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In the last class, we started looking at equations of state. We started out with the ideal gas
equation P V equals R T, which we realize present equation of state. And as we said
an equation of state is something that connects the pressure, specific volume, temperature
of a pure substance. Then we said that most gases are real, and therefore, we need more
accurate representations of those gases, then what is given by the ideal gas law that is
P V equals R T. And the first improvement that we saw was the virial equation of state;
which we said could be written either in terms of series in pressure or a series in specific
volume. Then we had post this problem to become comfortable with the use of virial equation,
and I had given you some hints and initial part of the solution.
So, let me present the complete solution here. This was the problem regarding isopropanol
and we said that, we are looking at an isopropanol at 200 degree C and 10 bars. And we need to
estimate the volume of the vessel necessary to maintain the above conditions for one mole
of isopropanol. Therefore, you have a pressure, temperature here, you have pressure here.
You are asked to find this specific volume of the molar volume, volume for one mole.
And the virial coefficients in terms of the volume expansion were given C 2 and C 3 were
given, and then after a couple of hints.
We went ahead with the solution ahead, mentioned that we could start out with just three terms,
in the virial expansion for specific volume, Z equals 1 plus C 2 by V plus C 3 by V squared,
this should work in most cases except when the pressures are high and so on. For our
purposes we said, we could write this as this Z equals P V by R T, and therefore, multiplying
by R T by p on both sides, we have V in terms of V an implicit equation which would be good
for iterations and this requires an iterative solution.
And I had given you the procedure for iteration and asked you to work things out. What I will
do now is to present a few more steps, and again ask you to work things out, because
you need to get a little more comfortable with an iteration process. I am not very sure,
how many of you have solved problems that had iterative solutions.
So, we said that the procedure was to guess a value for V, and substitute it into the
right hand side and find the value of the right hand side. And this can be compared
with a guessed value. One way of comparing is to take the difference between the calculated
value and the guessed value. And if the calculated value on the right hand side is close enough
to the guessed value, then the guessed value is a needed a value. We said close enough,
we need to be little careful. Otherwise, the calculated value is used as
the guessed value for the next iteration. And then you use that volume here, substitute
that volume here to find out the right hand side, and then compare again. And the process
is continued until the difference between the calculated, and the assumed value becomes
acceptably small. This is where we left off last time; I hope you would have worked things
out. If you had some difficulty, please pay attention to what follows.
For the first guess, we need to start with the first guess, and where do we get that
first guess from. If it is a gas, this equation is valid for a gas; we could use the ideal
gas value. Whatever, we get from P V equals R T as the first guess we would not be way
off. Therefore, V equals R T by p from the ideal gas equation. And if we substitute the
values in this particular case, in this particular exercise, R as we all know as 8.314 joules
per mole per Kelvin, T was 200 degree C, but we know that, we need to use T and Kelvin.
So, add a 273 to it to get 473 Kelvin and pressure was 10 atmospheres. We need it in
a consistent set of units therefore, we multiplied by 10 of 5 Newton per meter squared. And if
we substitute R T by P here, the numbers for that we get the ideal gas, let us say V 1
in terms of our iterative solution as 3.933 in to 10 power minus 3, you know 8.314 in
to 473 divided by this is 10 power 6. So, 3.933 in to 10 power minus 3 meter cubed
per mole. Hopefully, you got this if you knew where to look for. Substituting this value
in to the right hand side of the earlier equation, you know we had set of the iterative volume
equation. There, you know 3.933 into 10 power minus 3 V plus into 1 plus C 2 by V plus C
3 by V square C 2 was minus 3.9 into 10 power minus 4 V is 3.933 into 10 power minus 3 C
3 was minus 2.6 into 10 power minus 8 and V squared is this.
So, if we substitute the values here we end up with 3.573 into 10 power minus 3 meter
cubed per mole. This is V 2 which is the calculated value starting with V 1 the ideal gas value.
On the face of it these seems small, but, let us consider this further. We need to take
the difference between the guessed value, which was the ideal gas value and the calculated
value, which is this 3.537 to 10 power minus 3. In other words the difference between V
2 and V 1 in this case which turns out to be 0.396 in to 10 power minus 3. You know
this is 3.96 into 10 power minus 4, which is almost 11 to 12 percent of this value.
That is not acceptable, that is very high. 11 to 12 percent is very high, is unacceptably
large and therefore, we need to continue the iteration. In other words, we substitute V
2 into the R H S. Calculate the value and calculate the value of the R H S that is,
and then calculate the difference between the guessed value and the calculated value.
Till the difference is acceptably small.
For the iterations, it is easier to express the equation in the form of n and n plus 1.
In other words, you know we said V 2 was we get in terms of V 1 and so on. So, we can
generalize that and say V n plus 1 R T by P into 1 plus C 2 by V n plus C 3 by V n square.
If we do this it gives us a nice way of looking at the iterate iterations. As well as to write
a program to let a computing device do the iterations for us, we do not have to do it
manually. And the iterations can be terminated, when V n plus 1 minus V n the difference between
V n plus 1 and the guessed value V n is less than an acceptably small value, may be about
1 percent or 2 percent of the actual value.
