Concepts of Thermodynamics
Prof. Suman Chakraborty Department of Mechanical Engineering
Indian Institute of Technology, Kharagpur
Lecture – 57
Thermodynamic Relationships We have discussed about various facets of
laws of thermodynamics so far but what we have not discussed so far about the ways in
which we can characterize or determine properties, because the manifestation of various laws
of thermodynamics essentially culminates in terms of calculating the differences in certain
properties like entropy, enthalpy, internal energy, like that. Many of these properties
are not directly measurable. So, if they are not directly measurable, they
could be indirectly predicted by means of certain other properties which could be related
to the non-measurable properties and these kinds of relationships are called as thermodynamic
property relationships or in general Thermodynamic Relationships. So, in thermodynamic relationships, we use
essentially partial differential calculus or calculus of partial derivatives to predict
the derivatives of measurable or non-measurable properties in terms of—one in terms of the
other. So, to do that, we will essentially establish certain mathematical backgrounds.
Let us say that there is a function z which is a function of x and y, function of
two variables. Why we take such a function of two variables, what is the thermodynamic
motivation? For a simple compressible pure substance, if you know two independent intensive
properties, you can identify the state. So, these two variables can essentially identify
the thermodynamic state. So, this x and y generically could be pressure, volume, temperature,
internal energy, enthalpy, whatever; if they are independent, a property can be described
as a function of that. So, you can write… So, let us call this as M and let us call
this as N. Assuming the second order partial derivative to be continuous, we can write…right?
So, if the second order partial derivative is continuous, then it doesn’t matter whether
you differentiate with respect to y first or with respect to x first, so, that means,
from—so, this is M and this is N. So, from here, we can conclude…
So, when you are differentiating with respect to y, we are essentially fixing x and when
we are differentiating with respect to x partially, we are essentially fixing y. So, this is an
important outcome. So, we will write it at one corner of the
board because we will use it for our derivations. So, if z is a function of x and y and d z
is M d x plus N d y, then… Okay? So, this is the first thing. The second
important interesting question is how does the chain rule work for partial derivatives?
The chain rule of derivatives—ordinary derivatives, that dy dx into dx dy is equal to 1 right?
That kind of a chain rule, how does it work for partial derivatives? So, we start with…so, dz…
Now, you can also write x as a function of y and z, right? So, this x is a function of
y and z. So, dx, this you can write… Okay? So, you can write dz… Okay? So, this
part of the board is not visible, please show it in the camera, yes. So, then, you can write—so,
you can compare this term with this term and you can club these two together.
So, if you compare these two terms, the first observation that you have is del z del x at
constant y…okay? So…so, it is intuitive, but it follows from rigorous
derivations that the inverse of—the derivative of x with respect to z is 1 by the derivative
of z with respect to x. Also, by comparing the coefficients of dy, the boxes which are
marked with yellow color. So, you can have…this is 0. So, coefficient of dy; this is the coefficient
of dy in the right hand side and coefficient of dy in the left hand side is 0.
So, you can write… And del z del y is 1 by del y del z, from
the previous expression. So, from here, we can arrive at partial derivative of z with
respect to x into partial derivative of x with respect to y, partial—into partial
derivative of y with respect to z; intuitively, by chain rule, this would have been 1, right?
But it is actually minus 1. So, this result is very important and we will use this carefully.
So, with the mathematical background established by this one and these two equations, we will
now consider how to apply these for thermodynamic relationships. So, out of the thermodynamic
relationships, the combination of first law and second law gives us two important ones:
the two Tds relationships. So, we’ll start with that. So, you have Tds is equal to d u plus pdv.
So, you can cast it in this way, dz is equal to M d x plus N d y, by write—noting that
du is equal to Tds minus pdv. Okay? Then, you have Tds is equal to d h minus vdp; that
means, dh is equal to Tds plus vdp. That is also of the form M d x plus N d y.
Now, to develop two more forms like this, we will use the definition of two new functions,
one is called as Helmholtz function and another is called as Gibbs function. So, the Helmholtz
function: a is equal u minus T s. What is this also called is the Helmholtz free energy.
We have done exergy analysis and you have seen that the energy that is freely available
to make the most out of a change of a thermodynamic state is governed by either u minus T s or
h minus T s depends on whether it’s a closed system or a flow process.
So, this is called as free energy because this is energy that is freely available for
a spontaneous transition to take place from one state to another state. So, you can write
d a is equal to d u minus Tds minus sdT; du minus Tds is, from this equation, is minus
pdv. So, you can write from here, da is equal to minus pdv minus sdT. Then, similarly, what
we do for—with the internal energy, if we use enthalpy instead of this, this is called
as Gibbs function or Gibbs free energy. So, you can write dg is equal to dh minus
Tds minus sdT. So, dh minus Tds is vdp. So, this is vdp minus sdT. So, we have dg is equal
to vdp minus sdt. So, this four thermodynamic property relationships are of the form dz
is equal m dx plus n dy. So, you can derive these kind of four these kind of expressions
from these four equations and let us do that in a moment. So, this one will imply…
So, we are just using this del m del y is equal to del n del x. So, here, x is s, y
is v, m is t n is minus p. Okay? So, these four equations, let us note
down again here in the board separately because we will use it for certain derivations. These are called as Maxwell’s relationships.
