24:52 --> 00:25:00
So this turns out to be about
1.98 time 10 to the minus 4.

327
00:25:00 --> 00:25:04
And if we solve for molecular
weight, then we wind up

328
00:25:04 --> 00:25:13
with 907 grams per mole.
 

329
00:25:13 --> 00:25:16
So that's using one of the
colligative properties

330
00:25:16 --> 00:25:17
that you saw last time.
 

331
00:25:17 --> 00:25:27
Now let's, though, finish up
by calculating the additional

332
00:25:27 --> 00:25:30
pressure that we'll find.
 

333
00:25:30 --> 00:25:42
So what's the osmotic
pressure of the solution.

334
00:25:42 --> 00:25:45
So now we're going to assume
that we have that solution

335
00:25:45 --> 00:25:48
inside, in a situation like
this where there's pure water

336
00:25:48 --> 00:25:50
outside that can go in.
 

337
00:25:50 --> 00:25:53
And there's going to be excess
pressure because of the water

338
00:25:53 --> 00:25:55
rushing into the solution.
 

339
00:25:55 --> 00:26:01
So what happens?
 

340
00:26:01 --> 00:26:05
And let's use a value
of the density.

341
00:26:05 --> 00:26:15
0.995 grams per
centimeters cubed.

342
00:26:15 --> 00:26:24
So, pi is RT times c.
 

343
00:26:24 --> 00:26:35
That's 0.08314 bar
per Kelvin mole.

344
00:26:35 --> 00:26:48
Times 300 K times
c. c is nB over V.

345
00:26:48 --> 00:27:00
So it's 10 grams divided by
907 grams per mole, over 1010

346
00:27:00 --> 00:27:08
grams over 0.995 grams per
milliliter, I'm equating

347
00:27:08 --> 00:27:12
with centimeter cubed.
 

348
00:27:12 --> 00:27:17
And that turns out to be
1.09 times 10 to the minus

349
00:27:17 --> 00:27:22
5th mole per millileter.
 

350
00:27:22 --> 00:27:25
Let's get that into liters.
 

351
00:27:25 --> 00:27:31
So we'll multiply by 1000
milliliters per liter.

352
00:27:31 --> 00:27:39
So it's 1.09 times 10 to
the 5th moles per liter.

353
00:27:39 --> 00:27:40
OK, yeah.
 

354
00:27:40 --> 00:27:44
STUDENT: [INAUDIBLE]
 

355
00:27:44 --> 00:27:49
PROFESSOR: Let's see.
 

356
00:27:49 --> 00:27:51
Oh, don't forget that, right.
 

357
00:27:51 --> 00:27:56
STUDENT: [INAUDIBLE]
 

358
00:27:56 --> 00:27:58
PROFESSOR: Ah, yes, yes, yes.
 

359
00:27:58 --> 00:28:05
And I think that's the
case in the notes also.

360
00:28:05 --> 00:28:06
Yep.
 

361
00:28:06 --> 00:28:09
Thank you.
 

362
00:28:09 --> 00:28:13
OK, so the concentration,
just to calibrate us

363
00:28:13 --> 00:28:15
here, it's a pretty low
concentration, right?

364
00:28:15 --> 00:28:20
It's about 10 to the
minus 5 moles per liter.

365
00:28:20 --> 00:28:27
The excess pressure
is 0.27 bar.

366
00:28:27 --> 00:28:29
In other words, about a
quarter of an atmosphere.

367
00:28:29 --> 00:28:30
Uh-oh, what's happening?
 

368
00:28:30 --> 00:28:34
STUDENT: [INAUDIBLE]
 

369
00:28:34 --> 00:28:35
PROFESSOR: Where are we?
 

370
00:28:35 --> 00:28:38
STUDENT: [INAUDIBLE]
 

371
00:28:38 --> 00:28:40
PROFESSOR: Sorry,
where exactly?

372
00:28:40 --> 00:28:48
STUDENT: [INAUDIBLE]
 

373
00:28:48 --> 00:28:50
PROFESSOR: Oh, yes.
 

