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PROFESSOR NELSON: Well, today
we're going to continue with

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the lecture that was started
last time that had the

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scintillating and descriptive
title "some thermodynamic

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processes".
 

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And we're going to continue
along those lines, and really

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there a couple of things that
I want to do to follow up

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what you saw last time.
 

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One is that you've got
introduced and led through

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a couple of elementary
thermodynamic cycles.

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And I want to do a little
bit more of that today.

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And one of the -- there
are really two reasons.

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One is that as you saw last
time doing a thermodynamic

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cycle can be extremely useful,
because if you have one path

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along which some change occurs,
but along which it's not

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necessarily easy to calculate
all of the thermodynamic

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changes along such a path, it
may be irreversible, lots

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of variables may change.
 

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It can be easier instead of
looking at the complicated path

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to devise a cycle that involves
several easier steps, each one

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of which is facile for the
calculation of what happens to

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the thermodynamic properties.
 

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So in that case, it's much
easier to go through several

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elementary step, and then the
one complicated path you know

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because going around the entire
cycle state functions, you

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know that they won't
have any change.

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So in that sense, going through
a thermodynamic cycle can be

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useful because it helps you
calculate thermodynamic

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qualities.
 

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But the other reason to go
through the thermodynamic

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cycles and really to develop
great facility with them is

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because there are just an awful
lot of things in nature and

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things that we build that
run in cycles, where we

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want to calculate the
thermodynamics, right.

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If we build an engine for a car
or anything else, it almost

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always is going to have some
key element that's operating

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in a cycle, otherwise it
won't keep going, right.

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But it's not just engines
that we might build.

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It's biological engines, right.
 

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Things that either produce or
consume biological fuel to do

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different sorts of processes.
 

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Also, of course of key
importance to understand what

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the, in that case, biological
thermodynamics are.

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And again, there'll be key
elements that run in cycles

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and understanding those
can be extremely important

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in understanding how
cellular function works.

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OK, so what I'd like to do is
go through one cycle, and then

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I'll suggest another for you
to work through on your own.

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So consistent with the title
of the lecture, we've got a

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section from your last lecture
notes entitled "some

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thermodynamic cycles".
 

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And what we're going to do is
go through a cycle, please

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note these office hours and
locations, which I think

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wasn't specified before.
 

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So we're going to go through a
thermodynamic cycle, and here's

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what I want to calculate
when we do this.

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Delta u, delta H, familiar
state functions, changes

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in their values, q,
w, heat and work.

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And I also want to introduce
one new function.

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I won't give it a name yet,
but it's going to have the

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unusual form dq over T.
 

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OK, we'll call it
special function.

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It's so special that we'll
call this quantity dS.

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OK, we'll just go through this
on a whim, and this is going to

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be foreshadowing something
that'll take on tremendous

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importance in subsequent
lectures, but I'm going to use

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the opportunity now to just
introduce the behavior of it

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in going around a cycle.
 

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If that's for me, tell
them I'm busy please.

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OK, so let's look at the
cycle we'll go through.

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We're going to look at pressure
volume changes, similar

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to what you saw before.
 

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This time, we're going to have
initial and final volumes, V1

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and V2 and different pressures
where we might end up.

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And we're going to look
at two ways to go to

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a final volume V2.
 

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One of them is going to end up
at pressure p2 and the other is

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going to end up at pressure p3.
 

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One of them is going to be a
reversible isothermal path.

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This is going to be our path
we'll label A, reversible

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isothermal which is to
say constant T, right?

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And we're going to start at
temperature T1, so since it's

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constant temperature, this is
going to end up at

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temperature T1.
 

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And then we're also going
to consider a reversible

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adiabatic path.
 

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This'll be our path B,
reversible adiabatic.

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And then we'll close this
circle, this cycle.

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This is going to end up at
a different temperature

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by the way.
 

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You saw this last time in
a slightly different way.

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Last time what you saw is we
compared isothermal and

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adiabatic paths that ended up
at the same final pressure,

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and what you saw is that
therefore, they ended up in

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different final volumes.
 

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So this is just a little
bit different from that.

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So this is going to end up
at a different temperature,

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we'll call it T2.
 

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And then we're going
to close the cycle.

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And so this is a constant
volume path then.

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This is path C,
constant volume.

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OK?
 

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So now let's go around the
cycle and just compare notes

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on what happens to the
thermodynamic quantities

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as we do that.
 

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So here's path A,
it's isothermal.

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And I didn't specify, but let's
make sure to do so, we've got

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a, it's going to
be an ideal gas.

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OK, so A is constant
temperature.

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What does that mean for
delta u in path A?

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Zero right?
 

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Ideal gas, no temperature
change, right has to be zero

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because da is Cv dT for an
ideal gas. dT is zero, there's

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no temperature change.
 

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OK, how about delta
H, zero Cp dT for an

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ideal gas. dT is zero.
 

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All right, now,
it's reversible.

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So that means we can
immediately write down an

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expression for the work.
 

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Since it's reversible,
it's negative p dV right?

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And since it's an ideal gas
then we can replace p, right,

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with RT over V, right?
 

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We want to do that because we
have too many variables here.

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We've already got dV we'll get
rid of p as an additional

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variable and replace it with
V which is already in here.

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So it's minus R T1
dV over V, right?

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I can put in T1 because we're a
constant temperature. so now we

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can just integrate straight
away and find that the work in

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path A it's just the integral
of that quantity minus integral

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of R T1 going from V1 to V2 dV
over V. so it's minus

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R T1 log V2 over V1.
 

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00:10:02 --> 00:10:06
All right, so let's just
do a reality check here.

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00:10:06 --> 00:10:08
The way I've got this
drawn, the volume is

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going up in the process.
 

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It's an expansion.
 

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So is this system doing work
on the surroundings, or are

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the surroundings doing
work on the system?

