Good afternoon. Last class, we discussed the
behavior of ideal gases which obey the equation of state as p into v the pressure and the
volume equals to mass into the characteristic gas constant and temperature. In another form,
if you express in terms of moles, the amount of the gas then p times v becomes equal to
the number of moles times the universal gas constant is constant for all gases and the
temperature. Following this equation of state as the first line of the definition of an
ideal gas, we derived a number of corollaries; that is enthalpy and internal energy are functions
of temperatures. At the same time, we assumed that the specific
heat, and constant pressure and volume are constant for an ideal gas. Sometimes, for
an ideal gas, we will also tell that specific heat at constant pressure and volume may vary
with the temperature, but usually, we assume for an ideal gas, these are constant and those
cases, the ideal gases are called calorically perfect gas.
We derived the expression relating to pressure volume, pressure temperature and volume temperature
for an isentropic process. With the use of these equations that enthalpy and internal
energies or functions of temperatures and the equation of state, we derived certain
equations relating to work transfer in the displacement mode or that is pdv and all these
things. Also, the entropy change in a process, how it relates the pressures and temperatures
at the two equilibrium states executed by an ideal gas in a process.
Today, I will discuss first what are real gases? We also realize that the gases in reality
may not always obey the ideal gas states. The ideal gas equations are fairly accurate
for all real gases when they are at a extremely superheated or ratified state which can be
theoretically stated that the limiting value of the pressure and volume as pressure tends
to 0, or volume tends to infinity, becomes equal to R bar T.
For real gases that means the gases in reality, this is rather the equation that p, let us
consider the molar volume pv and as p tends to 0 or molar volume tends to infinity. We
get R bar T, or in terms of the specific volume, same thing limit pv as p tends to 0 or v tends
to infinity is RT, where R is the characteristic gas constant. Therefore, we see under one
limiting conditions, this equation that pv is equal to R bar T and pv is equal to RT
are valid, but not in all ranges of pressure and temperatures. It has been found from experiments that for
real gases where the pressures are very low and it is not at a very high superheated condition,
the pressure volume temperature relationship is not followed by that of the ideal gases.
At the same time, it has been found experimentally that a unique relationship between pressure,
volume and temperature is very difficult to be obtained in a wider range of pressure and
temperature. That is why, sometimes it is like empirical equations. When we develop
empirical equations for any process, the performance of a system that output parameter has a function
of input parameters or operating parameters. These relationships are valid within the range
of the experiments, including certain range of operating parameter.
Similarly, different functional relationships of pressure and volume are obtained at different
ranges for pressure and volume and temperature for real gases. It is not like that an ideal
gas at all ranges of pressure, volume and temperature, we can substantially take that
equation of state pv is equal to RT or R bar T depending upon whether it is specific volume
or molar volume. But there are equations which are only valid within a range of certain pressure
and volume. Usually, in practice, when we deal with real gases and solve for problems
relating to work and heating directions in a process, change of entropy in a process,
change of internal energy in a process or evaluating enthalpy, we are supplied with
the equation of states; within certain range, this equation is valid and within certain
range this equation is valid. So, we do not have to remember any equations.
There are different equations proposed by different scientists, which are valid for
certain gases. They specify these are the gases within these ranges and thus they obey
these equations; that means, you have to know simply that equation of state is a functional
relationship between pressure, volume and temperature. It is usually obtained from experiments
and these are being supplied if I know this relationship. Then we can apply this equation,
just is a mathematical operation, to evaluate all other thermodynamic parameters; the heat
work interactions, change in entropy and all these things.
One such equation of state was proposed by a scientist Van Der Waals and this is known
as Van Der Waals gas. Van Der Waals gas or Van Der Waals equation of state is a gas which
obeys this equation of state. It is beyond the scope of this study that will deal with
in details with all types of real gases and what are the probable equations and all these
things, because there are plenty of empirical equations in different ranges. but I will
tell you that how usually the real gases are being specified in terms of their equation
of state. One very important real gas is Van Der Waal
gas. It has been found that in a wider range of pressure and temperature, many real gases
follow a particular equation of state. Let me first write this equation of state, then
I will explain. The equation of state is like this; if p is the pressure in terms of the
molar volume, write p plus a by v square into v minus b is equal to R bar T, where v is
the molar volume. R bar is the universal gas constant whose value is same for that of ideal
gas, where a and b are parametric constant defining these Van Der Waal equation.