What I would like you to do is, this is this will converge in about 5 iterations. So, I
would like you take another 10 to 15 minutes. And actually work out the iterations, it is
going to take some time, because you are going to substitute numbers and so on. But, it is
good to get comfortable with the procedure. If you have already done it as a part of the
home work given yesterday. Go forward of by about 10 to 15 minutes to the next part of
the solution and the lecture, but, otherwise please take the time now. Take about 15 minutes
and do the iterations. Please go ahead.
You would have had enough time, to do the calculations. See whether you got this value
at the end of about 5 iterations, 3.486 in to 10 power minus 3 meter cubed for mole.
This is when the right hand side value is different from the left hand side value or
the guessed value only by a small amount and acceptably small amount. And thus we have
reached the volume of the vessel, that was needed just actually what we started out looking
for in the exercise. Also there was another part to the exercise. That was to find out
the variation between the value given by the virial equation, the volume given by the virial
equation and the volume given by the ideal gas, in other words, to find out the percentage
difference. So, to have found out the percentage difference,
take the difference divided by the actual volume and actually we need to multiply it
by 100. We will do that in a little while, for now V minus V ideal gas the whole divided
by the volume will turn out to be minus 0.128. Therefore, the difference between the volumes
calculated using the virial equation and the ideal gas equation is large. It is about 13
percent 12.8 percent. This you should keep in mind.
The next equation of state that we are going to consider in this course is called the cubic
equation of state. There are many cubic equations; we will probably see 1 or 2. The equations
of state that we have considered thus far, which we reviewed in the beginning of this
class which was the ideal gas equation and the virial equation. These equations described
only the gas phase, which is reasonable general for a pure substance, but, they look only
at the gas phase. Whereas, cubic equations of state, have an
ability to describe either the gas phase or the liquid phase with equal ease or they can
describe I would not get in to the ease of description. Then can describe the both the
gas phase and the liquid phase. In other words, the combination of P V T, if know P and T,
how do you get V? Or if you know P and V, how do you get T? It is good in predictions
for both the gas and the liquid phase and it is close to experimental values, somewhat
close. And what might be surprising, it is that you
have already seen one of the cubic equations in your earlier classes. I am sure you would
have seen the Vander Waals equation. Which is actually given by Vander Waals, the full
name is Johannes Diderick Vander Waals. He gave it way back in 1873 and even if you do
not remember the name you may remember the equation, this is P equals R T by V minus
b minus a by V square. Let us call this equation 3.6. You know this
cubic equation has the ability to describe the liquid phase because, the interactions
between molecules are considered in the cubic equations. A little better, and a and b as
you already know in the Vander Waals equation are constants for a given pure substance.
Therefore, you would have a table that lists a and b for various different pure substances,
there is one that lists at the end your text book itself.
The constants a and b can be calculated from the critical pressure and the critical temperature
values. I hope we can recall what P c and T c R remember that those are the pressure
and temperature values that correspond to the critical point, you know there was it
was a point in the P T diagram; and it was a point on the curve, the top part of the
P V curve in a P V diagram. And that is pretty much the place, where the critical phase starts.
And P c and T c values are available in tables, they are available in one of the appendices
in your text book also. This Smith Vanness and Abbot text book, there are values of P
c and T c, the critical pressure and the critical temperature, that are given for various pure
substances. Now, let me present this first. This a in
the Vander Waals equation can be written as 27 R square; this is a gas constant - R squared,
T c squared critical temperature squared divided by 64 P c; we will call this equation 3.7.
And b equals R T c by 8 P c - equation 3.8. Therefore the constants a and b can be calculated
from the critical pressure and the critical temperature values.
Therefore, I make some more comments on that let me present another popular cubic equation
of state, which is called the Redlich-Kwong equation of state. The equation itself is
P equals R T by V minus b, somewhat similar to the earlier cubic equation minus a divided
by T power 0.5 into V into V plus b. We will call this equation 3.9. a in this case, excuse
me is 0.42748 R squared T c power 2.5 by P c, we will call this equation 3.10. b equals
0.08664 R T c by P c, we will call this equation 3.11. Now, you might wondering, how did I
really write yeah did say that you can write in terms of T c and P c. How did people get
this? You know so, assumingly accurate number 0.42748 and so on and so forth.
And that is what you are going to do as a next exercise. The basis for the above equations,
that is the bases for the equations that gave you a and b for both the Vander Waals equation
and as well as the Redlich-Kwong equation. That definitely needs to be known as a part
of the course. Let me give you, how to go about doing it and give you some time to do
it. And then let us let us see what to do. In the, if you recall the P V diagram for
a pure substance pressure versus specific volume for a pure substance, at the critical
point the slope is actually 0. What is slope it is the derivative of P with respect to
V D P D V that is actually 0. Therefore, you could put D P D V equal zero and also it an
inflection point these slope changes from one direction to the other exactly at that
point. And that gives you another nice mathematical condition, which is D square P by D V square
equals 0. So, you have this D P D V equals 0 and D square P by D V squared equal 0 and
you have the equation of state which they describe P V T behavior. So, please go ahead
and try to get the expressions for a and b using these conditions. Take about fifteen
minutes to do it which should be till the end of this class. Please try that out. Go
ahead please.
You would have been able to work those things out, get expressions for a and b. If you did
not or even, otherwise please take a look at chapter three in your text book Smith Vanness
Abbot. There it would have been worked out in some detail; you would need to fill in
the missing steps. We will continue the remaining in the next class.