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Ghost Writer
14th June 2003, 13:26
Kinetic theory of gasses-

Bernoulli's theory is what is considered the Kinetic theory of gasses. It states that in any given confined space molecules that are moving at different speeds and direction will crash into the walls of the confined space creating pressure. The more gas there is in the space the more pressure there is. -Comrade RAF, Ph.D.

Comrade RAF is correct in some respects, and has given a simple definition of the kinetic theory of gases, but in no way has he covered all avenues. In fact, he left out the most important aspect of this fundamental principle. Although RAF's definition is an important one, relating the kinetic energy to pressure, more important is the relation between kinetic energy the temperature of the system. To be fair to the argument, I will mention that kinetic theory is the analysis of matter, at an atomic level, where the momentum of the system occurs randomly.

In addition, RAF suggests that this is Bernoulli's theory. This is only partially correct, as Bernoulli was probably the first to begin understanding air pressure from an atomic perspective. He created the famous billiard ball analogy used so often to break this subject down for incoming freshmen needing General Physics to fill a requirement. However, James C. Maxwell advanced the field with his distribution of speeds. Even still, our understanding of kinetics has greatly improved with the advent of quantum mechanics, thanks to work done by Schrodinger and Heisenburg.

To begin to understand the kinetic theory of gases we must make a few simple assumptions. These are the postulates used for this particular theory.

1.) There are numerous molecules moving randomly, with varying speeds

2.) The molecules are far apart (the average distance is greater than the diameter of the molecule)

3.) All molecules behave classically (meaning they only interact when they collide, and intermolecular forces can be ignored)

4.) Collisions between the molecules or the wall of the vessel are assumed to be completely elastic

These assumptions help describe Boyle's Law, which noted that volume was inversely proportional to pressure.

Now if we use a box model with sides of area (A) and length (l), we can quantify the pressure of the gas based on kinetic theory, like RAF suggests. For now we will only look at on side of the box. Using Newton's second law we can calculate the force:

eq. 1
F = dp/dt (force equals change in momentum)

or simply put

eq. 2
F = del(mv)/del(t) (force equals the change in mass times the change in velocity all over the change in time)

Assuming elastic behavior, a collision will only create a change in the x direction, giving us an initial state of mv(x), and a final state of -mv(x).

Thus, the change is equal to:

eq. 3
del(mv) = mv(x)-(-mv(x)) = 2mv(x)

Velocity is equal to the distance over the change in time:

eq. 4
v(x) = 2l/dt or dt = 2l/v(x)

Getting back to the second law, the force due to one molecule is given by:

eq. 5
F = del(mv)/del(t) = (mv(x)^2)/(l)

This one molecule may collide with all sides of the vessel, but the x component of the velocity will not change, because of the elastic behavior we are attributing to the system. Furthermore, the molecule may collide with other molecules, changing its v(x), but the acquired or lost momentum of the other molecule will make up for this loss or gain. It is easy to see how the very important statistical principle of averaging will allow us to calculate the overall average force on the vessel.

eq. 6
F = (m/l)*(v(x1)^2+v(x2)^2+...+v(xn)^2)

eq. 7
avg(v^2(x)) = (v(x1)^2+v(x2)^2+...+v(xn)^2)/N

where N is the number of molecules

eq. 8
F = (m/l)*N*avg(v^2(x))

Using pythagoras theorem

eq. 9
avg(v^2) = avg(v^2(x))+avg(v^2(y))+avg(v^2(z))

Since the velocities are random we can say that there is no preference between one component of the vector or the other and thus:

eq. 10
avg(v^2(x)) = avg(v^2(y)) = avg(v^2(z))

From both eq. 9 and eq. 10 we get the relation:

eq. 11
avg(v^2) = 3(v^2(x))

Substitution into eq. 8 we get:

eq. 12
F=(m/l)*N*(avg(v^2)/3)