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Somebody say it real loud.
 

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Yes, the system is working
on the surroundings, right?

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It's pushing out against them.
 

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Work is defined as the work
that the surroundings

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do on the system.
 

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So this is the negative
number, right?

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V2 is bigger than V1.
 

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00:10:42 --> 00:10:43
This is positive.
 

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This whole thing is negative.
 

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All right, q has to be just the
opposite of this because we've

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already figured out that
there's no change in u.

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This is just q plus w.
 

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There's w, q has to be R
T1 log of V2 over V1.

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OK, so that means heat is being
imparted to the system, right,

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from the surroundings in a
manner that compensates exactly

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the amount of work done.
 

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OK, All right, so these are the
thermodynamic quantities that

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you're familiar with already.
 

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Let's just quickly look
at our special function.

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It's not going to be hard to
calculate because it's a

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constant temperature path, so
that T in the bottom

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is just T1, right.
 

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You've already
looked at q, right.

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So dq A over T1 it's just
R log V2 over V1 right.

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So there's our special function
for this particular path.

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OK, so that's path A.
 

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Now let's compare to path B.
 

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So if path B is an adibat,
what's zero? q, it's adiabatic,

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that means there's no heat
exchanged between the system

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and the surroundings.
 

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So what happens, q B is zero.
 

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There is a change in
temperature, right?

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We've got, if we say it's a
mole of gas, it's going from

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p1, V1, and T1 to one mole of
our gas at p3, V2, and T2.

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So unlike before now
temperature is going to change.

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So that means that delta
u isn't zero this time.

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du, it's an ideal gas.
 

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So this is Cv dT and of
course we can just integrate

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00:13:51 --> 00:13:53
this straight away.
 

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So delta u B is Cv times
T2 minus T1, right. and

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similarly dH is Cp dT right.
 

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So delta H in pathway B is
just Cp T2 minus T1 okay?

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All right, what we've
already got that q is zero.

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Delta u is q plus w.
 

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Delta u isn't zero.
 

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So this must be equal
to work, right?

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And now we can look at our
special function, but since

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this is zero right dq B is
zero, along path B, so our

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special function, dS, so
we're going to go, dq B

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over T is equal to zero.
 

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OK?
 

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OK, finally let's look at our
third path, this constant

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volume path, that's going to
connect T1 and T2, right?

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Constant volume, what's zero?
 

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STUDENT: [INAUDIBLE]
 

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PROFESSOR NELSON: A
little more noise here.

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What's zero?
 

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STUDENT: Work.
 

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PROFESSOR NELSON: Work, great.
 

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So, path C constant volume
means our work in past

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00:16:01 --> 00:16:05
C is zero, right?
 

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00:16:05 --> 00:16:11
Now, our temperature is going
to change from T2 to T1, and we

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00:16:11 --> 00:16:16
just saw what happens to u and
H when that happens going from

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T1 to T2, and it's
no different.

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00:16:17 --> 00:16:20
It's an ideal gas going
along these path.

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00:16:20 --> 00:16:25
So we can immediately
write delta u C is Cv

211
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times T1 minus T2.
 

212
00:16:28 --> 00:16:40
Delta Hc C is Cp times
T1 minus T2, right?

213
00:16:40 --> 00:16:42
Work is zero.
 

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00:16:42 --> 00:16:47
Delta u is w plus
q, work plus heat.

215
00:16:47 --> 00:16:49
This is zero, this isn't.
 

216
00:16:49 --> 00:16:55
This must be heat q B, right.
 

217
00:16:55 --> 00:16:58
So again, there are our
familiar thermodynamic

218
00:16:58 --> 00:17:04
quantities, but now let's also
look at our special function.

219
00:17:04 --> 00:17:08
It's not going to be zero this
time because we have non zero

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00:17:08 --> 00:17:10
heat exchange between
the system and the

221
00:17:10 --> 00:17:12
environment, right.
 

222
00:17:12 --> 00:17:26
So are integral of dq C over T,
right, which is our heat, but

223
00:17:26 --> 00:17:29
in this case we can do this
calculation easily because

224
00:17:29 --> 00:17:36
since work is zero, we can
equate dq with du, right.

225
00:17:36 --> 00:17:44
So we can write this as our
integral from T1 to T2, du over

226
00:17:44 --> 00:18:06
T. du is just Cv dT, right.
 

227
00:18:06 --> 00:18:15
So this is just Cv
log T2 over T1.

228
00:18:15 --> 00:18:18
OK?
 

229
00:18:18 --> 00:18:26
So now we've
completed the cycle.

230
00:18:26 --> 00:18:28
So let's just compare.
 

231
00:18:28 --> 00:18:33
Let's compare what happened
in path A to what happened

232
00:18:33 --> 00:18:35
in paths B and C.
 

233
00:18:35 --> 00:18:43
Yes?
 

234
00:18:43 --> 00:18:43
STUDENT: [INAUDIBLE]
 

235
00:18:43 --> 00:18:46
PROFESSOR NELSON: Ah
sorry, yes of course.

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00:18:46 --> 00:18:56
Thank you.
 

237
00:18:56 --> 00:19:04
Yes, oh yes, so.
 

238
00:19:04 --> 00:19:05
STUDENT: [INAUDIBLE]
 

239
00:19:05 --> 00:19:07
PROFESSOR NELSON: And it's
right because, thank you

240
00:19:07 --> 00:19:11
again, it's because we're
going from T1 to T2.

241
00:19:11 --> 00:19:17
So I'm getting confused
by the oh sorry, we're

242
00:19:17 --> 00:19:20
going from T2 to T1.
 

243
00:19:20 --> 00:19:24
So it's in reverse of the
order of the subscripts

244
00:19:24 --> 00:19:27
and I let this confuse me.
 

