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Good morning. We are discussing about the properties of pure substances for last few
classes, phase diagram. Today, we will be closing that topic by discussing
one aspect of this which is known as Clausius-Clapeyron equation. We have recognized that saturation
states are specified by the couple of pressures and temperatures, both for transformation
from liquid to vapor and from Solid to liquid. What is the relationship between these two
so that one knows the pressure can automatically know the temperature, based on which the pressures
and temperatures are determined and tabulated. For an example, in steam table for water,
we know that when saturation pressure is fixed temperature. For example, if it is one atmospheric
pressure then the saturation temperature is 100 degree Celsius. Similarly, if the pressure
is ten atmospheric then what is the saturation temperature?
What is that relationship based on which the values are tabulated was given by two scientists
Clausius and Clapeyron. So, this we will discuss today.
Let us again recall the pv diagram. This is the p and v as usual. We are bothered only
on the liquid vapor, it does not matter. We can consider this for solid, liquid also.
Similar thing will apply and this is the Ts diagram; I am drawing it like this. So this
is liquid, liquid plus vapor, this is the triple point line Solid, vapor, Solid.
If we denote the pressure, for example, at the given pressure, this is the constant pressure
line. So, this is the pressure. The saturation temperature for this pressure is the isotherm
which passes through this, coincides with this pressure line in the vapour dome; this
is the isotherm. If this be the pressure the corresponding temperature is this isotherm.
Similarly, here also, if these be the isotherm, let these be the isotherm T. What is the pressure?
Corresponding pressure is this one, this is p; that means, p and T, here represents the
saturation pressure and temperature. Now, what are the relationship between this p and
T? That means if we increase the pressure, the temperature will increase.
Let us recognize one thing that in this type of phase change, the specific volume changes
as this change of phase takes place the specific entropy changes. Similarly, the specific enthalpy
changes which is found in hs diagram. But now, if was define the Gibbs function, specific
Gibbs function, according to his definition g is equal to h minus Ts. If we differentiate
dg then we get dg is equal to dh minus Tds minus sdT.
Dh minus Tds is vdp minus sdT . This relationship we also used earlier, just specific this function.
That is why it is v specific volume, s specific entropy. During this phase change, since pressure
and temperature remains constant which means dp and dT is 0. Therefore, dg is 0.
Another thermodynamic property that is the Gibbs function, or the specific Gibbs function
that remains constant along with pressure and temperature during a phase change; change
of phase from one state of matter to other state of matter. This is one very important
conclusion. If this is so then we can tell that this if we denote it as Gibbs function
of the liquid, and if we denote it as Gibbs function of the vapor, since we are using
g, we are not using f and g; otherwise, it will be confused.
For all properties, we have seen even in this steam table, the subscript f is used for the
liquid and subscript g is used for the tri-saturated vapor. But here we are using g itself as the
specific Gibbs function. I am using superscript l for the liquid and that for the vapor. So,
from this relationship we can take that at a particular pressure and the corresponding
temperature, this g of l of liquid state is equal to g of v that is vapor state.
Consider a small infinite small change in temperature. To find the relation, we have
to do like that an infinite small change in temperature or pressures. Here, the pressure
first; that means, if this is P all the time this change of pen. So, this is p and this
is p plus dp. Similarly, the isotherm which passes through this that means if the saturation
temperature also suffers a small change that means T plus dT. Similarly, the isobar which
passes through this coincides with this constant pressure thing; this isobar is p plus dp.
That means, we change the pressure from p plus dp. What is the corresponding change
in the temperature? Let this be T plus dT. Let us consider that
the Gibbs function in the liquid phase, as a change from g l to g l plus dg l and the
vapor phase Gibbs function changes from this to g v plus dg v. Again, they are equal, because
at this point, this g and this g that means Gibbs function for the saturated liquid and
dry saturated vapor are equal; that means these two are equal. Therefore, dg l which
simply means that change in the Gibbs function for the liquid state will be same as that
for the vapor state so that they again become equal if the states are saturation states.
So how to write then, okay you keep this please Sorry you keep this okay okay okay. I will
write okay. Therefore, we get dg l is dg v. If we write
this expression, dg l is vdp l or dp. dp is the same minus v l minus s l dT is equal to.