374
00:28:50 --> 00:28:53
Yes, sorry.
 

375
00:28:53 --> 00:28:54
Yeah.
 

376
00:28:54 --> 00:28:57
10 to the minus 2
moles per liter.

377
00:28:57 --> 00:28:58
All right.
 

378
00:28:58 --> 00:29:01
And so the pressure that
results is a modest

379
00:29:01 --> 00:29:01
pressure, right?
 

380
00:29:01 --> 00:29:04
It's about a quarter of an
atmosphere of excess pressure.

381
00:29:04 --> 00:29:13
2.7 time 10 to the 4 pascals.
 

382
00:29:13 --> 00:29:17
So let's just calculate, given
this small additional pressure,

383
00:29:17 --> 00:29:22
how high is the liquid going to
go, above the level

384
00:29:22 --> 00:29:26
of the solution.
 

385
00:29:26 --> 00:29:28
Anybody want to guess?
 

386
00:29:28 --> 00:29:29
You know, a centimeter.
 

387
00:29:29 --> 00:29:30
10 centimeters.
 

388
00:29:30 --> 00:29:32
100 centimeters?
 

389
00:29:32 --> 00:29:33
Give me an order of magnitude.
 

390
00:29:33 --> 00:29:35
Who guesses one centimeter?
 

391
00:29:35 --> 00:29:36
You got three choices.
 

392
00:29:36 --> 00:29:39
One, 10 and 100, and
you have to make one.

393
00:29:39 --> 00:29:41
Who guesses one centimeter?
 

394
00:29:41 --> 00:29:43
Who guesses 10 centimeters?
 

395
00:29:43 --> 00:29:46
Who guesses 100 centimeters?
 

396
00:29:46 --> 00:29:49
OK.
 

397
00:29:49 --> 00:29:53
Now, fortunately we
don't do problem sets.

398
00:29:53 --> 00:29:58
Or other scientific
problems by popular vote.

399
00:29:58 --> 00:29:59
Rather, we just solve them.
 

400
00:29:59 --> 00:30:01
Whenever possible.
 

401
00:30:01 --> 00:30:15
So, what's the height of
a column of solution?

402
00:30:15 --> 00:30:20
Well, that pi, remember,
that's equal to rho

403
00:30:20 --> 00:30:22
times g times h, right?
 

404
00:30:22 --> 00:30:24
And we know pi now.
 

405
00:30:24 --> 00:30:30
So h is pi over rho times g.
 

406
00:30:30 --> 00:30:39
So it's 2.7 times 10 to the 4
pascal over rho, that's 0.995

407
00:30:39 --> 00:30:43
grams per centimeters cubed.
 

408
00:30:43 --> 00:30:54
Times 1 kilogram over 1000
grams times 100 centimeters

409
00:30:54 --> 00:30:56
per meter cubed.
 

410
00:30:56 --> 00:31:00
I'm just getting this into
units that will give me, that

411
00:31:00 --> 00:31:06
will be in meters, times 9.8
meter per a second squared.

412
00:31:06 --> 00:31:13
And when I multiply all
that out, it's 2.8 meters.

413
00:31:13 --> 00:31:18
So that puny little pressure
ends up being meters high.

414
00:31:18 --> 00:31:21
Taller than any of us, by far.
 

415
00:31:21 --> 00:31:25
Sorry, it's up to there.
 

416
00:31:25 --> 00:31:26
And that's actually
pretty typical.

417
00:31:26 --> 00:31:31
And what that means is, it
actually makes osmotic pressure

418
00:31:31 --> 00:31:35
a very, very sensitive
measurement method.

419
00:31:35 --> 00:31:39
If you want to determine a
molecular weight, for example.

420
00:31:39 --> 00:31:41
Because it's so high, that
means, and of course you

421
00:31:41 --> 00:31:43
can measure that height
pretty accurately.

422
00:31:43 --> 00:31:46
Certainly within
about a millimeter.