245
00:19:27 --> 00:19:31
There it is.
 

246
00:19:31 --> 00:19:37
There it is.
 

247
00:19:37 --> 00:19:39
There.
 

248
00:19:39 --> 00:19:44
Thank you, any other details
which should be pointed out?

249
00:19:44 --> 00:19:46
All right, then let's proceed.
 

250
00:19:46 --> 00:19:49
So what we're going to do is
our comparison of what happened

251
00:19:49 --> 00:19:54
on pathway A to what happened
in combined pathways

252
00:19:54 --> 00:19:59
B and C, right?
 

253
00:19:59 --> 00:20:06
So here are results for
pathway A, right, for

254
00:20:06 --> 00:20:09
delta u zero delta H zero.
 

255
00:20:09 --> 00:20:18
I didn't actually explicitly
write it or did I?

256
00:20:18 --> 00:20:20
Let's just write it again.
 

257
00:20:20 --> 00:20:26
Here's our work in path
A, and here's our heat

258
00:20:26 --> 00:20:27
exchange in path A.
 

259
00:20:27 --> 00:20:31
Now let's look at
the sum of B and C.

260
00:20:31 --> 00:20:48
So B plus C, here is delta
uB Cv times T2 minus T1.

261
00:20:48 --> 00:20:57
Delta uC Cv times T1 minus
T2, the sum is zero right.

262
00:20:57 --> 00:21:01
That's the some of delta u in
these two paths, and if we

263
00:21:01 --> 00:21:05
look at delta u in just
path A, it's zero.

264
00:21:05 --> 00:21:07
And it's going to be the
same without delta H.

265
00:21:07 --> 00:21:12
The only difference is it'll
be Cp instead of Cv, but

266
00:21:12 --> 00:21:14
there it is for pathway B.
 

267
00:21:14 --> 00:21:17
There it is for a pathway C.
 

268
00:21:17 --> 00:21:20
So the state functions that
we're familiar with are doing

269
00:21:20 --> 00:21:23
what we expect they ought
to be doing, right?

270
00:21:23 --> 00:21:28
If you go around in a cycle,
starting and ending at the same

271
00:21:28 --> 00:21:30
place, the state functions
have to stay the same.

272
00:21:30 --> 00:21:32
They only depend on the
state the system is in.

273
00:21:32 --> 00:21:36
They don't depend on the
path, and that's what

274
00:21:36 --> 00:21:38
we've shown, right.
 

275
00:21:38 --> 00:21:41
Similarly if we go from this
point to this point through

276
00:21:41 --> 00:21:44
path A or if we do it through
the combination of these two

277
00:21:44 --> 00:21:48
paths, and the change in u
and H in any state function,

278
00:21:48 --> 00:21:49
those have to be the same.
 

279
00:21:49 --> 00:21:52
It turns out they're zero
in both cases, and that's

280
00:21:52 --> 00:21:56
what we've seen, right?
 

281
00:21:56 --> 00:22:01
Now let's compare what
happens to work and heat.

282
00:22:01 --> 00:22:04
So here are expressions
for work and heat.

283
00:22:04 --> 00:22:10
They're opposites
for the pathway A.

284
00:22:10 --> 00:22:13
Here's pathway B and C.
 

285
00:22:13 --> 00:22:17
In C the work is zero.
 

286
00:22:17 --> 00:22:24
In B it isn't, it's Cv
times T2 minus T1, right.

287
00:22:24 --> 00:22:31
The work isn't equal to the
work in pathway A, right,

288
00:22:31 --> 00:22:35
because you know work is not a
state function it depends on

289
00:22:35 --> 00:22:39
the path right, and there are
different amounts of work done

290
00:22:39 --> 00:22:44
on the system, or done by the
system on the surroundings in

291
00:22:44 --> 00:22:46
these two different processes.
 

292
00:22:46 --> 00:22:47
They're both expansions.
 

293
00:22:47 --> 00:22:50
They'll both have net work
done on the surroundings, but

294
00:22:50 --> 00:22:53
not the same amount, right.
 

295
00:22:53 --> 00:22:54
And of course it's going
to be the same with heat.

296
00:22:54 --> 00:22:58
We've already seen
that delta u is zero.

297
00:22:58 --> 00:23:00
So we know that in each case
the heat is going to be the

298
00:23:00 --> 00:23:03
opposite of the work, but the
work isn't the same in these

299
00:23:03 --> 00:23:08
two different ways of getting
from here to here, right.

300
00:23:08 --> 00:23:10
So let's just see
it explicitly.

301
00:23:10 --> 00:23:12
Here's our qA.
 

302
00:23:12 --> 00:23:18
Here's heat exchanged in
pathway A and in pathway B heat

303
00:23:18 --> 00:23:27
is zero, and in pathway C, here
is qC it's Cv T1 minus T2.

304
00:23:27 --> 00:23:29
So again, for both heat
and work we don't

305
00:23:29 --> 00:23:32
get the same result.
 

306
00:23:32 --> 00:23:36
Now let's look at our
special function, right.

307
00:23:36 --> 00:23:39
So here's path A.
 

308
00:23:39 --> 00:23:42
We found that it's
R log V2 over V1.

309
00:23:42 --> 00:23:46
 


310
00:23:46 --> 00:23:49
Pathway B is zero.
 

311
00:23:49 --> 00:23:53
Pathway C it's Cv
log T1 over T2.

312
00:23:53 --> 00:24:08
All right, let's just look at
that a little more carefully.

313
00:24:08 --> 00:24:12
R log V2 over V1.
 

314
00:24:12 --> 00:24:20
 


315
00:24:20 --> 00:24:31
B plus C, Cv log of
T1 over T2, right.

316
00:24:31 --> 00:24:37
Just some weird combination
of functions, right?