We can write for this following this equation v of v dp minus s v dT. This dp, dT is not
0. This dp, dT indicates a change in pressure and temperature like this. There is a change
in pressure and temperature for which there is a change in the liquid phase Gibbs function
which must follow this general thermodynamic relation. That means dg is equal to v dp minus
sdT. Whenever there is a change in pressure and temperature, there will be a change in
Gibbs function. Therefore, dg is this and dg v is this.
From which we get, dp by dT will be equal to s v minus s l divided by v v minus v l.
This s v minus s l is the change in the entropy, specific entropy for change of phase which
can be written as latent heat divided by the saturation temperature into v v minus v l.
Sometimes, the latent heat may be written in terms of the enthalpy change hfg. Sometimes,
it is neither by L nor by hfg. This is one of the very important property relations.
Sorry not property relations, I am extremely Sorry. This is one of the very important relations not property
relations. I am Sorry. dp by dT that is the change in pressure with respect to temperature
is L by T into v v minus v l.
Again, I write this expression. Therefore, we get is that dp by dT is equal to L or hfg
divided by T. That is the change in this specific entropy. This equation is known as Clausius-Clapeyron
equation. Therefore, this equation gives the relationship between the saturation pressure
and the saturation temperature. From this differential equation, we can integrate, but
integration is not very easy. This is because of the fact. This v v is a function of temperature
usually; v l is very small with respect to v v. Sometimes, we can neglect this, because
the specific volume of the liquid is very small compared to that of the dry saturated
vapor. So, hfg is also a function of temperature.
So, until and unless we know the functional relationship of hfg v v with temperature or
pressure, we cannot integrate it. Therefore, this equation can be integrated, provided
we know this variation. Usually, these variations are given in terms of empirical equations
so that one can integrate this equation and express an explicit functional relationship
of p with respect to temperature. This is the relationship of the saturation pressure
corresponding to the saturation temperature. This pressure is sometimes known as vapor
pressure; vapor pressure versus temperature relation. Sometimes, we tell colloquially
that what are the vapor pressure temperature relations? What is the relationship between
the saturation pressures to its corresponding saturation temperatures at which a phase change
takes place? In case of liquid vapor transitions, v v is
always higher than v l, hfg is a positive quantity, T is a positive quantity, dp by
dT is positive. Pressure changes with the temperature.
In case of Solid, liquid transitions that means melting or freezing this equation can
be written as hfs that means sf. I can write latent heat for melting usually this nomenclature
is used. In that case, v v will be v l that is the final phase minus this is the initial
phase that means v s. Now, this will be positive provided on melting, this specific volume
increases. So, for the substances where melting increases the specific volume; that means,
the system expands on melting. dp by dT is positive; means an increase in pressure will
increase the saturation temperature. For substances where v l is less than v s;
that means, this system contracts on melting. In that case, dp by dT is negative. An increase
in pressure will reduce the saturation temperature. What happens in case of water?
In this case, dp by dT is negative and when v l is greater than v s, dp by dT is positive.
So, this is the case for water. In case of Solid, liquid just I give this example, but
we are not much bothered on this aspect. We are only concerned with the boiling, where
the vapor is transformed into liquid. This is so small. Sometimes, it can be written
as hfg by Tv v. For all liquids, v v is much much higher than the v l.
Now, I come to the very simple old and known things, ideal gases.
This will be mostly the recapitulation of the things which you had read at your school
levels with addition of few materials but brushing of your concept.
For any compressible system, a functional relationship between pressure, volume and
temperature are the three fundamental properties of a system. All these three are directly
measurable and sensible also. We can sense pressure and temperature. A functional relationship
between these three properties is known as equation of state. An ideal gas if we ask what is the definition
of an ideal gas? It can be defined in various ways, but all
the definitions are same. One is a corollary of the other. We can start from the equation
of state point of view. An ideal gas is a gas which obeys this relationship as the equation
of state at all pressure, volume and temperatures. At any state points, the relationship will
be pV is equal to m RT, where m is the mass of the gas and R is a constant known as characteristic
gas constant whose value is constant for a particular gas, but varies from gas to gas.