423
00:31:46 --> 00:31:50
And so even relatively modest
concentrations, it's still

424
00:31:50 --> 00:31:52
easy to see the effect
and measure the impact.

425
00:31:52 --> 00:31:58
And use that to calculate the
properties of the solute.

426
00:31:58 --> 00:32:02
OK, any questions?
 

427
00:32:02 --> 00:32:07
Alright.
 

428
00:32:07 --> 00:32:10
Now we're going to
start something new.

429
00:32:10 --> 00:32:17
So, so far what you've seen,
and I hope come to appreciate,

430
00:32:17 --> 00:32:22
is the whole structure of
thermodynamics is built on

431
00:32:22 --> 00:32:27
empirical macroscopic
observation and deduction.

432
00:32:27 --> 00:32:29
We observe things.
 

433
00:32:29 --> 00:32:33
We formulate these broad
thermodynamic laws.

434
00:32:33 --> 00:32:34
All macroscopic.
 

435
00:32:34 --> 00:32:37
It doesn't depend on any
microscopic model at all.

436
00:32:37 --> 00:32:40
And in fact much of it was
formulated before there was

437
00:32:40 --> 00:32:42
a good microscopic
model of matter.

438
00:32:42 --> 00:32:45
Before the atomic theory of
matter, and molecules, and so

439
00:32:45 --> 00:32:50
forth was well worked out.
 

440
00:32:50 --> 00:32:54
And it's incredible how
powerful that whole

441
00:32:54 --> 00:32:56
formalism is, right?
 

442
00:32:56 --> 00:33:01
In some sense, although it may
seem like neglecting what we

443
00:33:01 --> 00:33:06
now know to be an important
part of nature, actually part

444
00:33:06 --> 00:33:09
of its power is its empiricism.
 

445
00:33:09 --> 00:33:09
Right?
 

446
00:33:09 --> 00:33:13
There are these very small
number of fundamental laws from

447
00:33:13 --> 00:33:16
which everything else follows.
 

448
00:33:16 --> 00:33:18
And at this point, of course,
there's just enormous

449
00:33:18 --> 00:33:21
confidence in that
small number of laws.

450
00:33:21 --> 00:33:24
Because of their being
verified in so many context.

451
00:33:24 --> 00:33:30
But, at the same time, we
do know about the atomic

452
00:33:30 --> 00:33:31
theory of matter.
 

453
00:33:31 --> 00:33:33
We know there are
atoms and molecules.

454
00:33:33 --> 00:33:38
So it ought to be possible, at
least in principle, to start

455
00:33:38 --> 00:33:42
from a purely microscopic
approach to nature.

456
00:33:42 --> 00:33:48
And just based on figuring out
the microscopic properties.

457
00:33:48 --> 00:33:51
And then saying, well, OK, my
macroscopic stuff is just

458
00:33:51 --> 00:33:55
a collection of those
microscopic entities.

459
00:33:55 --> 00:33:59
I should also be able to figure
out macroscopic thermodynamics.

460
00:33:59 --> 00:34:02
I should be able to start
from my microscopic picture

461
00:34:02 --> 00:34:06
and get to macroscopic
thermodynamic results.

462
00:34:06 --> 00:34:08
And in fact, that is possible.
 

463
00:34:08 --> 00:34:11
And the theoretical formulation
for it is what's called

464
00:34:11 --> 00:34:13
statistical mechanics.
 

465
00:34:13 --> 00:34:17
Called that because, of course,
you won't be surprised to learn

466
00:34:17 --> 00:34:21
that it's going to require
a statistical treatment.

467
00:34:21 --> 00:34:23
We're going to be dealing
with moles of material.

468
00:34:23 --> 00:34:26
But now we're going to be
trying to think about their

469
00:34:26 --> 00:34:28
microscopic properties.
 

470
00:34:28 --> 00:34:33
And so we'll be dealing not
with n number of moles, but

471
00:34:33 --> 00:34:37
with 10 to the 24th number
of molecules or atoms.

472
00:34:37 --> 00:34:40
And you know we won't be
able to keep track of every

473
00:34:40 --> 00:34:44
one of their individual
states, all the time.

