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PROFESSOR NELSON: Well, today we're going to continue with
the lecture that was started last time that had the
scintillating and descriptive title "some thermodynamic
processes".
And we're going to continue along those lines, and really
there a couple of things that I want to do to follow up what
you saw last time.
One is that you've got introduced and led through a
couple of elementary thermodynamic cycles.
And I want to do a little bit more of that today.
And one of the -- there are really two reasons.
One is that as you saw last time doing a thermodynamic
cycle can be extremely useful, because if you have one path
along which some change occurs, but along which it's
not necessarily easy to calculate all of the
thermodynamic changes along such a path, it may be
irreversible, lots of variables may change.
It can be easier instead of looking at the complicated
path to devise a cycle that involves several easier steps,
each one of which is facile for the calculation of what
happens to the thermodynamic properties.
So in that case, it's much easier to go through several
elementary step, and then the one complicated path you know
because going around the entire cycle state functions,
you know that they won't have any change.
So in that sense, going through a thermodynamic cycle
can be useful because it helps you calculate
thermodynamic qualities.
But the other reason to go through the thermodynamic
cycles and really to develop great facility with them is
because there are just an awful lot of things in nature
and things that we build that run in cycles, where we want
to calculate the thermodynamics, right.
If we build an engine for a car or anything else, it
almost always is going to have some key element that's
operating in a cycle, otherwise it
won't keep going, right.
But it's not just engines that we might build.
It's biological engines, right.
Things that either produce or consume biological fuel to do
different sorts of processes.
Also, of course of key importance to understand what
the, in that case, biological thermodynamics are.
And again, there'll be key elements that run in cycles
and understanding those can be extremely important in
understanding how cellular function works.
OK, so what I'd like to do is go through one cycle, and then
I'll suggest another for you to work through on your own.
So consistent with the title of the lecture, we've got a
section from your last lecture notes entitled "some
thermodynamic cycles".
And what we're going to do is go through a cycle, please
note these office hours and locations, which I think
wasn't specified before.
So we're going to go through a thermodynamic cycle, and
here's what I want to calculate when we do this.
Delta u, delta H, familiar state functions, changes in
their values, q, w, heat and work.
And I also want to introduce one new function.
I won't give it a name yet, but it's going to have the
unusual form dq over T. OK, we'll
call it special function.
It's so special that we'll call this quantity dS.
OK, we'll just go through this on a whim, and this is going
to be foreshadowing something that'll take on tremendous
importance in subsequent lectures, but I'm going to use
the opportunity now to just introduce the behavior of it
in going around a cycle.
If that's for me, tell them I'm busy please.
OK, so let's look at the cycle we'll go through.
We're going to look at pressure volume changes,
similar to what you saw before.
This time, we're going to have initial and final volumes, V1
and V2 and different pressures where we might end up.
And we're going to look at two ways to go to a
final volume V2.
One of them is going to end up at pressure p2 and the other
is going to end up at pressure p3.
One of them is going to be a reversible isothermal path.
This is going to be our path we'll label A, reversible
isothermal which is to say constant T, right?
And we're going to start at temperature T1, so since it's
constant temperature, this is going to end up
at temperature T1.
And then we're also going to consider a reversible
adiabatic path.
This'll be our path B, reversible adiabatic.
And then we'll close this circle, this cycle.
This is going to end up at a different
temperature by the way.
You saw this last time in a slightly different way.
Last time what you saw is we compared isothermal and
adiabatic paths that ended up at the same final pressure,
and what you saw is that therefore, they ended up in
different final volumes.
So this is just a little bit different from that.
So this is going to end up at a different temperature, we'll
call it T2.
And then we're going to close the cycle.
And so this is a constant volume path then.
This is path C, constant volume.
OK?
So now let's go around the cycle and just compare notes
on what happens to the thermodynamic
quantities as we do that.
So here's path A, it's isothermal.
And I didn't specify, but let's make sure to do so,
we've got a, it's going to be an ideal gas.