This relationship is very simple relationship and has to be valid at all state points. It
has been found in practice that all the gases in reality do not exactly obey this equation;
this functional relationship between pv and T. Gases only obey with a fair accuracy. This
relationship, when its pressure is extremely low, p tends to 0, very low or temperature
is extremely high. I will not write infinity. This will be odd, because at very high temperature,
the gases are ionized and are converted into another state known as plasma state which
is not a neutral state, ionized state. So, at very high temperature, if the gases are
highly superheated or rarefied, for example, we consider air is at a tremendous superheated
condition or a rarefied condition. I expressed this thing earlier, while discussing the phase
diagram what is the superheated or rarefied state.
Air in a superheated state, is a high degree of superheat, because if we consider air at
the atmospheric pressure then if we think its temperature, for example, today, just
now, I have been told that temperature is 23 degree Celsius. But what is the temperature
at which air is condensed or liquid air boils up at one atmospheric pressure. Though liquid
air is not a pure component, it consists of nitrogen and oxygen, but their boiling points
are very close. We can take a rough idea that it is some minus170
degree Celsius or so. Approximately, how much degree of superheat it is? It is at a highly
superheated condition corresponding to its existing pressure. At the same time, if we
consider the 23 degree Celsius as the existing temperature at which the air can be liquefied.
Yes, it can be liquefied. If we compress the air and to do that we have
to compress the air up to 50 or 100 atmospheric pressure or even more so that air is at a
rarefied state. Therefore, at a very superheated or rarefied state, a gas obeys that means
when the pressure is very low and temperature is very high obey the equation of state. This
is experiments, but theoreticians on the other hand have derived this equation from kinetic
theory of matters, provided we are allowed to assume certain postulate. They have to
make certain assumptions. What are those assumptions? Number one assumption is that, I am not writing
all these things it is boring. Molecules are point masses that they do not have any volume.
The volume of the molecules is neglected in comparison to the volume of the geometrical
system; that is number one. Number two is that no cohesive force. Cohesive
force is the force of attraction or repulsion between the molecules of the same kind. In
fluid mechanics, we have learnt that there are two types of forces. Intermolecular forces,
there exists, one is the force of cohesion which is the intermolecular forces; forces
of attraction and repulsion in between the molecules of the same kind. For a given system
which comprises a large number of molecules, there are internal forces, attraction and
repulsion forces of the molecules which largely depend upon the molecular density. The free
path of the molecule, the distance between the two molecules are in main free path. How
do did we define statistical average of the distance between two successive collisions?
Therefore, these intermolecular forces are known as cohesive force.
Another force is the adhesion force for that adhesion which is the intermolecular forces
between the molecules of the two different kinds between two different systems. So, the
molecular cohesive forces or molecular attractive or repulsive forces are 0, because of which
molecules are capable of moving in a straight line rectilinear motions. So, if we assume
these things then from the kinetic theory of matter, one can derive this relationship
pV is equal to m RT. The kinetic energy of the molecule is a function
of temperature which gives the concept of temperature from statical thermodynamics which
was given by Maxwell and Boltzmann. This is beyond the scope of this classical thermodynamics
which we are dealing with now. If we know this information that is sufficient
that an ideal gas from kinetic theory of matter had these assumptions, that the molecules
do not have any volume and the internal forces between the molecules; that is, cohesive forces
in molecules are absent. Then molecule moves perfectly in a straight line; the rectilinear
motion. One can reduce this, but we are not much bothered of these assumptions in the
kinetic theory. We are only bothered in classical thermodynamics with this equation pV is equal
to m RT.
Now, Avogadro gave a hypothesis after this. Now, pV is equal to m RT. We know after the
Avogadro hypothesis was discovered, there is another way of expressing the mass of a
substance. This is the m whose unit is kg in SI unit, but another way of expressing
the amount of a substance as a fraction of its molecular weight. If I define n as a fraction
of the molecular weight of the system; that means, not in terms of its actual mass in
kg, but in terms of fraction. For example, if there is 32 kg of oxygen.