474
00:34:44 --> 00:34:48
We may be able to do a terrific
calculation of the quantum

475
00:34:48 --> 00:34:51
mechanics or the classical
mechanics that describe the

476
00:34:51 --> 00:34:54
states that they could be in.
 

477
00:34:54 --> 00:34:56
But there's no way in practice
that we're either going to

478
00:34:56 --> 00:35:01
experimentally or theoretically
keep track of all of that.

479
00:35:01 --> 00:35:03
So we're going to, at
some point, have to

480
00:35:03 --> 00:35:05
introduce statistics.
 

481
00:35:05 --> 00:35:09
To take what we know about the
microscopic properties and try

482
00:35:09 --> 00:35:11
to go from there to the
macroscopic results.

483
00:35:11 --> 00:35:14
The thermodynamics that
we've seen so far.

484
00:35:14 --> 00:35:34
And that's what statistical
mechanics is all about.

485
00:35:34 --> 00:35:42
So let's try to introduce
a little bit of

486
00:35:42 --> 00:35:44
statistical mechanics.
 

487
00:35:44 --> 00:35:48
Where we're going to go from
what we know about microscopic

488
00:35:48 --> 00:35:51
properties all the way to
macroscopic thermodynamics.

489
00:35:51 --> 00:35:58
That's our objective.
 

490
00:35:58 --> 00:36:04
So let's start by just trying
to calculate energies of

491
00:36:04 --> 00:36:08
individual molecules, or
individual particles.

492
00:36:08 --> 00:36:14
And what's the probability that
some molecule, one of the

493
00:36:14 --> 00:36:17
oxygen molecules somewhere
in this room, is in a

494
00:36:17 --> 00:36:18
certain energy state.
 

495
00:36:18 --> 00:36:20
Right?
 

496
00:36:20 --> 00:36:24
And what our strategy is going
to be, is to determine what are

497
00:36:24 --> 00:36:27
the probabilities that
molecules are in certain

498
00:36:27 --> 00:36:29
states with certain energies.
 

499
00:36:29 --> 00:36:33
And if we can determine that
for all the possible states,

500
00:36:33 --> 00:36:36
then we can average over those.
 

501
00:36:36 --> 00:36:40
So without keeping track of
every individual molecule, we

502
00:36:40 --> 00:36:43
could then calculate, an
average energy, which is

503
00:36:43 --> 00:36:45
what you would measure
thermodynamically when you look

504
00:36:45 --> 00:36:48
at the whole collection and
measure what we call u, right?

505
00:36:48 --> 00:36:51
Of course, we're really
averaging over disparate

506
00:36:51 --> 00:36:54
energies of lots of different
atoms or molecules.

507
00:36:54 --> 00:36:58
They don't all have the
same molecular energy.

508
00:36:58 --> 00:37:01
And we don't try to measure
their individual energies.

509
00:37:01 --> 00:37:03
So that's what we'd
like to calculate.

510
00:37:03 --> 00:37:07
And so we'd like to be
able to know what are all

511
00:37:07 --> 00:37:10
these probabilities of
different energy states.

512
00:37:10 --> 00:37:14
And then from that,
statistically averaging,

513
00:37:14 --> 00:37:17
what are the macroscopic
average energies.

514
00:37:17 --> 00:37:19
And other macroscopic
quantities.

515
00:37:19 --> 00:37:22
So, let's start there.
 

516
00:37:22 --> 00:37:41
Probability that a molecule,
must be specific as possible

517
00:37:41 --> 00:37:50
is in state i with energy Ei.
 

518
00:37:50 --> 00:37:53
And right now we're not
even going to specify

519
00:37:53 --> 00:37:55
the nature of the state.
 

520
00:37:55 --> 00:37:59
We could be worrying about
translational energy.

521
00:37:59 --> 00:38:02
What state it is, how fast it's
whizzing around the room.

522
00:38:02 --> 00:38:05
We could worry about its
vibrational or rotational

523
00:38:05 --> 00:38:07
energy, or electronic state.
 