OK, so A is constant temperature.
What does that mean for delta u in path A?
Zero right?
Ideal gas, no temperature change, right has to be zero
because da is Cv dT for an ideal gas. dT is zero, there's
no temperature change.
OK, how about delta H, zero Cp dT for an ideal gas.
dT is zero.
All right, now, it's reversible.
So that means we can immediately write down an
expression for the work.
Since it's reversible, it's negative p dV right?
And since it's an ideal gas then we can replace p, right,
with RT over V, right?
We want to do that because we have too many variables here.
We've already got dV we'll get rid of p as an additional
variable and replace it with V which is already in here.
So it's minus R T1 dV over V, right?
I can put in T1 because we're a constant temperature. so now
we can just integrate straight away and find that the work in
path A it's just the integral of that quantity minus
integral of R T1 going from V1 to V2 dV over V. so it's minus
R T1 log V2 over V1.
All right, so let's just do a reality check here.
The way I've got this drawn, the volume is
going up in the process.
It's an expansion.
So is this system doing work on the surroundings, or are
the surroundings doing work on the system?
Somebody say it real loud.
Yes, the system is working on the surroundings, right?
It's pushing out against them.
Work is defined as the work that the surroundings do on
the system.
So this is the negative number, right?
V2 is bigger than V1.
This is positive.
This whole thing is negative.
All right, q has to be just the opposite of this because
we've already figured out that there's no change in u.
This is just q plus w.
There's w, q has to be R T1 log of V2 over V1.
OK, so that means heat is being imparted to the system,
right, from the surroundings in a manner that compensates
exactly the amount of work done.
OK, All right, so these are the thermodynamic quantities
that you're familiar with already.
Let's just quickly look at our special function.
It's not going to be hard to calculate because it's a
constant temperature path, so that T in the
bottom is just T1, right.
You've already looked at q, right.
So dq A over T1 it's just R log V2 over V1 right.
So there's our special function for
this particular path.
OK, so that's path A. Now let's compare to path B. So if
path B is an adibat, what's zero? q, it's adiabatic, that
means there's no heat exchanged between the system
and the surroundings.
So what happens, q B is zero.
There is a change in temperature, right?
We've got, if we say it's a mole of gas, it's going from
p1, V1, and T1 to one mole of our gas at p3, V2, and T2.
So unlike before now
temperature is going to change.
So that means that delta u isn't zero this time.
du, it's an ideal gas.
So this is Cv dT and of course we can just integrate this
straight away.
So delta u B is Cv times T2 minus T1, right. and similarly
dH is Cp dT right.
So delta H in pathway B is just Cp T2 minus T1 okay?
All right, what we've already got that q is zero.
Delta u is q plus w.
Delta u isn't zero.
So this must be equal to work, right?
And now we can look at our special function, but since
this is zero right dq B is zero, along path B, so our
special function, dS, so we're going to go, dq B over T is
equal to zero.
OK?
OK, finally let's look at our third path, this constant
volume path, that's going to connect T1 and T2, right?
Constant volume, what's zero?
STUDENT: [INAUDIBLE]
PROFESSOR NELSON: A little more noise here.
What's zero?
STUDENT: Work.
PROFESSOR NELSON: Work, great.
So, path C constant volume means our work in past C is
zero, right?
Now, our temperature is going to change from T2 to T1, and
we just saw what happens to u and H when that happens going
from T1 to T2, and it's no different.
It's an ideal gas going along these path.
So we can immediately write delta u C is Cv
times T1 minus T2.
Delta Hc C is Cp times T1 minus T2, right?
Work is zero.
Delta u is w plus q, work plus heat.
This is zero, this isn't.
This must be heat q B, right.
So again, there are our familiar thermodynamic
quantities, but now let's also look at our special function.
It's not going to be zero this time because we have non zero
heat exchange between the system and the
environment, right.
So are integral of dq C over T, right, which is our heat,
but in this case we can do this calculation easily
because since work is zero, we can equate dq with du, right.