So, its n will be one, because the molecular weight of oxygen is 32. Similarly, 2 grams
of hydrogen is one; that means, in that fraction 2 grams of hydrogen and 32 grams of oxygen
are expressed in the same number, because these are the multiples of the molecular weight.
If we express the mass in a different unit, what is that unit? Just scale it by the molecular
weight 1 divided by M; multiply it by one by M. This unit is known as mole, number of
mole. This is mass in kg, and if we do that and substitute these in the equation of state,
we get n. I am Sorry. Here we use R bar that we will
be get. no Sorry no no no no no I am Sorry R all right. The R bar I will use in universal
gas constant. If we substitute M is equal to n then M into R, I take in a bracket. So,
this will be the modified equation of state when equation of state is expressed in terms
of the number of moles of the substance. In equation of state, I do not want to use the
mass, but I will use the mass in terms of the moles; that means, divided by molecular
weight. So, M into R into nT. Now, Avogadro told that all gases at same pressure and temperatures
with same number of moles have the same volume. If we have different gases whose number of
moles is same that means their masses are proportional to their respective molecular
weight. If they are at same pressure and temperature, they have equal volume which is not true for
mass. If we have got different gasses at different
mass at same pressure and temperature, the volume will not be same. Density cannot be
same, but if the number of moles is same for all gases at same pressure and temperature.
They have the same volume. This is because of the same identity of the molecules; that
means, molecules are alike. This is the hypothesis which was given by
Avogadro and because of this MR is same for all gases. Now MR, I write as R bar which
is known as universal gas constant. Therefore, we see that if we replace M in terms of n,
we can write pv is equal to n R bar T. R bar is nothing but molecular weight into R and
which is a universal gas constant for all gases. These values, since it is universally
constant for all gases should be remembered. This value is, we do not go more than two
places of decimal, it will be enough to remember kilo Joule per kg mole Kelvin.
If I expect Something to be remembered, I should not go to an accuracy of four places
of decimal which have been found out or theoretically or experimentally, it will be sufficient,
because its unit will be Joule because this p into v is Joule in terms of kilo Joule per
kg mole. R bar T, how do you find out kg mole that is kg by the molecular weight. This is
the universal gas constant. If we write again, the two equations in terms of the specific
volume, one is RT another is pv bar. I will use always these two equations. One is pv
is equal to RT, small v is the specific volume, volume per unit mass. This is volume per unit
mole, just one minute I will listen to all question.
It is volume per unit mole and R bar is the universal gas constant. So, the two equations
are same; they are not independent equation. They are expressed in different form, where
v is the specific volume. Here, v bar is the molar volume.
Okay please question Excuse me sir
yes kg mole Kelvin. Do not use mole Kelvin wrong,
that wrong you from now onwards you write kilo Joule per kg mole Kelvin. Usually some
books write kg mole. Because kg mole has got a meaning that kg by molecular weight is the
kg mole, gram by molecular weight gram mole. Do not use those nomenclatures given in some
books like mole and kilo mole per kg mole, per gram mole. No kilo mole, it is kg mole,
because mole has something to do with kg. So, mole molecular weight is dimensionless,
So, kg by molecular weight is kg mole. We can write kilo mole but I prefer it is
kg mole or gram mole. This is my perception. I will better go with this.
One thing to be again remembered that for all real gases, limit of p into v bar as p
tends to 0, or v bar tends to infinity, rarefied state becomes equal to R bar by T only as
a limit, whereas an ideal gas will obey this, not as a limiting condition at all condition.
Next are two important corollaries of ideal gases. This is the definition of ideal gas
starts with the equation of state.
Now, two corollaries are that the specific enthalpy h or total enthalpy we can take.
I will deal with specific values only and specific internal energy are functions of
temperature. Here I tell you one thing. You will appreciate it afterwards just as an information,
afterwards you will appreciate. When the reaction part is not coming into
picture, it is always better with the specific values rather than the molar values. But the
molar properties come into picture when reactions are occurring; that means, when thermodynamics
of the reactive systems come into picture. Because we will see, the significance of dealing
with mole in case of reaction and the significance of dealing with mass, the specific property
when reaction is not there. That is why, now, I will deal with the specific value. So, h
and u are functions of temperature only. But probably you know by this time we have already
derived it in course of our other studies but again I will do that.