524
00:38:07 --> 00:38:09
For now, let's not
even specify it.

525
00:38:09 --> 00:38:12
Let's just say the only thing
we're really specifying about

526
00:38:12 --> 00:38:19
the state is it has some energy
that we presumably know.

527
00:38:19 --> 00:38:27
Now, what we'd like is to
know a functional form

528
00:38:27 --> 00:38:29
for the probability.
 

529
00:38:29 --> 00:38:32
And we don't know one a priori.
 

530
00:38:32 --> 00:38:38
But let's just think
about two molecules.

531
00:38:38 --> 00:38:42
And the probability that one
of them has energy i, and

532
00:38:42 --> 00:38:45
another one has energy Ej.
 

533
00:38:45 --> 00:38:48
It also, let me say, would be
sufficient to think about even

534
00:38:48 --> 00:38:51
one molecule and say, we'll
let's think about independent

535
00:38:51 --> 00:38:52
parts of its energy.
 

536
00:38:52 --> 00:38:54
Let's say, translational
energy in this direction.

537
00:38:54 --> 00:38:56
And in an orthogonal direction.
 

538
00:38:56 --> 00:39:02
What matters is, we're thinking
about two independent energies.

539
00:39:02 --> 00:39:05
Could be separate molecules
that aren't interacting.

540
00:39:05 --> 00:39:08
Or independent degrees of
freedom on the same molecule.

541
00:39:08 --> 00:39:12
But let's just consider it.
 

542
00:39:12 --> 00:39:25
So a molecule is in state i,
and that another molecule is

543
00:39:25 --> 00:39:35
in state j, with energy Ej.
 

544
00:39:35 --> 00:39:38
So what we're going to
calculate is the joint

545
00:39:38 --> 00:39:42
probability that the two
molecules are in those states.

546
00:39:42 --> 00:39:46
So we want a probability.
 

547
00:39:46 --> 00:39:52
This one, we'll call Pi(Ei).
 

548
00:39:52 --> 00:39:58
This would be Pj of Ej.
 

549
00:39:58 --> 00:40:00
And this probability together
will be Pij(Ei + Ej).

550
00:40:00 --> 00:40:05
 


551
00:40:05 --> 00:40:09
I could just write Ei comma
Ej, but the energy's add.

552
00:40:09 --> 00:40:14
And that's crucial.
 

553
00:40:14 --> 00:40:17
Now, since the molecules are
completely independent, this

554
00:40:17 --> 00:40:29
should just be the product of
the separate probabilities.

555
00:40:29 --> 00:40:34
Two completely independent
events, or things whose

556
00:40:34 --> 00:40:37
probability we want to
determine together.

557
00:40:37 --> 00:40:41
So, that suggests a really
simple functional form.

558
00:40:41 --> 00:40:48
Because what we're saying here
is we want a function of a plus

559
00:40:48 --> 00:40:54
b, which is equal to the same
sort of function of a times the

560
00:40:54 --> 00:40:55
same sort of function of b.
 

561
00:40:55 --> 00:40:59
In other words, when we
multiply these functions,

562
00:40:59 --> 00:41:05
we get a function of the
sum of those arguments.

563
00:41:05 --> 00:41:11
What functional form does that?
 

564
00:41:11 --> 00:41:12
Exponential.
 

565
00:41:12 --> 00:41:17
Sure.
 

566
00:41:17 --> 00:41:23
So in other words, e to the
a plus b power equals e to

567
00:41:23 --> 00:41:27
the a times e to the b.
 

568
00:41:27 --> 00:41:35
So it suggests that there's
an exponential probability.

569
00:41:35 --> 00:41:37
Suggests it.
 

570
00:41:37 --> 00:41:49
Now, there's also some
physical insight that we

571
00:41:49 --> 00:41:51
can apply to the problem.
 

572
00:41:51 --> 00:41:57
First of all, we expect that
states with higher energy in

573
00:41:57 --> 00:41:59
general are going to be less
probable than states

574
00:41:59 --> 00:42:02
with lower energy.
 