Therefore, we see this relationship between pvT. Even if we draw this curve in a pvT plane
as a surface or in pv for different values of T, the parameters a and b are there. That
means different gases obeying Van Der Waals equation will have different values of a and
b which are typical to the particular gas obeying the Van Der Waals. That means, Van
Der Waal gas represents a class of gases over a certain range of pressure, volume and temperature
which are being distinguished from one another by the different values of a and b.
This has been verified experimentally and it has been proposed from experimental results,
but Van Der Waals initially deduces this equation of state from kinetic theory of matter. The
way the ideal gases are developed by the kinetic theory of matter with the assumptions that
when the molecules move in a rectilinear path, there is no force of attraction and the volume
of the molecules are negligible with respect to the volume of the container in which they
exist. From that he relaxed. Actually, he made the
restrictions for these two assumptions; one is the volume of the molecules were taken
into account; that is, the volume of the molecules were not considered to be negligible. At the
same time, the molecular cohesive forces were taken into account, because in real gases,
the molecular cohesive forces will be there. So, he developed in this fashion. I am not
going to make a full proof of it that the pressure which is exerted now. Since the volume
of the molecules are taken into account, it is not negligible compared to the volume of
the container. So, molecules get less space for their movement.
If this be so the number of collisions that the molecules make, for example, if the wall
for which we define the pressure as the change of momentum, because of this collisions that
number of collisions increases. Therefore, the pressure exerted by a real gas on the
wall becomes more than that of the ideal gases. This factor Van Der Waal showed that a part
of it is taken to be as a kinetic pressure which showed that it should be proportional
to R bar T by v bar is the pressure for an ideal gas. It is proportional to this value,
where b is a volume parameter. But b indicates the volume excluding the molecular motions.
If we exclude the molecular motion, what is the volume that is b which means that v minus
b is actually the volume that is the restricted volume which is in access to the molecules
for its motion and he proved that it is in linear proportion to that. That means, v bar
minus b is less restricted volume. Therefore, the kinetic pressure that means the pressure
which it should give is equal to this; that means, this is more than the ideal gas values.
Apart from that actual pressure will be less by this kinetic pressure by an amount which
takes care of the molecular cohesiveness. We have read in fluid mechanics that molecules
are always being pulled by other molecules; that is, the force of attraction, by its neighboring
molecules. That means, if you concentrate our attention on a particular molecule and
we will see a molecule on the bulk which is far from the free surface, far from the wall,
is being pulled by all neighboring molecules from outside so that a molecule is almost
in a balance. But if we go very near to the wall or very near to the free surface, we
see there is a misbalanced force. we may know that the molecules are being pulled by more
number of molecules. If we concentrate our attention on a single
molecule, by more number of molecules from the bulk of the fluid, because there are very
fewer molecules near the free surface or near the wall. Therefore, if we consider a molecule
near the wall or near the free surface, there is a net attractive force of the molecule
towards the bulk of the system or bulk of the fluid or bulk of the gas. In this case,
it causes a reduction; rather, it causes a reduction in the pressure, the way the surface
is being stretched. That means always there is a surface in stretched condition. As we
have read in fluid mechanics, to create a surface we require energy, because a molecule
has to be brought to the surface against the inward attraction of the molecule.
Similar is the case, if a molecule has to be brought to the wall, work has to be done.
That means what happens when a molecule comes very close to the wall? Net pull on the molecule
is there towards the bulk of the gas for which a reduction in the pressure that is from the
kinetic pressure is there. He proved that it is directly proportional to the molar density.
By common sense, we can guess the number of moles per unit volume is more, so force will
be more so that the pressure will be less. Therefore, it is inversely proportional to
the molar volume. If we substitute pk we will get a relationship
like that. It is not a detailed proof, but thought that this will be sufficient enough
to know that the origin or the basic genesis from which this comes. That means a by v square
term accounts for the molecular cohesiveness or the molecular cohesive forces which were
neglected in case of an ideal gas. b is the parameter or term which takes care of the
volume of the molecules. In case of an ideal gas, p plus a divided by v bar square and
v bar minus b both are 0 and pv is equal to R bar T. This equation of state is known as
Van Der Waals equation. Many real gases obey this Van Der Waals equation of state over
a certain range of pressure and temperature. The next thing is that we see the Van Der
Waal gas. If we again write this that p plus a divided by v square into v minus b is equal
to R bar T. In this case, if we solve for v, v will give three roots. It is cubic in
various equations; that means, for a given pressure and temperature, v will be having
three roots. But it has been found that if we solve this, amongst these three roots only
one root is real at high temperature and at low temperature. There are three real
roots, but out of that one root is in a metastable state; only this information will be sufficient.
