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In the last class, we looked at the relationships between some of those properties, thermodynamics
properties using some simple aspects of mathematics. We also saw the Gibbs-Duhem equation, which
is a central equation in thermodynamics. Since, it gives the simultaneous variation with temperature,
pressure and chemical potential, which are the fundamental intrinsic properties of a
system. We will continue looking at useful relationships between various variable, and
also I will tell you how it comes in useful in the tutorial part. The next thing that
we are going to look at in the same vein is Maxwell’s relations.
To get a Maxwell’s relations, let us first consider a theorem and calculus that is applicable
to exact differentials. We do not have to worry too much about whether we are dealing
with exact different differentials or not in thermodynamics, because we deal with state
functions in thermodynamics mostly, and most of those state functions can be written as
exact differentials. The theorem says if f can be considered as
a function of some variables which are given here as x 1, x 2 and so on till x k, then
df f being an exact df being an exact differential can be written as dou f by dou x 1 at evaluated
at all these other variables kept constant which is indicated by x j as I had mentioned
earlier d x1plus dou f dou x 2 x j which means x1 x 3 and other variables x a x k are held
constant. That is what this x j means dx 2 and so on until dou f by dou x k all other
x js remaining constant dx k. So, this is a fundamental theorem that is applied to applicable
to exact differentials.
Now, let us take those partial differential dou f by dou x 1 x j at constant x j dou f
by dx 2 at constant x j and so on and replace them with another symbol for easy manipulation.
Let us say that y 1 equals dou f by dou x 1 at constant all other x js. If we do that,
then, we can write the previous equation as df equals y 1 dx 1 plus y 2 dx 2 and so on
till y k dx k. Let us call this equation 224. The theorem that was mentioned initially with
this also known as reciprocity relationship says that when we can do this that is df being
expressed as y1 dx 1plus y 2 dx 2 and so on till y k, dx k.
dow y i by dou x n x j at constant x j equals dou y n by dou y n dou x i at all other x
j s remaining constant. For example, if you take i to be 1 and n to be 2 dou y 1 by dou
x 2 at all other x j s remaining constant equals dou y 2 by dou x 1 at all other x j
s remaining constant. It is applicable for any i and any n that is not different that
is different from i. For example, it could be dou y 3 by dou x 10 at all other x j s
remaining constant equals dou y 10 by dou x 3 at all other x j s remaining constant.
If we apply this theorem, also called the reciprocity relationship to our basic equation
that we had given earlier, basic differentials that we had given earlier, If you recall equation
212 which is also given here dU T equals T d S T minus P d V T plus summation of mu i
d n i. Remember that this consist of many different
terms that I given by the sum sign. If we apply the reciprocity relationship here it
will give as dou T by let us chose this be other variable dou V T at all other things
remaining constants such as S T and all n is remaining constant. This equals minus dou
P which is related to V T earlier dou P dou S T at V T and all other n i s remaining constant.
So, this gives a nice relationship between the thermodynamic variables just by using
the reciprocity relationship of writing differentials. So, will call this equation 226. Let me show
this a few more times so, that it becomes simpler to remember. Let us consider the next
relationship d j P equals minus S T dT plus V T dP plus summation of mu i d n i as we
have already seen in equation 215. If we utilize the reciprocity relationship here one of them
then minus comes from here minus of dou S T dou dP at constant T and all other n i equals
dou V T dou T at constant P and all other n i. By now you must be getting comfortable
with writing reciprocity relationships from total differentials let us call this equation
227.
Let us take another example I will give you all four or all more than four. In the first
case d H T we have seen as T d T S plus V T dP plus summation of mu i d n i and this
was equation 217 before. If we write apply the reciprocity relationship here dou T by
dou T dou P at constant S all other n i equals dou V T dou S T at constant P all other n
i equation 228. Let us consider d a T now, we already have equation 218 as d a T equals
minus S T d T minus P d V T plus summation of mu i d n i. That must become second nature
to you. dou S T dou V T note both are minus here. So, you do not have to worry about it
if you take these two. dou S T dou V T at constant T all other n i equals dou P T dou
T at constant V T all other n i. We will call this equation 229.
Let us consider equation 215 again. Earlier for all the four equations we for a close
system we looked at just the first two terms for writing the reciprocity relationship.