575
00:42:02 --> 00:42:06
If it's cold, not much thermal
energy around, and you say, how

576
00:42:06 --> 00:42:09
much energy does a particle in
an equilibrium collection of

577
00:42:09 --> 00:42:14
particles have, at thermal
equilibrium, if there's not

578
00:42:14 --> 00:42:17
much thermal energy around, you
don't expect it to be very

579
00:42:17 --> 00:42:19
likely that a particle has
a huge amount of energy.

580
00:42:19 --> 00:42:21
That one individual
particle has it.

581
00:42:21 --> 00:42:22
Could happen, right?
 

582
00:42:22 --> 00:42:26
From some random collisions
that just happened to

583
00:42:26 --> 00:42:28
bulk up the energy of
that one particle.

584
00:42:28 --> 00:42:31
But on a balance it's not
going to be very probable.

585
00:42:31 --> 00:42:34
Lower energies will
be more probable.

586
00:42:34 --> 00:42:37
And the other thing is, even in
thinking about something like

587
00:42:37 --> 00:42:39
that, temperature comes in.
 

588
00:42:39 --> 00:42:40
Immediately.
 

589
00:42:40 --> 00:42:42
Right?
 

590
00:42:42 --> 00:42:45
If you raise the temperature a
lot, surely we have to expect

591
00:42:45 --> 00:42:48
many more molecules will
start to become energized.

592
00:42:48 --> 00:42:50
Will have more energy.
 

593
00:42:50 --> 00:42:55
So, lower probability
for higher energy.

594
00:42:55 --> 00:42:57
But the probability of higher
energy should go up if

595
00:42:57 --> 00:42:59
the temperature goes up.
 

596
00:42:59 --> 00:43:03
In other words, the ratio
of energy to temperature

597
00:43:03 --> 00:43:06
should be involved here.
 

598
00:43:06 --> 00:43:20
So, P of Ei should go
down as Ei goes up.

599
00:43:20 --> 00:43:28
And should depend on Ei over T.
 

600
00:43:28 --> 00:43:30
Right?
 

601
00:43:30 --> 00:43:35
So, we can start with
this functional form.

602
00:43:35 --> 00:43:40
And by the way, there's no
reason we can't have, in

603
00:43:40 --> 00:43:42
general we would have some
constants here, right?

604
00:43:42 --> 00:43:45
We haven't changed
anything to do that.

605
00:43:45 --> 00:43:49
And now let's just use the
little bit of physical

606
00:43:49 --> 00:43:54
intuition that we're thinking
about here, to refine

607
00:43:54 --> 00:43:55
this just a little bit.
 

608
00:43:55 --> 00:44:04
Let's write, we expect that Pi
of Ei is some exponential to

609
00:44:04 --> 00:44:12
the minus C Ei over T, where
C is some constant

610
00:44:12 --> 00:44:13
greater than zero.
 

611
00:44:13 --> 00:44:17
So this will have the property,
then, that as energy goes up,

612
00:44:17 --> 00:44:18
the probability goes down.
 

613
00:44:18 --> 00:44:20
But it's scaled by temperature.
 

614
00:44:20 --> 00:44:22
If I raise the temperature,
that makes the higher

615
00:44:22 --> 00:44:30
energy states get
more and more likely.

616
00:44:30 --> 00:44:41
Well, this is our functional
form for probability of

617
00:44:41 --> 00:44:44
a molecule being in a
state with energy Ei.

618
00:44:44 --> 00:44:47
And the only difference between
this and what's written

619
00:44:47 --> 00:44:51
conventionally is the way
the constant is labeled.

620
00:44:51 --> 00:45:00
So really what we have is Pi of
Ei is proportional to e to the

621
00:45:00 --> 00:45:13
minus Ei over kB T, or just
k T, where kB is called

622
00:45:13 --> 00:45:17
the Boltzmann constant.
 

623
00:45:17 --> 00:45:21
And it's just equal to R
over Avogadro's number.