Another root which is real, but it is not feasible. That root will lie on a curve which
gives a negative compressibility. Always, for all substances, if we increase the pressure,
volume decreases. Therefore, one of the roots, one is metastable state and another root represents
a physically impossible value. Therefore, only if you solve this, we always get one
real and physically possible value of v. So, this is mostly all about the Van Der Waals
equation. After that I will tell this virial form of
equation of state of a real gas. Like Van Der Waals equation there are number of equations
which we can express as I have told for different gases obey at different pressures and temperatures.
But one scientist Kamerlingh Onnes is a German scientist. He proposed that all
real gases should be expressed in this fashion; some constant A into one plus B dash p plus
C dash p square plus D dash p to the power three in an infinite power series. Either
these two A’s are same, or one plus B divided by bar plus C divided by v bar square plus
D divided by v bar cube in an infinite power series. I can use B C D here or dash here
doesn’t matter this is the usual convention that I am following.
If we can express this in power series then we can closely approximate all real gases
by properly evaluating these values of B dash C dash D dash and B C D. Here, we see this
A, B dash, C dash and D dash or A, B, C and D are all constants. They are functions of
temperatures; that means, at a constant temperature these are giving numerical values. So, A,
B dash, C dash and D dash which are the coefficients of this power series; either in p or in v
bar, they are functions of temperature. If we recall this limit pv bar as p tends to
0 or v tends to infinity always satisfies this relationship and we get A becomes equal
to R bar T. Therefore, this equation can be expressed
as pv bar R bar T is equal to one plus B dash p plus C dash p square plus D dash p cube
plus like that. Similarly, in terms of molar volume, B divided by v bar plus C divided
v square plus D divided v cube plus this in finite series. Therefore, we see for an ideal
gas, B dash, C dash, D dash B, C, D becomes 0. pv dash by RT if we represent as z then
we can write z is equal to one plus B dash p plus C dash p square plus D dash p cube
not A B dash C dash is zero sorry I am sorry I have told wrong I am sorry.
For real gas, B dash, C dash, D dash and B, C, D is 0; that is hypothesis that is not
wrong what I want to mean is that when p tends to 0 and v tends to infinity, obviously these
terms become 0. But for an ideal gas I am correct that these coefficients are 0. This
pv bar by RT is known as compressibility factor; this z is known as compressibility factor.
This compressibility should not be confused with the compressibility of the fluid mechanics,
but this gives almost a same concept physically, why because this is something to do in with
the compressible that is the compression. Here, you see this is known as compressibility
factor. First of all, we see mathematically that z is compressibility factor whose value
for ideal gas is 1, but for real gases are different from 1. Why it is compressibility
factor? Because of the compression, a gas at a ratified state satisfies the ideal gas
laws, but when you compress it at high pressure then it deviates from the ideal gas laws.
That is why z is defined as the compressibility factor when z is equal to 1 that represents
an ideal gas. This compressibility, usually a function of
pressure and temperature may be a function of volume and temperature. Let us draw a figure
z versus p at different values of T as parameter. I am showing a particular curve like this.
At different temperatures T1, T2, these are plotted. This is known as compressibility
chart; that means, if one can find out at a particular state point, what is the value
of z from the compressibility chart?, or compressibility table just like steam table. We know the steam
properties at a particular pressure or temperature either saturated state or superheated state
and corresponding tables we know all the properties. Similar way, if we know the compressibility
factors then you write the equation of state for any gas at z R bar T, where z is function
of p and T. That means this is a multiplying factor; that means, pv bar is equal to z R
bar T, where z is equal to 1 for an ideal gas. For a real gas, we have to find out the
value of z corresponding to that particular pressure and temperature. z is a function
of the state point, because state point will decide how much deviation is there from the
ideal gas. This is the definition of the compressibility and this is the compressibility chart.
