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Kinetic theory of Gases

by gemma786
Tags: gases, kinetic, theory
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gemma786
#1
Jul9-11, 03:35 AM
P: 29
Hi
I would like to ask is there any relation between compressibility factor of gases and temperature? My text book says that it always increases with increase in temperature but doesn't explains how?
I know that on increases the temperature it is harder for molecules to show attractive interaction. By taking this into consideration compressibility factor should never increase beyond unity on raising the temperature beyond Boyle's point, isn't it?
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Philip Wood
#2
Jul9-11, 09:27 AM
PF Gold
P: 936
Have you read the wiki article on compressibility factor? It has graphs and a table, and gives some simple explanations.
gemma786
#3
Jul9-11, 07:51 PM
P: 29
Thanks Philip Wood for replying.
My problem was not addressed there. How it could be possible that on increasing the temperature above Boyle's point compressibility factor increases beyond one with out increasing the pressure to any substantial amount? The factors which deviates a gases from from behaving as ideal one is intermolecular attractive force between there molecules and excluded volume occupied by molecules and gases should behave as an ideal one when these factors are negligible. On increasing temperature above Boyle's point i.e. the temperature at which they show ideality intermolecular attractive forces are negligible indeed.Then could CF go beyond one? Does excluded volume increases somehow?

klimatos
#4
Jul9-11, 07:59 PM
P: 409
Kinetic theory of Gases

Quote Quote by gemma786 View Post
Hi
I would like to ask is there any relation between compressibility factor of gases and temperature? My text book says that it always increases with increase in temperature but doesn't explains how?
I know that on increases the temperature it is harder for molecules to show attractive interaction. By taking this into consideration compressibility factor should never increase beyond unity on raising the temperature beyond Boyle's point, isn't it?
Attractive forces have nothing to do with it. The pressure of an ideal gas is the simple product of the number of impacts per square meter of surface and the mean impulse transferred per impact. The temperature of an ideal gas is a simple measure of its kinetic energy of translation normal to the sensing surface.

When you heat a gas in a closed container the mean molecular velocity increases, this increases both the frequency of impact and the mean impulse per impact. These increases are measured as an increase in pressure; and requires an increase in the force needed to reduce the volume .
Trevormbarker
#5
Jul9-11, 08:57 PM
P: 67
temperature is preportional to pressue. When you increase the temperature you are increasing the average kinetic energy or "speed" of each particle, each particle therefore hits the inside of its container more frequently, or the gas as a whole "collides" with the inside of the container more frequently which explains the increase in pressure when temperture goes up.
gemma786
#6
Jul9-11, 09:37 PM
P: 29
Thanks for the reply.
I think you guys have not understand what I am trying to say. Boyle's point and compressibility factor is about how well a real gas obey the stuffs that you are taking about. I want to know why gases do not follow ideal gas equation or CF is greater than one when temperature is higher than Boyle's temperature inspite of the fact that pressure is not increased to any substantial amount? I hope you understand.
Thanks
Philip Wood
#7
Jul10-11, 03:34 AM
PF Gold
P: 936
Sorry. I know how frustrating it is when replies fail to address the issue. I think I now understand the problem. The solution, in a nutshell, is that when well above the Boyle temperature, the initial (i.e. as P approaches zero) slope of the graph of PV against P doesn't go on increasing with increasing T, but approaches a constant value. On the van der Waals model this high temperature limit of the initial slope is simply b (proportional to the volume occupied by the molecules themselves). In fact, for a mole of gas, we have PV = RT + bP. So the compressibility factor is PV/RT = 1 + bP/RT so ([itex]\partial[/itex]/[itex]\partial[/itex]P)(PV/RT) = b/RT, which does go to zero for very large T, just as you argued it should.
gemma786
#8
Jul10-11, 04:33 AM
P: 29
Thanks Philip Wood.
The equation of state which you have mentioned could be rearranged as P(V-b)=RT, it means you are excluding volume b from original volume occupied by one mole of real gas at temperature T, but this process of excluding would be significant only when volume occupied by gases' molecule is not negligible and this happens at high pressure situation. But on keeping the pressure constant say one bar and increasing temperature above Boyle's point how could CF go beyond one? That's my question.
Have a look on the experimental data.http://en.wikipedia.org/wiki/Compressibility_factor
Philip Wood
#9
Jul10-11, 04:50 AM
PF Gold
P: 936
My high temperature formula for compressibility factor Z at low pressures does show the factor as being greater than 1, does it not? That's because the excluded volume isn't totally negligible. But I also argue that for very high temperatures the theory predicts that Z approaches 1. The Wiki data go up to 1000K, which isn't high enough for Z at 1 bar to have returned to 1. Or at least that's my story - it looks as if we don't have the high temperature needed to confirm or deny this experimentally.
gemma786
#10
Jul10-11, 08:06 AM
P: 29
Yeah, you were right Philip Wood in saying that excluded volume is not negligible under 1 bar pressure and 1000K temperature infact my calculation showed me that it could make a impact on the second decimal place of volume (measured in liters) which is sufficient in describing the CF mention on Wiki, only when we take the volume excluded by gas molecules proportional to fourfold of volume actually occupied by gas molecules.
But I am not sure about the fourfold factor!
Have any suggestion?
Philip Wood
#11
Jul11-11, 05:04 AM
PF Gold
P: 936
The effective volume of the container is the space in which the molecules can move, when we represent each by a central point. But the centres of molecules, considered as hard spheres of radius r, cannot come closer to each other than 2r. So a molecule excludes other molecules from a volume of (4/3)[itex]\pi[/itex](2r)3 = 8 x volume of molecule. But this excluded space belongs just as much to both molecules involved in a collision (assuming we only have two-molecule collisions). So the excluded volume per molecule is only 4 x volume of molecule.

To be honest, my understanding of this argument is fragile, in the sense that I'd probably not be able to answer objections to it very convincingly. The topic of excluded volume in a V der W's gas used to be a topic of keen interest and debate, and many clever physicists and applied mathematicians developed different approaches - and came up with somewhat different values. D. Tabor discusses some of these in Gases, liquids and solids, third edition, 1991.
IEVaibhov
#12
Jul11-11, 06:22 AM
P: 15
Keeping volume at about 1 bar, CF can not be greater than 1.
The real gas equation can be written in the given condition as (P + [a/(V^2)])*V = RT.
This gives us CF = 1 - (a/(RTV)). If you keep on increasing temperature, CF will tend to 1, but it will never become exactly 1.
Philip Wood
#13
Jul11-11, 06:36 AM
PF Gold
P: 936
IEValbhov. Your real gas equation contains the intermolecular force term (the one involving a) but takes no account of excluded volume. The original question was about what happened above the Boyle temperature. Above the Boyle temperature, the excluded volume term becomes more important than the intermolecular force term.
IEVaibhov
#14
Jul11-11, 06:44 AM
P: 15
Maybe, I haven't understood her doubt correctly.


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