624
00:45:21 --> 00:45:30
It's the gas constant per
molecule, rather than per mole.

625
00:45:30 --> 00:45:37
One way to try to rationalize
this is, you've probably seen,

626
00:45:37 --> 00:45:40
you've all seen Arrhenius
kinetics, right?

627
00:45:40 --> 00:45:44
Arrhenius rate laws?
 

628
00:45:44 --> 00:45:53
If you remember what that
looks like, you get

629
00:45:53 --> 00:45:56
this rate constant.
 

630
00:45:56 --> 00:46:00
Arrhenius rate constant is
some constant A times e

631
00:46:00 --> 00:46:05
to the minus Ea over RT.
 

632
00:46:05 --> 00:46:07
Remember that?
 

633
00:46:07 --> 00:46:10
You've all seen that before?
 

634
00:46:10 --> 00:46:15
And so what's happening
here, Ea is what is

635
00:46:15 --> 00:46:18
an activation energy.
 

636
00:46:18 --> 00:46:23
Remember, the idea is that
you've got reactants and

637
00:46:23 --> 00:46:26
products, and there's some
barrier you've got to get over.

638
00:46:26 --> 00:46:29
The activation energy, before
you can have reaction.

639
00:46:29 --> 00:46:32
So the rate depends
on surmounting that

640
00:46:32 --> 00:46:36
activation barrier.
 

641
00:46:36 --> 00:46:42
This is really coming from the
same idea as this, which is the

642
00:46:42 --> 00:46:45
probability of one of
the molecules having

643
00:46:45 --> 00:46:47
this much energy.
 

644
00:46:47 --> 00:46:51
Depends on e to the
minus energy over RT.

645
00:46:51 --> 00:46:53
This is per mole, so
this is per mole.

646
00:46:53 --> 00:46:55
The exact same relation.
 

647
00:46:55 --> 00:47:00
So the idea in the context
of kinetics is that the

648
00:47:00 --> 00:47:04
rate depends on how many
molecules can get up here.

649
00:47:04 --> 00:47:07
And what the probability is of
their getting enough energy

650
00:47:07 --> 00:47:10
to go over the barrier.
 

651
00:47:10 --> 00:47:12
And that energy, that
probability, is given

652
00:47:12 --> 00:47:13
by this expression.
 

653
00:47:13 --> 00:47:16
So in this form you've seen
this kind of dependence before

654
00:47:16 --> 00:47:29
in a very explicit way.
 

655
00:47:29 --> 00:47:36
Let me just take it one small
but important step farther.

656
00:47:36 --> 00:47:42
Which is that
proportionality constant.

657
00:47:42 --> 00:47:45
Couldn't we do better?
 

658
00:47:45 --> 00:47:49
Couldn't we say, couldn't
we figure out what exactly

659
00:47:49 --> 00:47:52
it is, not just what
it's proportional to?

660
00:47:52 --> 00:47:53
Well, let's try.
 

661
00:47:53 --> 00:48:00
So as we've written it, we've
got Pi of Ei is proportional

662
00:48:00 --> 00:48:03
to e to the minus Ei over kT.
 

663
00:48:03 --> 00:48:08
We could write that as equals
A e to the minus Ei over kT.

664
00:48:08 --> 00:48:12
This is just some
proportionality constant.

665
00:48:12 --> 00:48:16
But we can determine what that
constant is, because we know

666
00:48:16 --> 00:48:22
that if we sum over all the
possible states, the molecule

667
00:48:22 --> 00:48:25
has to be in some state, right?
 

668
00:48:25 --> 00:48:30
So that sum of all those
probabilities has to equal one.

669
00:48:30 --> 00:48:37
So sum over i of Pi of
Ei, is equal to one.

670
00:48:37 --> 00:48:44
Because the molecule must
be in one of the states.

671
00:48:44 --> 00:48:49
OK, now this is our expression
for all those p i's.

672
00:48:49 --> 00:48:53
So let's just write that way. a
is just a constant times the

673
00:48:53 --> 00:49:03
sum over all the i's of e
to the minus Ei over kT.