From the simple equation for pressure we know that the pressure is equal to the force over the area.

eq. 13
P = F/A = (1/3)*(((N*m*avg(v^2))/(A*l))

or

eq. 14
P = (1/3)*(((N*m*avg(v^2))/(V))

where V is the volume of the vessel

With a little rearrangement we get the ideal gas law:

eq. 15
PV = NkT

where k is the Bolztman constant (1.38x10^-23(J/K))

Comparing the two gives us a marvelous revelation about kinetic theory.

eq. 16
(PV/N) = (2/3)(.5*m*avg(v^2))

From basic physics we know that .5*m*v^2 is kinetic energy. Hence, we have relation between kinetic energy and temperature.

eq. 17
Ek = .5*m*avg(v^2) = (2/3)*k*T

Where T is the absolute temperature.

Other useful equations:

eq. 18
v(rms) = sqrt(avg(v^2)) = sqrt((3*k*T)/m)

Questions for Comrade RAF, Ph.D.:

1.) What is the average translational kinetic energy of a system, like the one described above, if the temperature of the gas is 375 deg. Celsius?

2.) At the beginning of the derivation, I described four assumptions, in two words what was I describing?

3.) 10 particles have the following speeds 122, 10, 25, 32, 54, 44, 32, 44, 88, and 47, in meters per second. Calculate the mean speed and the root mean square speed.

4.) What is the rms speed of air molecules (O2 and N2), at room temperature?

mass of a proton = 1.66x10^-27
mass of a nuetron = 1.67x10^-27

[/b]Hint: Consult your Periodic Table of the Elements.[/b]

5.) What implication does eq. 17 have for the thermodynamic property known as entropy?

(Edited by Ghost Writer at 1:34 pm on June 14, 2003)

Ghost Writer
14th June 2003, 13:48
Here is a good applet that demonstrates some important concepts not really discussed above. As you can see it takes Maxwell's distribution of speeds into consideration, average collision time, and irreversibility of thermodynamic processes.

Nice Animation (http://comp.uark.edu/~jgeabana/mol_dyn/KinThI.html)

Vinny Rafarino
14th June 2003, 15:53
As I already told you I am neither a mathmetition or a physicist. You asked for definitions. I gave you definitions. Correct definitions I might add. If you want to discuss theoretical Physics...By all means, lets discuss it as I offered. If you want to play around with eqations, have a bang up time little boy. If I found them even remotely interesting, I would have went to school for it.
When you are ready to discuss the topics I offerd to discuss with you let me know otherwise you're going to be posting equations all by your little lonesome boy, 'cos i can give a rat's arse about it.

Can you place the link to the website you copied and pasted this from?

(Edited by COMRADE RAF at 4:29 pm on June 14, 2003)

Anonymous
14th June 2003, 18:03
Quote: from COMRADE RAF on 3:53 pm on June 14, 2003
As I already told you I am neither a mathmetition or a physicist. You asked for definitions. I gave you definitions. Correct definitions I might add. If you want to discuss theoretical Physics...By all means, lets discuss it as I offered. If you want to play around with eqations, have a bang up time little boy. If I found them even remotely interesting, I would have went to school for it.
When you are ready to discuss the topics I offerd to discuss with you let me know otherwise you're going to be posting equations all by your little lonesome boy, 'cos i can give a rat's arse about it.

Can you place the link to the website you copied and pasted this from?

(Edited by COMRADE RAF at 4:29 pm on June 14, 2003)


Due you crack me up. Just because you memorize the words, does not mean you understand the physical principle.

Anonymous
14th June 2003, 18:34
Quote: from Ghost Writer on 1:48 pm on June 14, 2003
Here is a good applet that demonstrates some important concepts not really discussed above. As you can see it takes Maxwell's distribution of speeds into consideration, average collision time, and irreversibility of thermodynamic processes.

Nice Animation (http://comp.uark.edu/~jgeabana/mol_dyn/KinThI.html)


You kids these days. We'll in my day....