317
00:24:37 --> 00:24:42
But, of course, it's
not quite like that.

318
00:24:42 --> 00:24:44
Why?
 

319
00:24:44 --> 00:24:50
Because this path is
an adiabatic path.

320
00:24:50 --> 00:24:54
And you already saw last time
there was this relationship

321
00:24:54 --> 00:24:59
between the temperature and
volume changes along

322
00:24:59 --> 00:25:01
an adiabatic path.
 

323
00:25:01 --> 00:25:07
So let me just write that down.
 

324
00:25:07 --> 00:25:16
Adiabatic reversible path.
 

325
00:25:16 --> 00:25:26
What you saw is that T1
over T2 was V2 over V1

326
00:25:26 --> 00:25:32
to the power R over Cv.
 

327
00:25:32 --> 00:25:36
Isn't that something?
 

328
00:25:36 --> 00:25:39
So what does that mean?
 

329
00:25:39 --> 00:25:45
We take this Cv and put in
the exponent here, right.

330
00:25:45 --> 00:25:50
And we put this R up
in the exponent here.

331
00:25:50 --> 00:25:51
What are we going to discover?
 

332
00:25:51 --> 00:25:58
Those two things are equal
right, T1 over T2 to the power

333
00:25:58 --> 00:26:10
Cv is V2 over V1 to the power
R, so dS over path A equals dS

334
00:26:10 --> 00:26:17
over paths B plus C, okay?
 

335
00:26:17 --> 00:26:23
So what this suggests is that
whatever this funny special

336
00:26:23 --> 00:26:27
function is, at least for
this one cycle that we've,

337
00:26:27 --> 00:26:33
tried it's behaving like
a state function, right.

338
00:26:33 --> 00:26:38
It seems like its change
is independent of path.

339
00:26:38 --> 00:26:41
Start here, end up here, do it
through two different pathways,

340
00:26:41 --> 00:26:45
end up in the same
place, right.

341
00:26:45 --> 00:26:50
OK, now that's all the
foreshadowing I'm going to

342
00:26:50 --> 00:26:51
give it for right now.
 

343
00:26:51 --> 00:26:53
We will certainly come
back to this very special

344
00:26:53 --> 00:26:56
function shortly.
 

345
00:26:56 --> 00:26:59
Before I move on, I'm just
going to put on the board

346
00:26:59 --> 00:27:04
another cycle, and I'm
going to urge you to work

347
00:27:04 --> 00:27:05
through that on your own.
 

348
00:27:05 --> 00:27:10
It's worked through already,
let's see, I believe it's

349
00:27:10 --> 00:27:17
worked through in
the notes, yes.

350
00:27:17 --> 00:27:21
So if you need the help of the
notes it's in there, but I

351
00:27:21 --> 00:27:26
would urge you to work through
it on your own, and it's the

352
00:27:26 --> 00:27:49
following: let's start with
the same path A, all right?

353
00:27:49 --> 00:28:00
So we've got our reversible
adiabatic path right.

354
00:28:00 --> 00:28:06
And now, try working through
what happens if we close

355
00:28:06 --> 00:28:09
the cycle in a different
way, right, like this.

356
00:28:09 --> 00:28:18
So here's a path that's
constant pressure, and here is

357
00:28:18 --> 00:28:24
a path that's constant volume,
similar to the constant volume

358
00:28:24 --> 00:28:28
path that we did before, but
not between the same

359
00:28:28 --> 00:28:30
pressures, right.
 

360
00:28:30 --> 00:28:37
So we can label this one D and
this one E, and I urge you to

361
00:28:37 --> 00:28:41
just try going through that and
verifying for yourself that

362
00:28:41 --> 00:28:46
again the state functions will
behave the way state functions

363
00:28:46 --> 00:28:51
should, and you can see whether
this special function once

364
00:28:51 --> 00:28:54
again behaves like a
state function, right.

365
00:28:54 --> 00:28:57
And of course you should expect
to see that the work and the

366
00:28:57 --> 00:29:00
heat again won't behave like
state functions which they are,

367
00:29:00 --> 00:29:02
then you'll see you have
different results for those,

368
00:29:02 --> 00:29:07
depending on the path, OK?
 

369
00:29:07 --> 00:29:11
Any questions about going
through these cycles and

370
00:29:11 --> 00:29:16
using expressions like
this and so forth?

371
00:29:16 --> 00:29:20
Expressions like this you know
turn up in various places, like

372
00:29:20 --> 00:29:22
in the equations sheet that
appears at the end

373
00:29:22 --> 00:29:24
of exams, right?
 

374
00:29:24 --> 00:29:27
Which means that you
may not need to be

375
00:29:27 --> 00:29:30
committing it to memory.
 

376
00:29:30 --> 00:29:33
On the other hand it means that
you should, it's the sort of

377
00:29:33 --> 00:29:36
thing that you should be
familiar with so that the need

378
00:29:36 --> 00:29:40
to use it is something that
you'll conjure when you're

379
00:29:40 --> 00:29:43
working on a problem and
there's a reversible adiabat,

380
00:29:43 --> 00:29:46
and you have temperatures and
volumes that change that you

381
00:29:46 --> 00:29:53
might like to relate
to each other.

382
00:29:53 --> 00:29:56
Cycles that are suggested that
you go through and aren't gone

383
00:29:56 --> 00:29:59
through in class also sometimes
turn up on exams

384
00:29:59 --> 00:30:00
too, by the way.
 

385
00:30:00 --> 00:30:06
So that said, let's move on
to the next topic, which

386
00:30:06 --> 00:30:08
is thermochemistry.
 