Similar to the concept of fluid mechanics, as we have seen that it has been found that
for a Van Der Waal gas p plus a by v square into v bar minus b is equal to RT. Now, for
different values of a and b, it constitutes a different equation of state. Equation of
state will vary; the structure is same and functional relation is same. Because they
belong to a particular class; Van Der Waals equation of state, but the particular equation
will go on varying. If we have got ten gases following, for an example Van Der Waal gas
with different values of a and b, we will have to provide compressibility chart; for
ten gases ten compressibility charts. Try to understand z as a function of p and T,
but if we want to make that for a particular class of gases following a particular type
of equation of state by a single equation of state, there will not be any parameter
which is distinguishing from gas to gas. In that case also, we can express the compressibility
factor in terms of some normalized dimensionless pressure and dimensionless temperature which
will be unique just like following the principle of similarity. For an example, we come to
the fluid mechanics. We know that pressure drop is a function of flow rate for a given
fluid for a given diameter of the pipe. Pressure drop per unit length, for example, in a parallel
flow, the pressure drop per unit length is constant. So, pressure drop per unit length
is a function of flow rate for a given fluid; that means, for a given value of row mu and
for a given diameter. If I have to provide a figure, pressure drop
versus flow rate then we will have to mention this is valid for a pipe of this diameter
for a liquid of this density and this viscosity. If we have to show this similar relationship
delta p versus q for different pipe diameters at least for ten diameters, I will have to
produce ten curves. Again, for each diameter, if I have to show
the influence of density then I will have to produce so many curves for variation of
density. This is the relationship q, but it is not done. Then from the principle of similarity,
we have shown also from the theoretical equations, the Navier Stokes equation. It has been found
that this phenomena that pressure drop per unit length depends upon the flow rate, depends
upon the density, depends upon the viscosity and depends upon the diameter.
All these things clubbed together, we define implicit functional relations of all these
variables then apply the Buckingham’s pi theorem and show. Ultimately, this entire
physical phenomenon is guided by two governing parameters in non-dimensional forms. They
are known as pi terms. 1 is the friction factor and another is the Reynolds number.
So, density, viscosity, diameter and velocity; that means the flow rate in terms of the average
velocity of flow is clubbed in one dimensionless term. Reynolds number and the pressure drop
is normalized with half row v square is one parameter as the friction factor. So, we give
the value of friction factor versus Reynolds number which takes care of pressure drop flow
rate curve for all diameters and all fluids in a particular group. What is that group?
Newtonian fluid where in non-Newtonian fluid there is the different relations. Newtonian
fluid gives a group homologous series in fluid machines. Also, we will read that machines
of similar geometrical shapes are falling on a particular class these are known as homologous
series in a particular group. Similarly, here also, a number of gases following the Van
Der Waals equation can be expressed in terms of this dimensionless parameter so that one
can express one single equation of state. A single compressibility chart to work with
all gases of this particular kind with this genesis and theory the concept of law of corresponding
states came. Law of corresponding states, this is very
important. In thermodynamics, we tell that this corresponding word is same to the similar
points that the similar parameters. To do that, the basic physical concept is taken
that pv diagram, because there has to be some sort of similarity. Otherwise, the similarity
parameter that is pi parameters cannot be taken. For example, when I compare the two
pi flow problems with different diameters or different fluids then I see that there
is some similarity in the physics of flow. The two flows are governed by the pressure
forces and viscous forces. Similarly, when the flow in a surface waves
or in a free surface flow, the physics are same. The two flows different situation, different
liquids, different geometry are governed by the gravity force and the inertia of forces.
Similarly, some common similarity should be there. What is that it has been found that
this vapor dome is similar for all substances; that is, there is a similarity in the structure
of these vapor dome. That means if you draw the pv diagram, you see their structure is
same quantitative value maybe different. There is a critical point and the slope and all
these things are similar; that means, they reveal a geometrical similarity on this pv
diagram. Therefore, from this it was thought that similarity
parameters can be resorted to make the similarity principle valid there so that we can do what
we have told. For that, what we do? This critical point which is different for different gases,
even the Van Der Waals gases, there may be number of gases following the same equation
of state, but having different a, b values different critical point. Similarly, for ideal
gases also all gases are not saying, because their values of R the characteristic gas constant
may be differing. For ideal gases one thing is unique that if
we express in terms of the molar volume, they reduce to a single equation. R bar is same
for all gases which is not so for the ideal gases. Even if we express in terms of the
molar volume a, b parameter, for an example, Van Der Waal gas all gases will be having
such parameters; they are not same. Their critical points are also different, but we
can use these value pc, vc, Tc to normalize the corresponding dimensional parameter p,v,T.