I have no desk reference to a PN junction. Check out this primer I found:

http://britneyspears.ac/physics/pn/pnjunct.htm
http://britneyspears.ac/lasers.htm

Dhul Fiqar
14th June 2003, 18:54
My cat's breath smells like (5alpha,6alpha)-7,8-Didehydro-4,5-epoxy-17-methylmorphinan-3,6-diol diacetate (ester).

--- G.

Dhul Fiqar
14th June 2003, 18:56
BTW, I love threads with sad people dicksizing through google ;)

canikickit
14th June 2003, 18:57
What is your sig supposed to be Kelvin?

Norm. Wow, big deal, you did Science in college, nobody cares though.

Ghost Writer
15th June 2003, 08:18
I just wanted you to quit calling me stupid, RAF. You brought your educational background into this discussion, while you tried to insult my intelligence. Now, either disagree with me respectfully, or shut the fuck up.

Vinny Rafarino
15th June 2003, 08:21
Quote: from Ghost Writer on 8:18 am on June 15, 2003
I just wanted you to quit calling me stupid, RAF. You brought your educational background into this discussion, while you tried to insult my intelligence. Now, either disagree with me respectfully, or shut the fuck up.

You almost had me boy...I felt a twinge of guilt for a moment.

Go fuck yourself little one.

Ghost Writer
15th June 2003, 11:30
I teach you guys the fundamentals of the kinetic theory of gases, and I don't even get a thank you. Instead, I am treated with further disdain. What the fuck is wrong with you people.

By the way, does anyone know the answer to #5? It is an important one.

Dhul Fiqar
15th June 2003, 13:08
Well, then let me thank you for it. Not that I am interested in kinetic gases in any way, but it was a well structured and relatively non-confrontational post.

Just because your politics disgust me doesn't mean I am going to ***** at you about a totally unrelated topic, such as kinetic gases. And I'll give you this, you've no doubt forgotten more about kinetic gases than I will ever know.

But thenl, it's more of a question of spheres of interest.

--- G.

suffianr
15th June 2003, 13:10
I teach you guys the fundamentals of the kinetic theory of gases, and I don't even get a thank you. Instead, I am treated with further disdain. What the fuck is wrong with you people.

Thank you, GW. Your impressive expose on the kinetic theories has left many of us Philistines absolutely breathless...

And to think that I thought this was another on of those random Personal Ego-Booster-type threads...tsk, tsk, tsk... ;)

Dhul Fiqar
15th June 2003, 13:51
lol, touché. Geography homework anyone? :biggrin:

Ghost Writer
15th June 2003, 14:08
Well, then let me thank you for it. Not that I am interested in kinetic gases in any way, but it was a well structured and relatively non-confrontational post.

Just because your politics disgust me doesn't mean I am going to ***** at you about a totally unrelated topic, such as kinetic gases. And I'll give you this, you've no doubt forgotten more about kinetic gases than I will ever know.

But thenl, it's more of a question of spheres of interest.

I didn't ask anyone to lick my balls. Something other than fuck you would have been nice.

Ghost Writer
15th June 2003, 14:40
In case anyone is actually interested.

5.) If the kinetic energy of the system is given by:

Ek = .5*m*avg(v^2) = (2/3)*k*T

then as the temperature approaches absolute zero the energy of the system also approaches zero. That is why some people make the mistake of stating that all motion of matter ceases at absolute zero. This is not quite true, work in the field of quantum mechanics suggests that the kinetic energy actually approaches a very small non-zero number. This idea is closely related to the 3rd law of thermodynamics, which states that for perfect crystalline substances the absolute entropy of the system is zero at absolute zero. This allows us to calculate absolute entropies, a convenience for those who work with this particular state function. Entropy is a function of both temperature and pressure.