So we can write this as our integral from T1 to T2, du
over T. du is just Cv dT, right.
So this is just Cv log T2 over T1.
OK?
So now we've completed the cycle.
So let's just compare.
Let's compare what happened in path A to what happened in
paths B and C. Yes?
STUDENT: [INAUDIBLE]
PROFESSOR NELSON: Ah sorry, yes of course.
Thank you.
Yes, oh yes, so.
STUDENT: [INAUDIBLE]
PROFESSOR NELSON: And it's right because, thank you
again, it's because we're going from T1 to T2.
So I'm getting confused by the oh sorry, we're
going from T2 to T1.
So it's in reverse of the order of the subscripts and I
let this confuse me.
There it is.
There.
Thank you, any other details which should be pointed out?
All right, then let's proceed.
So what we're going to do is our comparison of what
happened on pathway A to what happened in combined pathways
B and C, right?
So here are results for pathway A, right, for delta u
zero delta H zero.
I didn't actually explicitly write it or did I?
Let's just write it again.
Here's our work in path A, and here's our heat exchange in
path A. Now let's look at the sum of B and C. So B plus C,
here is delta uB Cv times T2 minus T1.
Delta uC Cv times T1 minus T2, the sum is zero right.
That's the some of delta u in these two paths, and if we
look at delta u in just path A, it's zero.
And it's going to be the same without delta H. The only
difference is it'll be Cp instead of Cv, but there it is
for pathway B. There it is for a pathway C. So the state
functions that we're familiar with are doing what we expect
they ought to be doing, right?
If you go around in a cycle, starting and ending at the
same place, the state functions
have to stay the same.
They only depend on the state the system is in.
They don't depend on the path, and that's what
we've shown, right.
Similarly if we go from this point to this point through
path A or if we do it through the combination of these two
paths, and the change in u and H in any state function, those
have to be the same.
It turns out they're zero in both cases, and that's what
we've seen, right?
Now let's compare what happens to work and heat.
So here are expressions for work and heat.
They're opposites for the pathway A. Here's pathway B
and C. In C the work is zero.
In B it isn't, it's Cv times T2 minus T1, right.
The work isn't equal to the work in pathway A, right,
because you know work is not a state function it depends on
the path right, and there are different amounts of work done
on the system, or done by the system on the surroundings in
these two different processes.
They're both expansions.
They'll both have net work done on the surroundings, but
not the same amount, right.
And of course it's going to be the same with heat.
We've already seen that delta u is zero.
So we know that in each case the heat is going to be the
opposite of the work, but the work isn't the same in these
two different ways of getting from here to here, right.
So let's just see it explicitly.
Here's our qA.
Here's heat exchanged in pathway A and in pathway B
heat is zero, and in pathway C, here is qC
it's Cv T1 minus T2.
So again, for both heat and work we don't
get the same result.
Now let's look at our special function, right.
So here's path A. We found that it's R log V2 over V1.
Pathway B is zero.
Pathway C it's Cv log T1 over T2.
All right, let's just look at that a little more carefully.
R log V2 over V1.
B plus C, Cv log of T1 over T2, right.
Just some weird combination of functions, right?
But, of course, it's not quite like that.
Why?
Because this path is an adiabatic path.
And you already saw last time there was this relationship
between the temperature and volume changes along an
adiabatic path.
So let me just write that down.
Adiabatic reversible path.
What you saw is that T1 over T2 was V2 over V1 to the
power R over Cv.
Isn't that something?
So what does that mean?
We take this Cv and put in the exponent here, right.
And we put this R up in the exponent here.
What are we going to discover?
Those two things are equal right, T1 over T2 to the power
Cv is V2 over V1 to the power R, so dS over path A equals dS
over paths B plus C, okay?
So what this suggests is that whatever this funny special
function is, at least for this one cycle that we've, tried
it's behaving like a state function, right.
It seems like its change is independent of path.