How to prove that h and u are functions of temperature only? Let us write this expression
that dh. If we recollect the very important thermodynamic property of relations which
are most probably the most important and fundamental property relation is T ds plus v dp.
Now, what is Tds from the first Tds equation, second Tds equation? If we write from second Tds equation, we know that Tds,
I am writing in terms of the specific entropy. Therefore, it will be cp dT minus T into del
v this is the specific value v, cp because s is specific at constant pressure dp. This
is the second Tds equation. This is one of the very important relations Tds.
If we replace this, while finding out the Joule Kelvin coefficient; that is, the similar
way, dh is equal to cp dT plus v minus T into del v divided by del T at constant p into
dp. This cp can be written as del h divided by del T at constant pressure that means h
is a function of T and p that means as if dh is equal to, if I write h is a function
of T and p, I can write del h divided by del T at constant pressure into dT plus del h
divided by del p at constant temperature into dp. This we have already recognized.
This is nothing, but del h divided by del p at constant temperature. Now, what is this
value for ideal gas? If we take the ideal gas equation, pv is equal to RT. If we use
this to find this, this will be 0, because T del v divided by del T at constant p will
be v. If we differentiate del v divided by del T
at constant pressure it is R by p. del v divided by del T at constant pressure will be R by
p. RT by p means v, v minus v. Therefore, this becomes 0 which proves that del h divided
by del p at constant temperature is equal to 0. Therefore, h becomes a function of temperature
only. That means if I write dh is cp dT, from this equation, we can write dh is cp dT which
means this h is a function of temperature only. h cannot become a function of any other
quantity, because h can be expressed as p and T, but it is not a function. But how do
we explicitly prove that h is not a function of volume that we can do? There are several
ways.
One of the ways I am telling it is unnecessary that pv is equal to RT. So, if we keep the
temperature fixed then volume will vary provided there is a variation in pressure. Therefore,
at constant temperature, enthalpy cannot change with volume, if it does not change with pressure.
Therefore, enthalpy does not change with volume, if it does not change with pressure at constant
temperature, because this is the equation of state.
Therefore, we see enthalpy is a function of temperature only. The definition dh is equal
to cp dT which means dh divided by dT is equal to cp. Earlier, in first law, we recognized
that definition of cp is del h divided by del T especially, at constant pressure at
p. This is valid for all substances. But since h is not the function of pressure, this can
be simply written dh divided by dT. cp is dh divided by dT. Similarly, we can show that
u is also of function of temperature. How to do it?
In this case, we have to write du. We will start with du. What is this equation? Another
fundamental and most important and popular expression is du minus pdv. Tds is du plus
p dv. This is a probability relationship. Is it first law of thermodynamics?
yes sir How many of you are telling yes.
du is equal to Tds minus p dv is the first law of thermodynamics
no no no no no it is not the no it is not the first law of
thermodynamics but can it be first law of thermodynamics
For a reversible process, with only displacement work, but this is the probability relationship.
It may or may not be the first law of thermodynamics. It is first law of thermodynamics under a
special situation. When the process is reversible and had only displacement work, as the work
displacement mode of work as the work transfer du is equal to T ds minus p dv, because in
the viva portion these questions will be asked. Nobody will ask you to deduce the Clausius-Clapeyron
equations okay In a similar way, if we recall the first T ds equation. It is cv dT plus
T into del p divided by del T at constant v into dv.
We write it here, du is equal to cv dT plus T into del p divided by del T at constant
v minus p into dv. cv is equal to del u divided by del T at constant ; that means, du is expressed
as functions of T and p and this is nothing but del u divided by del v at constant T.
What is this quantity? If we evaluate this quantity for an ideal
gas, where pv is equal to RT. This becomes 0; that means, u is a function of temperature
only. These are very simple thing. Therefore, one of the important corollaries of the ideal
gas is u and always du is equal to cv dT. h and u are functions of temperatures.
Therefore, we come across this thing that for ideal gases, we can write du is equal
to cv dT and dh is equal to cp dT. If we make this is total internal energy then we make
the heat capacity cv, that both side multiply with much similar is the case. dH capital
H that is the total enthalpy is the heat capacity cp into dT.