We define that dimensionless pressure as pr is equal to p divided by pc; that means these
are used as the reference variable for normalizing the actual dimensional variables.
We define Tr is equal to T divided by Tc and vr is equal to v divided by vc and this non-dimensional
terms in science of thermodynamics are known as reduced properties. That means pr is the
reduced pressure, Tr is the reduced temperature and vr is the reduced volume and if you do
so, we will see that this equations take a unique form. I just give an example . Let
us do it for Van Der Waals gas. Van Der Waals equation is p plus a by v bar
square into v bar minus b is equal to R bar T. Along with that, we have to take care of
another physical fact that if we draw this pv diagram for example these are liquid vapor
as you know I am not drawing all these thing these are known.
So, one thing is that at the critical point, you know these are the isotherms. So, the
isotherms at the critical point have a 0 slope and 0 curvature. This is a physical fact,
which means mathematically, del p divided by del v; this is for all gases, del p divided
del v at the at Tc; that means, the slope of the isotherm that is T is equal to constant
which is Tc. Here is the isotherm that at Tc this slope is 0 and del square p divided
del v square at Tc is 0; that means, this is almost flat here which has got both a 0
slope and the 0 curvature. If we use this, we write p is equal to R bar
T divided by v bar minus v minus a by v square. Then if we find out del p divided del v at
at Tc is equal to minus R bar T by vc bar minus b whole square plus 2 a divided by vc
bar whole cube. This is one equation; another equation is the second derivative. Then second
derivative del square p divided by del v square at Tc second derivative; that means, again
it will be plus 2 R bar T divided by vc bar minus b whole cube minus 6 a by vc bar to
the power four is 0. So, that is another equation. If we solve these two equations that I am
not doing it here the simple algebra then we get
a, b and R bar in terms of pcv. This we have done at the school level understand but again
it will be recapitulation cannot help. We get b is equal to one-third vc molar volume
at the critical state and a is equal to 3 pc vc bar square and R bar is equal to 8 by
3 pcvc divided by Tc; this is zc that is compressibility at the critical point. If you substitute this
in the Van Der Waals equation of state then the equation of state becomes like this; 3
by vr bar square into 3 vr minus 1 is equal to 8 Tr. This Van Der Waals equation of state
is valid for all gases, following the Van Der Waals equation a, b is eliminated. In
that case, if we define z by zc as I reduce of course z is non-dimensional, but if we
scale it with zc and if we plot zr as a function of pr Tr, we get only one equation.
That is a series of curves in zr, pr plane with different values of Tr is unique that
means this families of curves are unique for all Van Der Waal gases. This is the theory
behind the law of corresponding state. This is expressed like that and corresponding states
means a state of a system gas. For example, Van Der Waals gas or any gas which are having
the same pr, vr, Tr are known as corresponding states.
Similar situations in fluid flow; that means the same Reynolds number, in two flows same
Froude’s number. All these things that means their actual velocity may be different, their
actual gravity condition may be different, but they are having the same Reynolds number;
that means, the situations are similar. Here also, the p, v, T maybe different at
two states, but for the two gases. If we normalize the pc, vc, Tc and if we see the reduced properties
are same, then they are told as corresponding states that means states are similar. For
corresponding states, we get the same value of zr. For example, a Reynolds number is fixed;
friction factor is fixed, but if we have got a higher velocity then we get a higher pressure
drop. If we have got a lower velocity, we get a lower pressure drop. That will be decoded
from the friction factor which is unique, because friction factor, if we decode in terms
of the pressure drop, we will see automatically, the flow for which velocity is high we will
get a more pressure drop; for flow whose velocity is less, there will be a less pressure drop,
but Reynolds number is same. That is if velocity is high, for that case,
d may be lower or rho mu may be lower. So, I adjusted that Reynolds number is same. This
is the concept of the similarity situations. Similarly, here, the same similarity situations
are just replaced by the term in thermodynamic science; that is, the law of corresponding
state. So, there is nothing much in the law of corresponding states.