387
00:30:08 --> 00:30:14
So, so far what you've seen in
most of the examples in the

388
00:30:14 --> 00:30:19
class are essentially
mechanical kinds of changes,

389
00:30:19 --> 00:30:22
pressure volume sorts of
changes, where the system is

390
00:30:22 --> 00:30:24
doing essentially a kind of
mechanical work, pressure

391
00:30:24 --> 00:30:28
volume work on the surroundings
or vice versa and heat is being

392
00:30:28 --> 00:30:31
exchanged, and you're
calculating the basic

393
00:30:31 --> 00:30:33
thermodynamics just like
we went through, of

394
00:30:33 --> 00:30:36
processes of that sort.
 

395
00:30:36 --> 00:30:40
Now I'd like to start
introducing something that's

396
00:30:40 --> 00:30:43
really central to chemical
thermodynamics, mainly

397
00:30:43 --> 00:30:45
chemistry, right.
 

398
00:30:45 --> 00:30:48
Let's start looking at chemical
reactions, right, and

399
00:30:48 --> 00:30:54
understanding what the changes
in thermodynamic properties are

400
00:30:54 --> 00:30:57
that occur when chemical
reactions occur, and the

401
00:30:57 --> 00:31:01
immediate application is to
calculate the change in

402
00:31:01 --> 00:31:04
enthalpy, which, as you've seen
it, at constant pressure which

403
00:31:04 --> 00:31:09
is the condition under which an
awful lot of chemistry is done,

404
00:31:09 --> 00:31:11
that's just the heat.
 

405
00:31:11 --> 00:31:13
Which means we're going
to calculate heats

406
00:31:13 --> 00:31:16
of reaction, right.
 

407
00:31:16 --> 00:31:17
You're running some reaction.
 

408
00:31:17 --> 00:31:19
It's in the atmosphere.
 

409
00:31:19 --> 00:31:21
It's at constant pressure.
 

410
00:31:21 --> 00:31:23
You know, you mix acid and
base together and the thing

411
00:31:23 --> 00:31:26
heats up like crazy, right.
 

412
00:31:26 --> 00:31:28
Or other things might
react spontaneously.

413
00:31:28 --> 00:31:31
You can feel something
cooling off right.

414
00:31:31 --> 00:31:34
I mean simple examples of these
happen when you, you know, if

415
00:31:34 --> 00:31:36
you buy cold or hot packs.
 

416
00:31:36 --> 00:31:38
You break the seal between
them and feel the thing get

417
00:31:38 --> 00:31:42
cold, for example, right.
 

418
00:31:42 --> 00:31:43
What's happening there?
 

419
00:31:43 --> 00:31:46
Well, the selection of
reactants has been done

420
00:31:46 --> 00:31:50
judiciously to provide
either heat or to provide

421
00:31:50 --> 00:31:51
something that's cool.
 

422
00:31:51 --> 00:31:56
And all of that is coming out
of the heat of reaction,

423
00:31:56 --> 00:31:59
whether it's positive
or negative.

424
00:31:59 --> 00:32:04
Of course the biggest practical
application is burning of fuel.

425
00:32:04 --> 00:32:08
You might use the heat to,
and convert it into pressure

426
00:32:08 --> 00:32:10
volume work, right?
 

427
00:32:10 --> 00:32:12
So if it's an internal
combustion engine,

428
00:32:12 --> 00:32:13
you'll do a reaction.
 

429
00:32:13 --> 00:32:15
There will be the
heat of reaction.

430
00:32:15 --> 00:32:18
There'll be a volume change,
depending on the conditions

431
00:32:18 --> 00:32:22
under which is occurs, and
you can extract work through

432
00:32:22 --> 00:32:24
the reaction and so forth.
 

433
00:32:24 --> 00:32:27
So let's just to see how
you run through some

434
00:32:27 --> 00:32:28
of those calculations.
 

435
00:32:28 --> 00:32:34
Now let me just ask
who here took 5.111?

436
00:32:34 --> 00:32:36
And who here took 5.112?
 

437
00:32:36 --> 00:32:43
OK, now you've seen
some thermal chemistry

438
00:32:43 --> 00:32:46
in 5.111 and 5.112.
 

439
00:32:46 --> 00:32:48
So I I'm going to do some
review of that, but I'm

440
00:32:48 --> 00:32:51
also going to call on the
thermochemistry that I'm pretty

441
00:32:51 --> 00:32:53
sure you're familiar with.
 

442
00:32:53 --> 00:32:57
So I won't go painstakingly
over every element

443
00:32:57 --> 00:32:58
of the notes here.
 

444
00:32:58 --> 00:33:02
Subsequently we'll go on into
additional topics that you

445
00:33:02 --> 00:33:24
haven't seen, and I'll
treat them in full detail.

446
00:33:24 --> 00:33:33
OK, so let's do some
thermochemistry.

447
00:33:33 --> 00:33:36
 


448
00:33:36 --> 00:33:40
What we want to do is we'd like
to be able to predict for any

449
00:33:40 --> 00:33:43
reaction what's the
heat of reaction.

450
00:33:43 --> 00:33:54
And so what we're going to
calculator is delta H.

451
00:33:54 --> 00:34:01
So for any kind of reaction
we should be able

452
00:34:01 --> 00:34:03
to do that, right?
 

453
00:34:03 --> 00:34:09
And we're going to treat
constant pressure situations

454
00:34:09 --> 00:34:17
certainly at first and
most of the time, right?

455
00:34:17 --> 00:34:31
Constant pressure, right,
reversible delta H is

456
00:34:31 --> 00:34:37
going to give us our
heat of reaction, OK.

457
00:34:37 --> 00:34:45
And then if we can also
determine delta u, then we know

458
00:34:45 --> 00:34:50
this, we know delta u is q plus
w, then we can determine

459
00:34:50 --> 00:35:07
work as well, right?
 

460
00:35:07 --> 00:35:14
And typically, we'll be
treating at least some cases

461
00:35:14 --> 00:35:16
where we're dealing with
ideal gases in which case we

462
00:35:16 --> 00:35:20
can easily get delta u.
 