Start here, end up here, do it through two different
pathways, end up in the same place, right.
OK, now that's all the foreshadowing I'm going to
give it for right now.
We will certainly come back to this very
special function shortly.
Before I move on, I'm just going to put on the board
another cycle, and I'm going to urge you to work through
that on your own.
It's worked through already, let's see, I believe it's
worked through in the notes, yes.
So if you need the help of the notes it's in there, but I
would urge you to work through it on your own, and it's the
following: let's start with the same path A, all right?
So we've got our reversible adiabatic path right.
And now, try working through what happens if we close the
cycle in a different way, right, like this.
So here's a path that's constant pressure, and here is
a path that's constant volume, similar to the constant volume
path that we did before, but not between the same
pressures, right.
So we can label this one D and this one E, and I urge you to
just try going through that and verifying for yourself
that again the state functions will behave the way state
functions should, and you can see whether this special
function once again behaves like a state function, right.
And of course you should expect to see that the work
and the heat again won't behave like state functions
which they are, then you'll see you have different results
for those, depending on the path, OK?
Any questions about going through these cycles and using
expressions like this and so forth?
Expressions like this you know turn up in various places,
like in the equations sheet that appears at the end of
exams, right?
Which means that you may not need to be
committing it to memory.
On the other hand it means that you should, it's the sort
of thing that you should be familiar with so that the need
to use it is something that you'll conjure when you're
working on a problem and there's a reversible adiabat,
and you have temperatures and volumes that change that you
might like to relate to each other.
Cycles that are suggested that you go through and aren't gone
through in class also sometimes turn up on exams
too, by the way.
So that said, let's move on to the next topic, which is
thermochemistry.
So, so far what you've seen in most of the examples in the
class are essentially mechanical kinds of changes,
pressure volume sorts of changes, where the system is
doing essentially a kind of mechanical work, pressure
volume work on the surroundings or vice versa and
heat is being exchanged, and you're calculating the basic
thermodynamics just like we went through, of
processes of that sort.
Now I'd like to start introducing something that's
really central to chemical thermodynamics, mainly
chemistry, right.
Let's start looking at chemical reactions, right, and
understanding what the changes in thermodynamic properties
are that occur when chemical reactions occur, and the
immediate application is to calculate the change in
enthalpy, which, as you've seen it, at constant pressure
which is the condition under which an awful lot of
chemistry is done, that's just the heat.
Which means we're going to calculate heats
of reaction, right.
You're running some reaction.
It's in the atmosphere.
It's at constant pressure.
You know, you mix acid and base together and the thing
heats up like crazy, right.
Or other things might react spontaneously.
You can feel something cooling off right.
I mean simple examples of these happen when you, you
know, if you buy cold or hot packs.
You break the seal between them and feel the thing get
cold, for example, right.
What's happening there?
Well, the selection of reactants has been done
judiciously to provide either heat or to provide something
that's cool.
And all of that is coming out of the heat of reaction,
whether it's positive or negative.
Of course the biggest practical application is
burning of fuel.
You might use the heat to, and convert it into pressure
volume work, right?
So if it's an internal combustion
engine, you'll do a reaction.
There will be the heat of reaction.
There'll be a volume change, depending on the conditions
under which is occurs, and you can extract work through the
reaction and so forth.
So let's just to see how you run through some of those
calculations.
Now let me just ask who here took 5.111?
And who here took 5.112?
OK, now you've seen some thermal chemistry
in 5.111 and 5.112.
So I I'm going to do some review of that, but I'm also
going to call on the thermochemistry that I'm
pretty sure you're familiar with.
So I won't go painstakingly over every element of the
notes here.
Subsequently we'll go on into additional topics that you
haven't seen, and I'll treat them in full detail.
OK, so let's do some thermochemistry.
What we want to do is we'd like to be able to predict for
any reaction what's the heat of reaction.
And so what we're going to calculator is delta H. So for
any kind of reaction we should be able to do that, right?
And we're going to treat constant pressure situations
certainly at first and most of the time, right?