674
00:49:03 --> 00:49:09
And so there it is. a is equal
to one over the sum over i,

675
00:49:09 --> 00:49:17
e to the minus Ei over kT.
 

676
00:49:17 --> 00:49:27
So now we can rewrite Pi of Ei
is equal to e to the minus Ei

677
00:49:27 --> 00:49:41
over kT over, the sum over i,
e to the minus Ei over kT.

678
00:49:41 --> 00:49:47
Now, just so you're not
confused, this i matters.

679
00:49:47 --> 00:49:48
That's the i we're
talking about.

680
00:49:48 --> 00:49:51
What's the probability of it
being in some particular

681
00:49:51 --> 00:49:53
state i with energy e i.
 

682
00:49:53 --> 00:49:55
This i is just a
dummy variable.

683
00:49:55 --> 00:49:57
So just to be explicit, we
could write this just as well

684
00:49:57 --> 00:50:03
as e to the minus Ei over
kT times the sum over j, e

685
00:50:03 --> 00:50:09
to the minus Ej over kT.
 

686
00:50:09 --> 00:50:11
It could be confusing,
but this is usually the

687
00:50:11 --> 00:50:13
way that it's written.
 

688
00:50:13 --> 00:50:16
Even though this dummy variable
here has nothing to do with the

689
00:50:16 --> 00:50:23
particular choice
that's made here.

690
00:50:23 --> 00:50:24
So now we know.
 

691
00:50:24 --> 00:50:29
And we can calculate what the
probability is for a molecule

692
00:50:29 --> 00:50:31
to be in any particular state
if we know the energy

693
00:50:31 --> 00:50:32
of that state.
 

694
00:50:32 --> 00:50:34
Now, notice we need to know
all the energies of all

695
00:50:34 --> 00:50:36
the other states, too.
 

696
00:50:36 --> 00:50:42
Because this thing, in other
words, this a, right, depends

697
00:50:42 --> 00:50:45
on summing over all of those.
 

698
00:50:45 --> 00:50:48
But if we know those,
then we can do it.

699
00:50:48 --> 00:50:49
We can do the whole
calculation.

700
00:50:49 --> 00:50:52
And for a good number of
cases, we realistically

701
00:50:52 --> 00:50:55
do know it all.
 

702
00:50:55 --> 00:51:00
So from the proportionality
relationship alone, of course,

703
00:51:00 --> 00:51:04
we can tell the ratio of
chances that you're in

704
00:51:04 --> 00:51:06
one state or another.
 

705
00:51:06 --> 00:51:11
Even without this a constant,
we already could've said well,

706
00:51:11 --> 00:51:15
if what's the ratio of, the
probability of being in state i

707
00:51:15 --> 00:51:21
with energy Ei to state
j with energy Ej.

708
00:51:21 --> 00:51:26
Well, it's just e to the
minus Ei over kT over e

709
00:51:26 --> 00:51:29
to the minus Ej over kT.
 

710
00:51:29 --> 00:51:37
It's e to the minus
Ei minus Ej over kT.

711
00:51:37 --> 00:51:39
But apart from just being able
to get the ratio, we can

712
00:51:39 --> 00:51:42
get the absolute number.
 

713
00:51:42 --> 00:51:44
What's the absolute probability
that a molecule is in any

714
00:51:44 --> 00:51:50
particular state of
a given energy.

715
00:51:50 --> 00:51:54
Next time what you'll see is
what can be calculated based

716
00:51:54 --> 00:51:57
on the results that you've
seen, just the results

717
00:51:57 --> 00:52:00
you've seen so far.
 

718
00:52:00 --> 00:52:05
From these quantities alone,
it turns out you'll be able

719
00:52:05 --> 00:52:07
to calculate every
single macroscopic

720
00:52:07 --> 00:52:12
thermodynamic quantity.
 

721
00:52:12 --> 00:52:14
So you'll see some
of that next time.

722
00:52:14 --> 00:52:15