463
00:35:20 --> 00:35:24
And then we'll be able to in a
very straightforward way get w.

464
00:35:24 --> 00:35:30
So it's a really powerful and
simple formalism, once we set

465
00:35:30 --> 00:35:34
up what is needed the
go through it, right?

466
00:35:34 --> 00:35:56
So let's just consider
any reaction.

467
00:35:56 --> 00:35:59
The one that I've got written
down for you is it's the

468
00:35:59 --> 00:36:02
reverse of the rusting
of iron, right?

469
00:36:02 --> 00:36:07
So iron left to own devices in
the atmosphere in the presence

470
00:36:07 --> 00:36:09
of a little bit of water, and
the atmosphere will start

471
00:36:09 --> 00:36:11
combining with the
water form iron oxide.

472
00:36:11 --> 00:36:13
There's an equilibrium
between the two.

473
00:36:13 --> 00:36:19
So we'll have an
iron oxide species.

474
00:36:19 --> 00:36:27
It's a solid.
 

475
00:36:27 --> 00:36:38
It's combining with hydrogen
which is a gas to give iron,

476
00:36:38 --> 00:36:49
also a solid and water, right?
 

477
00:36:49 --> 00:36:52
So the way I've written this
we're imagining the iron, the

478
00:36:52 --> 00:36:57
solid iron immersed in the
water as a liquid, could be

479
00:36:57 --> 00:37:00
calculated otherwise as well,
but in this case were imaging

480
00:37:00 --> 00:37:03
the water as liquid and going
in this direction then it's

481
00:37:03 --> 00:37:12
forming iron oxide and
evolving hydrogen gas, okay?

482
00:37:12 --> 00:37:21
And our heat of reaction or
enthalpy of reaction is

483
00:37:21 --> 00:37:26
defined as the enthalpy
at constant pressure.

484
00:37:26 --> 00:37:31
Isothermal conditions,
temperature won't change

485
00:37:31 --> 00:37:38
and reversible work.
 

486
00:37:38 --> 00:37:57
For Isothermal reaction
constant pressure

487
00:37:57 --> 00:38:00
reversible, right?
 

488
00:38:00 --> 00:38:02
Of course, they're all sorts of
conditions under which a

489
00:38:02 --> 00:38:05
reaction could be wrong in the
lab or outdoors or

490
00:38:05 --> 00:38:08
however, right.
 

491
00:38:08 --> 00:38:11
But this is the way we're going
to define delta H of reaction.

492
00:38:11 --> 00:38:14
We want to have that definition
clear because in fact we're

493
00:38:14 --> 00:38:17
going to, we might want
tabulate heats of reaction,

494
00:38:17 --> 00:38:20
right, and of course want to
know what the conditions are

495
00:38:20 --> 00:38:23
for the tabulated values apply.
 

496
00:38:23 --> 00:38:26
And we're going to want to
calculate them from other

497
00:38:26 --> 00:38:27
quantities, and again, we're
going to need to know

498
00:38:27 --> 00:38:30
each case what are the
relevant conditions?

499
00:38:30 --> 00:38:37
OK, so what happens?
 

500
00:38:37 --> 00:38:40
Well, we should be
able to get this.

501
00:38:40 --> 00:38:44
We should be able to
calculate delta H.

502
00:38:44 --> 00:38:46
It's a state function.
 

503
00:38:46 --> 00:38:53
If we know the enthalpy of the
products minus the enthalpy

504
00:38:53 --> 00:39:00
of the reactants, right.
 

505
00:39:00 --> 00:39:05
It's a state function.
 

506
00:39:05 --> 00:39:10
And we can we can do this in
principle except for one

507
00:39:10 --> 00:39:16
important detail, which is that
enthalpy, just like energy u

508
00:39:16 --> 00:39:19
isn't measured on an absolute
scale but on a relative

509
00:39:19 --> 00:39:21
scale, right?
 

510
00:39:21 --> 00:39:24
You know, if you want to
measure the potential energy of

511
00:39:24 --> 00:39:27
something in a gravitational
field, you have to define

512
00:39:27 --> 00:39:30
the zero somewhere, right,
because it's arbitrary.

513
00:39:30 --> 00:39:32
You can set it
anywhere you want.

514
00:39:32 --> 00:39:33
It's the same with enthalpy.
 

515
00:39:33 --> 00:39:36
Enthalpy is just u plus p V.
 

516
00:39:36 --> 00:39:41
So there is an arbitrary
set point that needs

517
00:39:41 --> 00:39:43
to be defined, right?
 

518
00:39:43 --> 00:39:47
Because what you actually
measure in the lab are changes

519
00:39:47 --> 00:39:51
in enthalpy, just like what you
measure when you look at energy

520
00:39:51 --> 00:39:54
change of some sort, you
measure the change

521
00:39:54 --> 00:39:55
in energy, right.
 

522
00:39:55 --> 00:39:59
The absolute number that you
assign to it is something

523
00:39:59 --> 00:39:59
that's arbitrary.
 

524
00:39:59 --> 00:40:03
You have to set
what the zero is.

525
00:40:03 --> 00:40:07
And so there's a
well-understood convention

526
00:40:07 --> 00:40:09
for what the zero is.
 

527
00:40:09 --> 00:40:17
What we define as zero is the
enthalpy of every element in

528
00:40:17 --> 00:40:20
its natural state at
room temperature and

529
00:40:20 --> 00:40:21
ambient pressure.
 

530
00:40:21 --> 00:40:26
In other words, we choose a
convention for the zero of

531
00:40:26 --> 00:40:30
entropy, so that then we can
write entropies of products and

532
00:40:30 --> 00:40:35
reactants always referring
to the same standard state.