These were originally introduced by Maxwell. Look at the genius of Maxwell; the same Maxwell
introduced the four laws or four rules of electromagnetism and those are Maxwell’s
equations in electromagnetics. So, you have four Maxwell’s equations in electromagnetics
and here you have Maxwell’s equation. In the third equation minus .
This is not there. So—and here, you have four Maxwell’s equations—relationships;
these are not actually called as Maxwell’s equations, sometimes these are called as—called
as Maxwell’s relationships in thermodynamics, so, these are—but we will call it Maxwell’s
equations because these are essentially equations. Now, in the Maxwell’s equation, you can
see certain things you know very striking in Maxwell’s equation; see, that, crosswise,
it is always T and s and on the other side, cross wise it is—cross wise it is always
p and v; so T s p v, T s p v, p v T s, p v T s. So, it’s, you know, sometimes—although
memorization is not, you know, very important thing for understanding our particular subject,
but in case you want to, you know, sensitize your memory a little bit on keeping these
in purview, this crosswise T s and p v can help you to some extent.
Now, you can see that these equations are very unique because these give certain derivatives
with respect to some non-measurable properties like entropy in terms of derivatives which
are measurable. So, I will give you an illustration of how we can make use of this properties
by using the third Maxwell relation. So, let us consider a phase change process
of a simple compressible pure substance. So, all these are valid for simple compressible
pure substance, because then only two independent properties are describing a state. So, the
fundamental premises with which we started our discussion is that two independent properties
are describing the state. So, it’s a simple compressible pure substance. So, that is already
assumed for all the discussions that we are making today.
So, let us consider simple compressible pure substance changing phase from state prime
to state double prime. Okay? So, if that be the case, you can write, from this third T
dl—third Maxwell’s equation—now, if there is a change in pressure—change in
phase, p is only a function of T. So, this partial derivative becomes ordinary derivative
and whether volume is fixed or not is not important, because p phase change is a unique
function of T phase change. So, this becomes dp dT during the phase change process.
So, this—so, during the phase change process, the change of this is nothing but the change
in entropy divided by the change in volume; so…during phase change. Now, as I have mentioned
that we will try to express the non-measurable parameters in terms of measurable parameters;
entropy is not directly measurable, but we can write—we can use the Tds relation, Tds
is equal to dh minus vdp. During the phase change, pressure doesn’t
change, right? And temperature is constant. So, 0 during phase change. So, if you now
integrate this from state prime to double prime, then the temperature is fixed. So,
T into s double prime minus s prime is equal to h double prime minus h prime. So, dp dT
phase change, in place of this. Okay? So, this—why this is important because from
the enthalpy of phase change, which is also sometimes called as latent heat, it is possible,
and from the temperature of the phase change and volume change, it is possible to get the
saturation pressure versus saturation temperature diagram.
Remember now that we had a diagram for water something like this. So, saturation—the
phase change pressure versus phase change temperature, this is the critical—triple
point. So, you have solid, liquid and vapor. So, interestingly, this is for water, this
line has a negative slope, can you explain through this? Imagine that the state prime
is solid and the state double prime is liquid. So, when water at a particular temperature
gets converted from liquid—solid to liquid. So, its volume, what happens? Its volume shrinks,
right? This is very typical to water—for not all fluids it happens.
So, this is negative, T is absolute temperature it is positive and for melting, this is latent
heat of melting, this is positive. So, positive divided by positive, this is negative. So,
that makes dp dT phase change for solid to liquid negative, right? So, you can see that
this kind of unique behaviour which we have earlier seen in properties of pure substances
can be explained by this thing. Now, as a second example, let us apply it
to—so, this equation is known as Clapeyron equation. Now, let us apply it to a specific
example of liquid to vapor, f to g, phase change. Saturated liquid to saturated vapor.
So, then, dp dT saturated is equal to h g minus h f by T into v g minus v f, right?
So, h f g by T into v g minus v f. Now, we make two very important assumptions.
What are the assumptions? These are very practical, number 1: v f is much much less than vg. This
is true, because the density of saturated liquid is much much more than the density
of saturated vapor. So, the specific volume is much less than that of specific volume
of saturated vapor, right? And the state 2 is—can be approximated as an ideal gas.
So, state g is approximately ideal gas; it is almost full—it is almost superheated
vapor, so, it can be approximated as ideal gas, without bad approximation, provided the
pressure is sufficiently low or temperature is sufficiently high.
If it’s the other way, that you know pressure is quite high and temperature is quite low,
that will not—that combination will not normally be there; then, that—this approximation
doesn’t work, so, this being very close to super heated region, ideal gas approximation
is not bad. So, you can write P into v g is approximately equal to RT, R of vapor. So,
this is water vapor. So, this P is P sat. So, then, you can write here as dp dT sat
is equal to h f g by T—in place of vg minus v f, it is v g approximately, and that is
R T, by P. So, then, we have just one more step. So, you can combine this two and write d of
ln p dT, saturation process, is equal to h fg by RT square. This 1 by P—d p by p is
absorbed in d of ln p. So, d of ln p dT is hfg by RT square. This helps you to construct
the saturation pressure versus saturation temperature diagram for water if you know
h fg at a given temperature. So, if you know the latent heat at a given temperature, latent
heat of evaporation, from that data, you can—you can construct the saturation pressure versus
saturation temperature. So, this is a very important relationship;
this is called as Clasius Clapeyron equation—Clasius clapeyron equation. So, we will consider more
of these thermodynamic relationships in the next lecture.
For the time being, thank you very much.