(Edited by Ghost Writer at 3:08 pm on June 15, 2003)

Anonymous
15th June 2003, 21:48
Quote: from Ghost Writer on 2:40 pm on June 15, 2003
In case anyone is actually interested.

5.) If the kinetic energy of the system is given by:

Ek = .5*m*avg(v^2) = (2/3)*k*T

then as the temperature approaches absolute zero the energy of the system also approaches zero. That is why some people make the mistake of stating that all motion of matter ceases at absolute zero. This is not quite true, work in the field of quantum mechanics suggests that the kinetic energy actually approaches a very small non-zero number. This idea is closely related to the 3rd law of thermodynamics, which states that for perfect crystalline substances the absolute entropy of the system is zero at absolute zero. This allows us to calculate absolute entropies, a convenience for those who work with this particular state function. Entropy is a function of both temperature and pressure.



(Edited by Ghost Writer at 3:08 pm on June 15, 2003)


Sorry it has been several years since I looked at this, how can you get an absolute value from a probabalistic distribution? This is kinetic theory right? So then the velocity spread is a distribution? RAF your comments?

(Edited by kelvin9 at 9:51 pm on June 15, 2003)

Ghost Writer
16th June 2003, 08:38
I'm sorry. I was trying to be brief, and I may not have been clear. Since the property entropy is generally the measure of randomness or disorder of a system (not to be confused with chaos), a zero value for kinetic energy implies a reduction of this measure of disorder, through the slowing of vibrations and interactions within the system to nothing, then the entropy is considered to be zero. Of course, this is qualified by the condition that the substance must appear in a pure crystalline form, since amorphous, mixed substances are less ordered. This is taken to be the theoretical framework for the 3rd law of thermodynamics, and it appears to be mostly correct.

With respect to actually calculating the absolute values of entropy, we apply the third law and integrate a series of terms, which include the heat capacities divided by the temperature, for all necessary phases. The number of terms depends on the phase of the system, at the absolute temperature of interest. We use a process path beginning at the solid state, with the lower limit of integration taken to be zero. If we were to undergo a phase change, from solid to liquid, we would take the upper limit to be the normal melting point temperature, and then add the standard heat of fusion over the normal melting point temperature. If we undergo a further phase change to that of the gas, the same applies, only the limits of integration change to represent the defined process path. Finally, we must integrate from the normal boiling temperature up to the temperature of interest. This calculation is done using the following equation. SI units for this property are given in [Joules/Kelvin].

http://www.che-lives.com/cgi/community/upload/absolute_value_of_entropy_2.jpg

Where:

S is the absolute entropy
(Cp)s is the solid state heat capacity
(Cp)l is the liquid state heat capacity
(Cp)g is the gaseous state heat capacity
T is the temperature
T(f) is the normal melting point temperature
T(v) is the normal boiling point temperature
del(H(f)) is the standard heat of fusion
del(H(v)) is the standard heat of vaporization

You are right in thinking that we have diverted from kinetic theory. Scientists have used a revelation obvious from the relation between Temperature and Kinetic Energy, and have applied it in the realm of thermodynamics to formulate a postulate known as the third law. Which states that when T = 0 K, S = 0. Although there has been divergence away from the idea that all matter ceases at 0 K, it is still mostly correct, and it can be applied to find absolute entropies. Measurements of heat capacities at extremely low temperatures have been made, and the evidence supports this postulate.

In short, it is close enough for our purposes, although it would not 100% accurate, if the number were to be reported to an infinite number of significant figures. None of the numbers used in science are that accurate. So, we have stumbled onto one of the important realizations of chaos theory.

(Edited by Ghost Writer at 10:10 am on June 16, 2003)

Anonymous
17th June 2003, 03:37
Quote: from Ghost Writer on 8:38 am on June 16, 2003
I'm sorry. I was trying to be brief, and I may not have been clear. Since the property entropy is generally the measure of randomness or disorder of a system (not to be confused with chaos), a zero value for kinetic energy implies a reduction of this measure of disorder, through the slowing of vibrations and interactions within the system to nothing, then the entropy is considered to be zero. Of course, this is qualified by the condition that the substance must appear in a pure crystalline form, since amorphous, mixed substances are less ordered. This is taken to be the theoretical framework for the 3rd law of thermodynamics, and it appears to be mostly correct.