533
00:40:35 --> 00:40:41
And then we calculate changes,
the convention is understood

534
00:40:41 --> 00:40:45
with respect to what
is the zero, right.

535
00:40:45 --> 00:40:48
And so our tabulated
values, they'll all work.

536
00:40:48 --> 00:40:52
They'll all refer to the
same standard state.

537
00:40:52 --> 00:40:55
And we'll be able to use the
formalism that we set up.

538
00:40:55 --> 00:41:37
So our reference point for H,
it's 298.15 Kelvin, one bar and

539
00:41:37 --> 00:41:44
in that standard state our
molar enthalpy is defined as

540
00:41:44 --> 00:41:59
zero for every element
in its stable form.

541
00:41:59 --> 00:42:05
OK, very important.
 

542
00:42:05 --> 00:42:15
OK, now, given that reference
point, all we need to do to get

543
00:42:15 --> 00:42:18
the value of enthalpy that
we're going to use for each

544
00:42:18 --> 00:42:25
reactant and each product is
calculate how much does the

545
00:42:25 --> 00:42:32
enthalpy change to form it from
the elements in their stable

546
00:42:32 --> 00:42:37
form at room temperature
and pressure, right.

547
00:42:37 --> 00:42:42
So we're going to replace the H
or we're going to put in for

548
00:42:42 --> 00:42:46
the H what we'll call our heat
of formation or delta H of

549
00:42:46 --> 00:42:49
formation starting from the
elements in their stable states

550
00:42:49 --> 00:42:51
at room temperature and
pressure, for each

551
00:42:51 --> 00:42:53
of these things.
 

552
00:42:53 --> 00:42:57
And we can do that because now
we're going to have instead of

553
00:42:57 --> 00:43:00
just sort of H, it's going to
be delta H of formation for

554
00:43:00 --> 00:43:03
each of these things, is going
to appear in our calculation

555
00:43:03 --> 00:43:06
of H, but that's OK.
 

556
00:43:06 --> 00:43:12
Delta H of formation means the
enthalpy of this compound minus

557
00:43:12 --> 00:43:17
the enthalpy of its constituent
elements in their most stable

558
00:43:17 --> 00:43:20
state at room temperature
and pressure.

559
00:43:20 --> 00:43:24
But we've defined the enthalpy
of those elements in their

560
00:43:24 --> 00:43:26
stable state at room
temperature and pressure

561
00:43:26 --> 00:43:29
as zero, right?
 

562
00:43:29 --> 00:43:33
So we're just subtracting, in
effect, zero, right, from the

563
00:43:33 --> 00:43:37
enthalpy of the product, but of
course it's important have that

564
00:43:37 --> 00:43:40
established because the heat of
formation is something you

565
00:43:40 --> 00:43:42
could measure, right?
 

566
00:43:42 --> 00:43:46
You could run the reaction,
take solid iron, gaseous

567
00:43:46 --> 00:43:51
oxygen, form iron oxide,
measure the heat of formation

568
00:43:51 --> 00:43:54
of it, tabulate it.
 

569
00:43:54 --> 00:43:55
We know it.
 

570
00:43:55 --> 00:43:58
We know it forever, right.
 

571
00:43:58 --> 00:44:02
And for any of the
other compounds.

572
00:44:02 --> 00:44:05
So in other words, by defining
that reference state, we can

573
00:44:05 --> 00:44:10
then figure out or measure
heats of formation of a

574
00:44:10 --> 00:44:13
vast number of compounds.
 

575
00:44:13 --> 00:44:14
We can tabulate them.
 

576
00:44:14 --> 00:44:17
We can know them, and then when
we have reactions that

577
00:44:17 --> 00:44:21
inter-convert different
compounds, we can calculate the

578
00:44:21 --> 00:44:23
heat of reaction is just the
difference between the heat of

579
00:44:23 --> 00:44:26
formation of the reactants, and
the heat of formation of

580
00:44:26 --> 00:44:30
the products, right.
 

581
00:44:30 --> 00:44:37
So let's just write that out.
 

582
00:44:37 --> 00:44:46
So first of all, for example,
right, molar enthalpy

583
00:44:46 --> 00:44:57
of hydrogen gas at
298.15 K is zero.

584
00:44:57 --> 00:45:03
Hydrogen gas it's in its most
stable state, right at room

585
00:45:03 --> 00:45:04
temperature and pressure.
 

586
00:45:04 --> 00:45:10
That little zero, that
little superscript

587
00:45:10 --> 00:45:17
means one bar always.
 

588
00:45:17 --> 00:45:17
OK.
 

589
00:45:17 --> 00:45:33
A molar enthalpy of or
whatever, iron, as a solid

590
00:45:33 --> 00:45:38
at 298.15 Kelvin is zero.
 

591
00:45:38 --> 00:45:42
Iron as an element is a solid.
 

592
00:45:42 --> 00:45:45
That's it's most stable state
at room temperature and

593
00:45:45 --> 00:45:52
pressure, right, and so on.
 

594
00:45:52 --> 00:45:55
And then we can figure
out heats of formation.

595
00:45:55 --> 00:45:58
Now in this particular
reaction, I've got

596
00:45:58 --> 00:46:03
hydrogen gas, iron solid.
 

597
00:46:03 --> 00:46:07
Those already are elements in
their most stable forms at room

598
00:46:07 --> 00:46:09
temperature and pressure.
 

599
00:46:09 --> 00:46:13
But this is a compound, right,
it has some non-zero heat of

600
00:46:13 --> 00:46:16
formation from the elements.
 

601
00:46:16 --> 00:46:19
So is water, right?
 

602
00:46:19 --> 00:46:23
So I need to find out the heats
of formation of the iron

603
00:46:23 --> 00:46:26
oxide and the water.
 