Now let us look at third term also, note that this consist of sum. So, each one is different
here. So, let us take one of those dou mu i dou T at all dou mu i dou T at constant
P all n i equals minus dou ST dou n i at constant T P and all n j s apart from this n i. Let
me repeat this dou mu i dou T at constant P all other n i equals minus, minuses here
dou S T dou n i constant T P all other n j s which is which are different from i. Let
us do this once again to get other useful relationships before that lets call this equation
230. dow mu i dou P at constant T all n i equals
dou V T dou n i at constant T P all n j s different from n i remaining constant. We
had considered these two terms while this equation. Let us call this equation 231. There
was a reason why we chose to compare these two and these two and actually use d j 2 to
do it. If you recall we have dG T with variation of T P and n i. So, which are easily measurable
variables? If they are easily measurable then we can do experiments with them quietly easily.
This gives the variation of chemical potential with temperature which is a very useful relationship
to have when the pressure and the number of moles of all species are held constant which
can be done experimentally. This gives the variation of chemical potential
with pressure when the temperature and all moles are held constant which can be again
done in an experiment. And this is given in terms of the other thermodynamic variables
which may be easier to determine. So, that is going to be some sort of a theme in this
particular module. That is expression of difficult to measure thermodynamic variables in terms
of easy to measure thermodynamic variables. That is a way we are going to use these equation
these equation n are valid for anything that you want to do. These relationships are there
for anything that you want to do. We are going to do one small aspect or one aspect of the
many different things that you can do with these relationships.
The equation 226 to 231 let me go back and show you 226. Just for recall 227, 226 was dou T dou V T constant S T n i equals
minus dou P dou S T constant V T n i. 227, minus dou S T dou P at constant T n i equals
dou V T dou T at constant P n i equation 227. 228, was dou P dou T constant S n i equals
dou V T dou S T constant P n i. Equation 229 was dou S T dou V T constant T n i equals
dou P dou T at constant V T n i. And as we spent some extra time here dou mu i dou T
at constant P n i equals minus dou S T dou n i at constant T P all other n j s and dou
mu i dou P at constant T n i equals dou V T dou n i at constant T P all other n j s.
So, these equations are called Maxwell’s relations. Very useful as I had already mentioned
its worth mentioning again. Temperature, pressure and the total volume are easily measurable.
Maxwell’s relations can help us that is one of the things they do you can do many
other things with them can help us express the other variables such as U T internal energy
S T enthalpy total enthalpy. All these are total quantities since that are more than
one mole of the substance of the pure substance. That way considering here H T is enthalpy
total enthalpy A T is total held moles free energy and G T is total Gibbs free energy.
All these can be written in terms of easily measurable T P and total volume and therefore,
by these measurements under suitable conditions we can estimate these thermodynamic variables
U T, S T, H T, A T and G T. We started is a slightly more general set of relations.
The Maxwell’s relations that we have written down so far are valid for any system any pure
substance irrespective of the size of the system or the number of the moles in the system.
Whereas, in many different books including your text book initially when this is introduced
you would find the Maxwell’s relations written for one mole. Which means the mu i d n i terms
any d n i related terms n i related terms will not be there because you have only one
mole and that is held constant.
Let us look at these relationships for completeness because you can directly use these relationships
under such conditions where you have one mole of the substance. dou T by dou V constant
S equals minus dou P dou S constant V. This was obtained by applying the reciprocity relationship
to the total differential return for one mole of the substance. dou T dou V at constant
S equals minus dou P dou S at constant V minus dou S dou P constant T equals dou V dou T
constant P dou T dou P constant S equals dou V dou S constant P and dou S dou V constant
T equals dou P dou T constant V. You want to compare this expression I would
like to do that I do not want to go back now. I would like do that if you just compare this
expression with the earlier Maxwell’s relations that we have written. You would see that these
are exactly the same as the first four Maxwell’s relations that we had written except that
you do not have any n i s occurring because we have one mole of the substance. These would
come and handy in some of your problems in the university exams and so on.
Now, we are going to do some tutorial work. I am going to present this problem to you
and then I will give you some time to work it out. Because you will understand the application
of these equations and get comfortable with the applications of these equations only by
working out problems. And you need to work them out first that would be the most effective.
So, you will work them out. I will give you time may be about 10 minutes. And may be after
some time, I will present a part of the solution so that you can start working at a faster
pace if that hint or part of the solution is going to help you.