Now after this very simple I will leave you today early do not know how much I can tell. Let us come to mixture of ideal gases. First,
I will tell mixture of ideal gases, though mixture of real gases excluded from these
syllables for you. Now, again, go back to ideal gas. This is some information about
the real gas and especially about the Van Der Waal gas. As an example, how we can reduce
it into in terms of the reduced properties and the concept of corresponding states, how
you can express the real gas in terms of a virial expansion coefficients. There are of course I have forgotten to tell
this coefficients b, c, d are known as virial and this expansion is known as virial expansion
or virial series. Then the definition of the compressibility and compressibility is 1 in
case of an ideal gas and the law of corresponding states. Let us go back to mixture of ideal gas. In
many situations, we have to deal with a mixture of ideal gases; that means, it is not a single
component system. For example, air itself is a mixture of ideal gases, but its composition
is throughout same that it behaves almost like a single component. That is why it is
a pure substance, but somewhere composition may change sometime. However, at any state,
mixtures of ideal gases also behave as an ideal gas. This is a hypothesis. This has
been found experimentally. This has been proved from kinetic theory of matter; that is, if
we assume all the ideal gas is hypothesis then mixture of ideal gases are also ideal
gas; that means if a mixture of ideal gas is there, for a fixed composition this gas
will behave as an ideal gas the way air does. So, mixture of ideal gases, the first thing
is they behave as an ideal gas. There are certain laws; for example, if we
have in a container for example there are number of gases with mass mA, mB and mC. There
are n number of gases with mass mA and corresponding number of moles are there nA nB. Like that
there are number of gases. When a mixture of gases is in equilibrium, means what? The
pressure and temperature are same. That means the entire system having a pressure p and
a temperature T. In this case what happens? When we mix two gases what happens to that
ideal gas hypothesis? The molecules do not have any volumes. That means there is an intermolecular
diffusion; both the gases diffuse into one another. What happens if there is a fixed
container? All the constituent gases occupy the same volume. Therefore what happens?
Each gas is under an expanded state; that means, its pressure is reduced, because it
has to expand to its volume. That means it goes in an intermolecular diffusion with the
other gases. Therefore, the pressure exerted by each gas individually is less than that
of the pressure which the mixture is exhibiting. That pressure is known physically as the partial
pressure which means that if any one of the gas, at the same temperature, the temperature
of the mixture is allowed to exist alone in that particular volume, because exactly it
is occupying the same volume. Then the pressure it could have exerted because other molecules
are there of the other constituting component is known as partial pressure.
If we deal with the mole then it will be better. That means I can write if the partial pressure
of A is pA and the volume is V of the container that means the mixture pAV is equal to number
of moles nA R bar T, where R bar is same for all constituting components ideal gas that
is one good thing. This is the mathematical equation which defines
the partial pressure and which I have told just now is the physical concept; how you
can conceive the partial pressure of a particular constituting ideal gas in a mixture of number
of ideal gases. Therefore, this is the basic equation for the partial pressure. Similarly,
if we define the partial pressure for the component B then we get another expression;
that is, pB V is equal to number of moles nB R bar T. nB am sorry very very very very good and ultimately,
all these things can write for the nth component is equal to nn R bar T. If I add this, we
get pA plus pB plus dot dot dot upto pn V is equal to nA plus nB plus dot dot dot up
to nn into R bar T. Now, what is nA plus nB plus dot dot dot? These are the total number
of moles of the gas, because they are the amount of thing. So, amounts are additive.
These are the extensive properties and therefore, this for the entire mixture as an ideal gas.
I can write its pressure into its total volume of the ideal gas; ideal gas as a whole, I
am now considering as a pure substance. Ideal gas as a whole is a pure substance if it has
got same composition which does not change. pV is equal to n R bar T. If I compare, the
pressure of the mixture that is sometimes known as total pressure is equal to the sum
of the partial pressure. Who is the person who first derived developed it?
Dalton is the scientist. That is why it is known as Dalton's law that sum of the partial
pressures equals the total pressure. If we see that for a given particular component
pAV is nA R bar T. For example, here, If we see for any component
kth component, let us write for kth component that is equal to nk R bar T. If we now substitute
R bar T as sigma n like this, that means nk divided by sigma n pV, then you cancel V.
The ratio of the partial pressure to the total pressure is the mole fraction; that means,
depending upon the mole fraction its partial pressure will be determined.
Ratio of the partial pressure to the total pressure is the ratio it depends upon the
number of moles, what pressure will exist. It will depend upon its number of moles as
compared to the total number of moles. This is one equation. Therefore, the sum of the
partial pressure is the total pressure for extensive properties. How do you write other
properties? For example, what is u? Before that, how do you define the characteristic
gas constant? Again refer the same figure. The number of components I am writing the
mA, mB like this mC mn, and p and T. If we write it down, the partial pressure point
of view pAV is equal to mA RAT, because all are ideal gases. But the equation of stating
terms of characteristic gas constant is varying. So, I am giving A sorry capital A sorry.