Constant pressure, right, reversible delta H is going to
give us our heat of reaction, OK.
And then if we can also determine delta u, then we
know this, we know delta u is q plus w, then we can
determine work as well, right?
And typically, we'll be treating at least some cases
where we're dealing with ideal gases in which case we can
easily get delta u.
And then we'll be able to in a very
straightforward way get w.
So it's a really powerful and simple formalism, once we set
up what is needed the go through it, right?
So let's just consider any reaction.
The one that I've got written down for you is it's the
reverse of the rusting of iron, right?
So iron left to own devices in the atmosphere in the presence
of a little bit of water, and the atmosphere will start
combining with the water form iron oxide.
There's an equilibrium between the two.
So we'll have an iron oxide species.
It's a solid.
It's combining with hydrogen which is a gas to give iron,
also a solid and water, right?
So the way I've written this we're imagining the iron, the
solid iron immersed in the water as a liquid, could be
calculated otherwise as well, but in this case were imaging
the water as liquid and going in this direction then it's
forming iron oxide and evolving hydrogen gas, okay?
And our heat of reaction or enthalpy of reaction is
defined as the enthalpy at constant pressure.
Isothermal conditions, temperature won't change and
reversible work.
For Isothermal reaction constant
pressure reversible, right?
Of course, they're all sorts of conditions under which a
reaction could be wrong in the lab or
outdoors or however, right.
But this is the way we're going to
define delta H of reaction.
We want to have that definition clear because in
fact we're going to, we might want tabulate heats of
reaction, right, and of course want to know what the
conditions are for the tabulated values apply.
And we're going to want to calculate them from other
quantities, and again, we're going to need to know each
case what are the relevant conditions?
OK, so what happens?
Well, we should be able to get this.
We should be able to calculate delta H.
It's a state function.
If we know the enthalpy of the products minus the enthalpy of
the reactants, right.
It's a state function.
And we can we can do this in principle except for one
important detail, which is that enthalpy, just like
energy u isn't measured on an absolute scale but on a
relative scale, right?
You know, if you want to measure the potential energy
of something in a gravitational field, you have
to define the zero somewhere, right, because it's arbitrary.
You can set it anywhere you want.
It's the same with enthalpy.
Enthalpy is just u plus p V. So there is an arbitrary set
point that needs to be defined, right?
Because what you actually measure in the lab are changes
in enthalpy, just like what you measure when you look at
energy change of some sort, you measure the change in
energy, right.
The absolute number that you assign to it is something
that's arbitrary.
You have to set what the zero is.
And so there's a well-understood convention for
what the zero is.
What we define as zero is the enthalpy of every element in
its natural state at room
temperature and ambient pressure.
In other words, we choose a convention for the zero of
entropy, so that then we can write entropies of products
and reactants always referring to the same standard state.
And then we calculate changes, the convention is understood
with respect to what is the zero, right.
And so our tabulated values, they'll all work.
They'll all refer to the same standard state.
And we'll be able to use the formalism that we set up.
So our reference point for H, it's 298.15 Kelvin, one bar
and in that standard state our molar enthalpy is defined as
zero for every element in its stable form.
OK, very important.
OK, now, given that reference point, all we need to do to
get the value of enthalpy that we're going to use for each
reactant and each product is calculate how much does the
enthalpy change to form it from the elements in their
stable form at room temperature
and pressure, right.
So we're going to replace the H or we're going to put in for
the H what we'll call our heat of formation or delta H of
formation starting from the elements in their stable
states at room temperature and pressure, for
each of these things.
And we can do that because now we're going to have instead of
just sort of H, it's going to be delta H of formation for
each of these things, is going to appear in our calculation
of H, but that's OK.
Delta H of formation means the enthalpy of this compound
minus the enthalpy of its constituent elements in their
most stable state at room temperature and pressure.
But we've defined the enthalpy of those elements in their
stable state at room temperature and pressure as
zero, right?