With respect to actually calculating the absolute values of entropy, we apply the third law and integrate a series of terms, which include the heat capacities divided by the temperature, for all necessary phases. The number of terms depends on the phase of the system, at the absolute temperature of interest. We use a process path beginning at the solid state, with the lower limit of integration taken to be zero. If we were to undergo a phase change, from solid to liquid, we would take the upper limit to be the normal melting point temperature, and then add the standard heat of fusion over the normal melting point temperature. If we undergo a further phase change to that of the gas, the same applies, only the limits of integration change to represent the defined process path. Finally, we must integrate from the normal boiling temperature up to the temperature of interest. This calculation is done using the following equation. SI units for this property are given in [Joules/Kelvin].

http://www.che-lives.com/cgi/community/upload/absolute_value_of_entropy_2.jpg

Where:

S is the absolute entropy
(Cp)s is the solid state heat capacity
(Cp)l is the liquid state heat capacity
(Cp)g is the gaseous state heat capacity
T is the temperature
T(f) is the normal melting point temperature
T(v) is the normal boiling point temperature
del(H(f)) is the standard heat of fusion
del(H(v)) is the standard heat of vaporization

You are right in thinking that we have diverted from kinetic theory. Scientists have used a revelation obvious from the relation between Temperature and Kinetic Energy, and have applied it in the realm of thermodynamics to formulate a postulate known as the third law. Which states that when T = 0 K, S = 0. Although there has been divergence away from the idea that all matter ceases at 0 K, it is still mostly correct, and it can be applied to find absolute entropies. Measurements of heat capacities at extremely low temperatures have been made, and the evidence supports this postulate.

In short, it is close enough for our purposes, although it would not 100% accurate, if the number were to be reported to an infinite number of significant figures. None of the numbers used in science are that accurate. So, we have stumbled onto one of the important realizations of chaos theory.

(Edited by Ghost Writer at 10:10 am on June 16, 2003)


I aint no "mathmetition", but a little off tread. Funny what you sometime can remember. I asked that same question regarding making an absolute calculation from a velocity distribution:

Ek = .5*m*avg(v^2) = (2/3)*k*T

I was being a smart*ss in class. The instructor snidley remarked "Um, like temp, the average velocity, is um like an average for the whole system not for one particle." It is quote obvious, but if your were there in one stroke he called me an as*hole, an idiot, and got a laugh an my expense. Something RAF would understand; the laugh at his expense, not the physics.

I not comfortable calling anyone stupid. I like to give people the benefit of the doubt. In his case I am fine calling him an idiot.

Anonymous
17th June 2003, 03:40
Quote: from canikickit on 6:57 pm on June 14, 2003
What is your sig supposed to be Kelvin?

Norm. Wow, big deal, you did Science in college, nobody cares though.


Steve made a comment about forgetting what history records about communism. I made several versions of his quote. In an effort to not miss quote him out of context, I just threw in the entire post.

Anonymous
17th June 2003, 03:43
Quote: from suffianr on 1:10 pm on June 15, 2003

I teach you guys the fundamentals of the kinetic theory of gases, and I don't even get a thank you. Instead, I am treated with further disdain. What the fuck is wrong with you people.

Thank you, GW. Your impressive expose on the kinetic theories has left many of us Philistines absolutely breathless...

And to think that I thought this was another on of those random Personal Ego-Booster-type threads...tsk, tsk, tsk... ;)





Hmmm, no. In an effort to stop calling him stupid we were discussioning education in another thread. RAF invited him to discuss theoritical physics elsewhere.