Similarly, pB V is equal to that means I am writing in terms of the mass; earlier, I wrote
nA R bar T, now I am writing mB RBT like this. If we add it now, we will get p into V sorry
I am not writing again because pA, pB, pC, is total pressure you know that is is equal
to sigma m R in to T. What is pV? pV is sigma m R of the mixture;
that means, I define the R of the mixture by this equation pV is equal to mixture mass
R mixture T, because if that mixture is an ideal gas which is at a pressure p and temperature
T then what is the definition of the R mixture? Then we have to use this equation which will
define the R mixture. This is the equation which defines the R mixture.
If we write this then we get m R into T that means R mixture is the weighted mean of mi.
For example, we write mi Ri, i is equal to 1. So, I am not writing in that fashion. So,
you know this is very simple that means it is a weighted average of m. This way, all
other extensive properties can be defined. That means internal energy of an ideal gas,
enthalpy of an ideal gas, these are all mass basis. That means if total mass is sigma m then specific
internal energies are defined; the total internal energy will be mA uA plus mB uB dot dot dot
is equal to total mass sigma mu is specific internal energy. That means per unit mass
that is cv into T. Therefore, u is sigma mu divided by sigma m. Similarly, uh is equal
to sigma mh by sigma m. So, all extensive properties can be written as weighted average
very simple mass basis. That means total energy is the contributed energy, but specific internal
energy is cv into T. Oh sorry very good am in a little don’t
know in a hurry all right so today have to finish it the number of classes.
I will solve certain examples first. I think this one will be very much appreciated I do
first this now The change of entropy in a mixture of ideal gases that will not describe
in terms of this nomenclature. These are very simple; just a school level thing the recapitulation,
but will solve this problem and then you will see how interesting it is. Probably, you know
these things already. How to calculate the change in entropy when
a mixing of different gases take place which is a very important thing, which makes the
definition that how do you define the entropy of a mixture of ideal gas? Let us consider
this example. A closed rigid cylinder is divided by a diaphragm into two equal compartments
each of volume 0.1m cube. Each compartment contains air no I am sorry will not go this
this is a simple example but later what I told let us first sorry I am sorry. A closed vessel is divided into three compartments
A, B and C. The compartments A contains 32 kg of oxygen, the compartment B contains 28
kg of nitrogen and the compartment C contains 44 kg of CO2. We can understand from masses
that there are one mole carbon dioxide. All the compartments are at a pressure of 0.1M
Pa same pressure the atmospheric pressure and at a temperature of 20 degree Celsius.
When the partitions are removed, the gases mix. Determine the change of entropy of the
universe. Assume the vessel is insulated. Problem is like this; there are three compartments:
one contains 32 kg of oxygen, another contains 28 kg of Nitrogen and the other contains 44kg
of CO2. Pressure for all these compartments p is same; that is, p is equal to 0.1MPa and
temperature is also same that is t is equal to 20 degree Celsius.
If I rupture this diaphragm, all these gases are in mechanical and thermal equilibrium.
What is the definition of thermodynamic equilibrium? Three equilibriums have to be maintained;
mechanic, thermal and chemical. So, what process is going on? There is no work transfer process,
there is no heat transfer process, but are they in chemical equilibrium. Any two? No,
because there is the difference in chemical potential which we can visualize simply by
the concentration gradient. In this compartment nitrogen concentration is 0, whereas oxygen
concentration is 100; reverse is the case here. Therefore, there is a concentration
gradient. Thermodynamic potentials are different. Thermodynamics of multi-component systems
are not being taught. That is why we cannot understand what is thermodynamic potential,
but at least, you can conceive. When they are ruptured, the gases will mix. There will
be intermolecular diffusion which is a process defined by the gradient of thermodynamic potential
or concentration gradient. That means if we have two gases at different chambers, if we
remove and if there is a concentration gradient that flows, just as I explained in the last
class that a water vapor flows, because water vapor from a surface it flows into the surrounding.
Because surrounding water vapor is less, there is a less concentration, whereas water vapor
concentration is more here, and that is why it flows. There is a molecular diffusion.