So we're just subtracting, in effect, zero, right, from the
enthalpy of the product, but of course it's important have
that established because the heat of formation is something
you could measure, right?
You could run the reaction, take solid iron, gaseous
oxygen, form iron oxide, measure the heat of formation
of it, tabulate it.
We know it.
We know it forever, right.
And for any of the other compounds.
So in other words, by defining that reference state, we can
then figure out or measure heats of formation of a vast
number of compounds.
We can tabulate them.
We can know them, and then when we have reactions that
inter-convert different compounds, we can calculate
the heat of reaction is just the difference between the
heat of formation of the reactants, and the heat of
formation of the products, right.
So let's just write that out.
So first of all, for example, right, molar enthalpy of
hydrogen gas at 298.15 K is zero.
Hydrogen gas it's in its most stable state, right at room
temperature and pressure.
That little zero, that little superscript
means one bar always.
OK.
A molar enthalpy of or whatever, iron, as a solid at
298.15 Kelvin is zero.
Iron as an element is a solid.
That's it's most stable state at room temperature and
pressure, right, and so on.
And then we can figure out heats of formation.
Now in this particular reaction, I've got hydrogen
gas, iron solid.
Those already are elements in their most stable forms at
room temperature and pressure.
But this is a compound, right, it has some non-zero heat of
formation from the elements.
So is water, right?
So I need to find out the heats of formation of the iron
oxide and the water.
And if I do that then I can find out the change in
enthalpy of this reaction.
It's just going to be the heat of formation of these three
moles of water, minus the heat of
formation of the iron oxide.
OK.
So how am I going to do that?
Well, I need to write the reaction that forms that
compound from it's elements, right?
And I want to tabulate that for an enormous number of
materials, right?
So for example, if I want to look at HBr, there's a simple
case, right, hydrogen bromine.
What's my heat of formation?
Delta Hf.
It's going to have our little zero, right?
How do I calculate it?
Well, I need to write the reaction that forms this from
its constituent elements.
What is it?
Well, 1/2 H2 as a gas, temperature, at one bar, plus
1/2 bromine, no actually bromine is a liquid at room
temperature and one atmosphere.
They form HBr which is a gas at room temperature and one
bar, right.
I can measure that.
And the heat of reaction for this, delta H of reaction is
equal to delta H of formation, of HBr as a gas at our
temperature and one bar, okay?
So that's how we determine our heats of formation.
We measure them for all these compounds.
And then we go back to reactions like this, and we
can just very simply determine the heat of reaction because
all we're doing is the following cycle.
We go from reactants to products and there's some heat
of reaction, and here's the cycle that we have.
We go to the elements in their standard
states, in both cases.
So we're really just doing our delta H is the negative sum of
delta H of formation for the reactants, all right?
Because here what we're doing is we're going to take apart
our reactants and form the elements from them, right?
So written this way, for example, I'm going to pull
these things apart, and I'm going to have solid iron, and
I'm going to have gaseous oxygen, and I know the
enthalpy of that reaction.
That's just given by the heat of formation, the enthalpy of
formation of iron oxide.
Here, now I'm going to take those same elements, and I'm
going to make the products, right.
I know it's going to work because I already had this
reaction, so all the, you know, I'm going to conserve
atoms in the right way.
So here, I'm going to have delta H, is just the sum for
all the products of delta heat of formation, right?
Here I'm going to put together all the products.
So this is a positive heat of formation.
This is the negative heat of formation, right?
In other words, I've got reactants,
and I've got products.
What's delta H of reaction?
It's delta H of formation of the products minus delta H of
formation of the reactants.
That's the important message, delta H of formation of the
products, minus delta H of formation of the reactants.
Any questions?
All right, tomorrow I'll go through the remaining details.
The truth is after this, it's all arithmetic, right?
We're just going to go through it and
execute it a little bit.
And then we'll move on and talk about how we'd actually
make some of these measurements of delta H and
compare them to what we calculate.
See you tomorrow.