Later I will give you the entire solutions. This exercise essentially shows a way to use
some of the relationships that we have a developed so far. The question is for a closed system
express in terms of easily measurable properties pressure, total volume and temperature the
variation of internal energy with volume for a process in which the temperature and the
number of moles are held constant. Essentially we are looking for dou U T the variation of
internal energy with volume. So, dou U T dou V T at constant temperature and n i. Please
go ahead you have about 10 minutes I will come back in some time in present
part of the solution.
. Let us look at a part of the solution now.
To do this, you know that we are looking for dou U T dou V T, let us begin with equation
212. Because this has the relationship for dU T and these would be dou U T and so on.
dou U T variation. We know that dU T equals T dS T minus P dV T plus some overall terms
i mu i d n i. The partial derivative of V at constant T and n i can be written as dou
U T dou V T T n i. T dou S T dou V T minus P dou V T dou V T T n i plus summation overall
i mu i dou n i dou V T at constant T n i. Continue further we are looking at dou U T
dou V T at constant T n i. Look at these terms here and see what you could do to get it in
terms of measurable quantities. Take another five minutes or so.
Let us continue with the solution. Look at the second and third terms on the right hand
side here and this P dou V T dou V T T n i plus summation of mu i dou n i dou V T at
constant T n i. The second and third derivative terms on the right hand side are 1 and 0 1
because dou V T dou V T. So, that goes to 1 and 0 because n i is a constant. So, any
variation with n i would be 0. Therefore, what remains of that equation is dou U T dou
V T at constant T n i this was the initial left hand side. The first term remains equals
T dou S T S dou V T at constant T n i and since this is gone to one whatever accompany
P here we just have a P remaining minus p. So, dou U T dou V T at constant T n i equals
T dou S T dou V T constant T n i minus P. Now if we use one of the Maxwell’s relations
which is equation 229 here dou S T dou V T this we know this is a little difficult to
measure. So, let us try to write this in terms of easily measurable quantities. dou S T dou
V T at constant T n i equals dou P dou T at constant V T n i. In this case P T is all
easily measurable. If we do that then you can write dou U T dou V T at constant T n
i which is what we need equals T. This we are going to replace with this dou
P dou T at constant V T n i minus P equation 232. Let me call this equation 232. Because
it will come in useful later which is actually the needed relationship? You have gotten this
in terms of all easily measurable thermodynamic variables T, P, V T, n i and so on.
To recapitulate the easily measurable once are P, T, V the molar volume or V T the total
volume. In addition the following are measurable and the data is available in the form of tables
or figures in books handbooks or papers for pure substances. Some of these are available
at the back of your text book Smith Vanness and Abbott. C P heat capacity at constant
pressure this is available the data is available. C V heat capacity at constant volume the data
is available, alpha which is expansivity. I will go into details of this in a little
while and kappa the compressibility is all available. So, if they are available we might
as well make use of them to estimate the difficult to measure thermodynamic variables. In addition
to these the latent heats are available as well as the heats of reaction are available.
Let us look these in a little more detail other measurable thermodynamic variables.
Heat capacity C P is defined as dou H dou T at constant P. This is for a pure substance.
We will call this equation 233. For any system C P is defined as dou H T dou T at constant
P. This is for one mole of a pure substance this is for any system. Now C V is defined
as dou U dou T, C P was dou H enthalpy dou T, C V is dou internal energy dou T at constant
volume. Since this is V at constant volume we call C V is specific heated constant volume.
This is for a pure substance we will call equation 234 and for any system you can write
you can define C V as dou U T dou T at constant V.
Expression of the not-so-easy to measure thermodynamic properties in terms of measurable thermodynamic
properties helps in estimation of the not-so-easy to measure once. I am just repeating this,
so that it gets a cross better. Now, using that exercise, let us those relations the
definitions for C P and C V and so on; let us do a problem; let me probably post the
problem now; and let you work on that we still have about 5, 6 minutes left in this lecture.
Please take that as a tutorial part of this lecture; please work that out, and when we
meet in the next class, I will give you the solution.
The problem is as follows: For a closed system consisting of one mole of the pure substance
mixings easier; one mole of the pure substance express in terms of more easily measurable
properties. The variation of enthalpy and entropy with temperature and pressure respectively.
What I mean is dou H T dou T enthalpy with temperature at constant pressure dou S T dou
T at constant pressure in variation of entropy with temperature at constant pressure, variation
of enthalpy with pressure at constant temperature, just the vice versa, and the variation of
entropy with pressure at constant temperature. Please go ahead and do this. We will start
with the hints in the next class.