So, they diffuse and they mix. So, there is a process which may not incur work transfer
or heat transfer, but there is a process, intermolecular diffusions because of this
chemical non-equilibrium to be existing there. My problem is that how to find out the delta
s universe; that means, we will have to find out in that case delta s O2, delta s N2, and
delta s CO2 and all of them contain one mole. Oh kilo mole very good. One not kilo mole
last class told that you do not call it kilo mole many book tell kilo mole one kg mole.
When they mixed total mixture is 3 kg mole. Now you tell me first that when there is an
entropy change of the oxygen, what property of the oxygen is changing, for which entropy
change is there, there is no change in temperature. Volume of the oxygen is changing from one
volume, but I do not know the volume divisions. Any other parameter you think here that is
correct; that is, the volume is changing, oxygen is expanding from this volume to the
total volume, but can you take any other parameter? Pressure, because we have got equations of
entropy change relating to both volume and pressure, because we have Tds is equal to
cpdT minus vdp or Tds is cvdt plus pdv. If we use this equation, we write the molar volume
or the total whatever way we write. Then if we use this equation then you get delta s
is cpln T2 divided by T1, T1 is the initial state minus R bar into ln p2 divided by p1,
where 2 and 1 are the final and initial state. If we use this, we will get cv here plus R
bar ln v2 by v1. But if I use this one then for this case, for all gases this part is
0, but all gases starting from an initial pressure of point 1MPa is attending a final
pressure which is the respective partial pressure in the mixture.
When the equilibrium mixing state is there, after mixing equilibrium state all gases have
got their individual partial pressures and that partial pressure is nothing but 0.1MPa
into 1 by 3. Therefore, this term is 1 by 3. Therefore, delta s of O2 is minus R bar
into ln 1 by 3 Minus very good because this is minus so this
will be plus so in both the cases there is a increase of entropy. Therefore delta s of
N2 is minus R bar into ln 1 by 3 and delta s of CO2 is minus R bar into ln 1 by 3, because
there is one mole for each. So, R bar into 1 R bar into 1. If the values are different,
we have to calculate dividing by the molecular weight what are the number of moles number
of moles, then sum up these and then find out the mole fractions.
For each gas, the ratio of the final to initial pressures may change. To make the calculation
simple, I have made this data so that it is easier to calculate. Concept is the same.
Here all one mole. So, everywhere it is 1 by 3. So, if we do it, delta S is minus 3
R bar into ln 1 by 3. It is the one mole one mole one mole. The
pressure temperature is same, volume will be same, but we do not have to consider the
volume. We can find out the volume, of course that is another thing that you can tell that
when the moles are same, pressure temperature same, volume is same, that is good thing and
then you can use that three; that is also correct. So then another problem, time is up but this
problem you solve this is a given as the your task homework I will give you the answer also
this is a homework its not there in the tutorial sheet you take it. A closed rigid cylinder is divided by a diaphragm
into two equal compartments each of volume point 0.1m cube; that means the volume is
same. Each compartment contains air. This is not a mixture different components same
component air at a temperature of 20 degree Celsius. The pressure in one compartment is
2.5 MPa and in other compartment is 1MPa; that means there is a difference of pressure.
They are in mechanical disequilibrium, but the partition is strong enough to sustain
that pressure difference so that no work transfer is taking place. The diaphragm is now ruptured.
So, there will be a work transfer in between so that the air in both the compartment mixes
to bring the pressure to a uniform value throughout the cylinder.
When they mix final state will represent to unique value of the pressure. Equilibrium
state means what? All properties will be uniform and constant throughout the cylinder which
is insulated; that means entire cylinder is insulated. There is a partition. Earlier problem
also, the vessel was insulated. There are two partitions and three compartments similarly
insulated. Find the net change of entropy for the mixing process. We have to find, here
before calculating this, we have to calculate what the final pressure is.
We use the equation of state for finding out the pressures and accordingly by the same
procedure, you can find out delta s A delta s B. Finally, I tell you the delta Suniverse
is equal to 0113 kilo Joule per kilogram kelvin. Delta s A you take it for your homework it
will come as 0.304 kilo Joule per kilogram kelvin. Here, the situation is little different;
delta s B will be negative that is minus 0.191 kilo Joule per kilogram kelvin so that the
algebraic sum of this two gives this one. Then thank you for today. So attendance will
take so next class will start the cycles the vapor power cycles and the