If we integrate this, we will get u is equal to integration of cv dT plus some arbitrary
constant. Similarly, h is cp dT plus some arbitrary constant. These constants are neither
functions of pressure nor function of volume, just simply a constant, numerical constant.
To express this u and h as a function of temperature, we have to integrate this equation. If we
know this cp and cv explicitly as functions of temperature. For an ideal gas, we also
assume cp and cv are constants. They do not become the functions of temperature or any
other properties. Sometimes, this is known as calorically ideal or calorically perfect
gas. For ideal gas, when we tell, we always assume that it is a calorically ideal gas
until and unless it is told. These are not very rigid things. Sometimes, we express that
this is an ideal gas, but with a variation of cp, cv with temperature.
Sometimes, when we realize this assumption make this cp, cv varying with temperature,
we tell the gas is semi ideal. So, several terminologies are there which has to be exclusively
mentioned. If nothing is mentioned exclusively, if it is ideal gas given that means we will
assume cp, cv are constants. But when cp and cv are constants, we tell that calorically
perfect gas. Sometimes, perfect gas is a synonymous term for ideal gas.
If we consider a gas to be calorically perfect or ideal then cp, cv are constants. We consider
for an ideal gas, another assumptions like pv is equal to RT as the equation of state
that cp,cv are constant. Then it becomes simply u is equal to cv T plus c1 is equal to cp
T plus c2. So, as long as the difference is considered, del u is equal to cv into delta
T. That means I can integrate this. If cv is constant, then I do not care with this
constant of integration. That means u2 minus u1 is equal to cv T2 minus T1. This thing
we have already used with the problems of first law h2 minus h1 is equal to cp into
T2 minus T1. If I want to ascribe the specific internal
energy, specific enthalpy at a particular state point, I have to know this c1, c2. These
c1, c2 are made by a convention for all ideal gases at T is equal to 0; both u is equal
to 0, h is equal to 0 so that one can write u is equal to cvT and h is equal to cpT. So,
these equations are valid.
Already we have proved another result that if pv is equal to RT is the equation of state,
v is the specific volume. Then the value of cp minus cv is R. This already we have proved
from thermodynamic relation; cp and cv. From two tedious equations, we found out the expression
of cp minus cv in terms of the derivative of other thermodynamic properties and utilize
this relationship as the equation of state for ideal gas. We have shown that cp minus
cv is equal to R. This is one of the important relations for an ideal gas.
Another important relationship for ideal gas which is often used in heat transfers studies.
That beta, which I defined earlier, is the volume expansibility or volume expansion coefficient.
When we read heat transfer then beta for an ideal gas is 1 by T then, you will not be
able to tell them your heat transfer teacher may ask that how you have read thermodynamics.
It may not be very much related to thermodynamic studies in free convection studies on heat
transfer. We always encounter with this parameter v; volume expansibility is 1 by T for ideal
gases. Do not consider that this definition is 1 by T for an ideal gas. I will now prove
by simple two line proof. Beta definition is 1 by v del v divided by del T at constant
pressure. We use pv is equal to RT and get these things
1 by T pv is equal to RT. Find out del V divided by del T from here at constant pressure and
get these things. We will get simply 1 by T volume expansibility. We see the dimension
is temperature, and it exactly becomes 1 by the absolute temperature in case of an ideal
gas.
Another very important relations for pv for a particular process in an isentropic process.
We know that displacement work W is integral of p dv per unit mass. But can we integrate
this? To integrate this, we have to know p as a function of volume. Therefore, we have
to know the path along which the process has executed this work transfer. That path is
specified by the functional relationship of pv that depends upon path to path. So, if
we know this relationship, we can integrate this. One of the paths is isentropic process.
Let us consider that isentropic process. s is equal to constant. Basically, isentropic
process is a reversible adiabatic process. Now, we write very two important equations;
T ds is du minus pdv and T ds is dh minus vdp. For ideal gas, this equation is valid
for all processes property relationship.
yes
First one is Tds is du plus second one negative. These two equations are property relations
valid for all. Now, if I write this as T ds is equal to cv dT plus pdv. Is it valid for
all systems?