604
00:46:26 --> 00:46:29
And if I do that then I can
find out the change in

605
00:46:29 --> 00:46:31
enthalpy of this reaction.
 

606
00:46:31 --> 00:46:36
It's just going to be the heat
of formation of these three

607
00:46:36 --> 00:46:42
moles of water, minus the heat
of formation of the iron oxide.

608
00:46:42 --> 00:46:48
OK.
 

609
00:46:48 --> 00:46:50
So how am I going to do that?
 

610
00:46:50 --> 00:46:54
Well, I need to write the
reaction that forms that

611
00:46:54 --> 00:46:57
compound from it's
elements, right?

612
00:46:57 --> 00:47:00
And I want to tabulate that
for an enormous number

613
00:47:00 --> 00:47:02
of materials, right?
 

614
00:47:02 --> 00:47:10
So for example, if I want to
look at HBr, there's a simple

615
00:47:10 --> 00:47:18
case, right, hydrogen bromine.
 

616
00:47:18 --> 00:47:21
What's my heat of formation?
 

617
00:47:21 --> 00:47:22
Delta Hf.
 

618
00:47:22 --> 00:47:28
It's going to have our
little zero, right?

619
00:47:28 --> 00:47:29
How do I calculate it?
 

620
00:47:29 --> 00:47:33
Well, I need to write the
reaction that forms this from

621
00:47:33 --> 00:47:36
its constituent elements.
 

622
00:47:36 --> 00:47:36
What is it?
 

623
00:47:36 --> 00:47:48
Well, 1/2 H2 as a gas,
temperature, at one bar, plus

624
00:47:48 --> 00:47:56
1/2 bromine, no actually
bromine is a liquid at room

625
00:47:56 --> 00:48:00
temperature and one atmosphere.
 

626
00:48:00 --> 00:48:06
They form HBr which is a
gas at room temperature

627
00:48:06 --> 00:48:11
and one bar, right.
 

628
00:48:11 --> 00:48:17
I can measure that.
 

629
00:48:17 --> 00:48:25
And the heat of reaction for
this, delta H of reaction is

630
00:48:25 --> 00:48:36
equal to delta H of formation,
of HBr as a gas at our

631
00:48:36 --> 00:48:42
temperature and one bar, okay?
 

632
00:48:42 --> 00:48:46
So that's how we determine
our heats of formation.

633
00:48:46 --> 00:48:53
We measure them for
all these compounds.

634
00:48:53 --> 00:48:58
And then we go back to
reactions like this, and we can

635
00:48:58 --> 00:49:12
just very simply determine the
heat of reaction because all

636
00:49:12 --> 00:49:26
we're doing is the
following cycle.

637
00:49:26 --> 00:49:36
We go from reactants to
products and there's some

638
00:49:36 --> 00:49:42
heat of reaction, and here's
the cycle that we have.

639
00:49:42 --> 00:49:57
We go to the elements in their
standard states, in both cases.

640
00:49:57 --> 00:50:09
So we're really just doing our
delta H is the negative sum

641
00:50:09 --> 00:50:16
of delta H of formation for
the reactants, all right?

642
00:50:16 --> 00:50:23
Because here what we're doing
is we're going to take apart

643
00:50:23 --> 00:50:27
our reactants and form the
elements from them, right?

644
00:50:27 --> 00:50:29
So written this way, for
example, I'm going to pull

645
00:50:29 --> 00:50:32
these things apart, and I'm
going to have solid iron, and

646
00:50:32 --> 00:50:36
I'm going to have gaseous
oxygen, and I know the

647
00:50:36 --> 00:50:38
enthalpy of that reaction.
 

648
00:50:38 --> 00:50:41
That's just given by the heat
of formation, the enthalpy

649
00:50:41 --> 00:50:43
of formation of iron oxide.
 

650
00:50:43 --> 00:50:50
Here, now I'm going to take
those same elements, and I'm

651
00:50:50 --> 00:50:51
going to make the
products, right.

652
00:50:51 --> 00:50:54
I know it's going to work
because I already had this

653
00:50:54 --> 00:50:56
reaction, so all the, you
know, I'm going to conserve

654
00:50:56 --> 00:50:59
atoms in the right way.
 

655
00:50:59 --> 00:51:03
So here, I'm going to have
delta H, is just the sum for

656
00:51:03 --> 00:51:10
all the products of delta
heat of formation, right?

657
00:51:10 --> 00:51:17
Here I'm going to put
together all the products.

658
00:51:17 --> 00:51:19
So this is a positive
heat of formation.

659
00:51:19 --> 00:51:22
This is the negative heat
of formation, right?

660
00:51:22 --> 00:51:24
In other words, I've
got reactants, and

661
00:51:24 --> 00:51:25
I've got products.
 

662
00:51:25 --> 00:51:27
What's delta H of reaction?
 

663
00:51:27 --> 00:51:32
It's delta H of formation of
the products minus delta H of

664
00:51:32 --> 00:51:35
formation of the reactants.
 

665
00:51:35 --> 00:51:42
That's the important message,
delta H of formation of the

666
00:51:42 --> 00:51:54
products, minus delta H of
formation of the reactants.

667
00:51:54 --> 00:52:00
Any questions?
 

668
00:52:00 --> 00:52:03
All right, tomorrow I'll go
through the remaining details.

669
00:52:03 --> 00:52:08
The truth is after this,
it's all arithmetic, right?

670
00:52:08 --> 00:52:11
We're just going to go through
it and execute it a little bit.

671
00:52:11 --> 00:52:14
And then we'll move on and talk
about how we'd actually make

672
00:52:14 --> 00:52:17
some of these measurements of
delta H and compare them

673
00:52:17 --> 00:52:19
to what we calculate.
 

674
00:52:19 --> 00:52:20
See you tomorrow.
 

675
00:52:20 --> 00:52:21