It is valid for ideal gas. A form of this equation only for ideal gases
and this is also cp dT minus vdp. It is valid for ideal gases. Now, I will write these two
property relations for ideal gases and consider an isentropic process that means the left
hand side is 0. Then if we divide one by other that means if we take this side that means
pdv is equal to cv dT and vdp minus and vdp is cp dT. We divide it then we get dp by p
plus cp by cv into dv. I know the result. So without seeing that I am writing you can
also write it it will come like this. If this is represented as gamma and integrates
it; that is, ln p plus ln v gamma, gamma lnv means lnv to the power gamma and in that case,
constant is also logarithm of some constant ln c. So, that pv gamma is constant. This
is a very important relationship for an isentropic process. In school probably, you have done
this.
pv gamma is constant is valid for an isentropic process executed by an ideal gas, where gamma
is constant and equals to cp by cv. Gamma is always greater than one, because cp is
always greater than cv that has been already proved.
Now if I equate this is one important thing, we can think it trivial it is not very important
but one you do it. It is better that remember this formula otherwise there will be no problem
in deriving this. If we have to evaluate this integral from
a point one to state point two, and if we have got a relationship of this Sometimes
for all other processes, so all other reversible processes, it is possible to express the relationship
as this pv n is equal to constant, where n has got any value. When n is equal to gamma
the process is isentropic. When n is equal to 1, the process is isothermal. pv isothermal
for an ideal gas, all for ideal gas and n is equal to 0 isobaric.
For any value of n which may not be equal to1, gamma, 0? It represents the reversible
process which may not be isentropic, may not be constant pressure, may not be constant
temperature, but it is a reversible process. When a reversible path of the process is prescribed
by a functional relationship of pv in the form of a power law that p v to the power
n is constant. Then one can integrate this. Let this is equal to constant c. Then one
can relate it constant, dv divided by v n. So, if we do this, then we get v c 1 minus
n from one to two. So, two v2 one minus n and c is p2 v2 n. So, this becomes p2 v2 minus
p1 v1 by 1 minus n, because v2 one minus n minus v1 this bracket. After putting the limit,
this will be like that. When it will be multiplied with c which is
p2 v2 n, here that is p2 v2, when we will multiply with that we will use p1 v1. So,
this is a very important expression. If n is equal to gamma, this will be one minus
gamma. So, depending upon the value of the index and the corresponding values of pv is
that initial and final state, the sign of W will be determined so that we can find out
whether the work is coming out or work is going into the system. Next is pdv for a polytrophic
process. Then we come to the entropy change of an ideal gas.
How to calculate the entropy change of an ideal gas? The entropy change of an ideal
gas is very simple. Again we start from the same equation that T ds is equal to cv dt
plus pdv, T ds is equal to cp dT minus vdp. Now, for the two points, pressure, volume,
and temperature, all these things may not be given. We may know pressure temperature,
we may know volume temperature or we know everything. So, we can use any pair. So, if
we use the temperature and volume, we use this ds is equal to cv dT by T. pv is equal
to RT. So, p by T is equal to R dv by v. Therefore, s2 minus s1 is cv ln T2 by T1 plus R ln v2
by v1. Any question all right it is very simple school level.
Similarly, we integrate the second one. We get cp ln T2 by T1 minus R ln p2 by p1. So,
the change in entropy between these two states given by T and v and p and T we can use, and
p and v we can make from these two equations. We can find out the relationship between p
and v. This way we can find out the expression for entropy change of an ideal gas. Therefore,
we have to remember for ideal gas all those corollaries while solving problems.
After this I will start the real gas, what is a real gas? The few real gases and how
we can express the equation of state for a real gas.
But I think today the time is limited the time much time is not there. So, I will start
the expressions for real gas. How it deviates from the ideal gas and how we can express
the equation of state of a real gas compressibility factor and all these things. I will discuss
in the next class. But before that you ask me any questions you
do have. Today is extremely simple what I have taught. But regarding just a minute till
five minutes we can spare if you feel So that we can discuss anything about the change of
phase diagram and all. Any questions? No question.