3 past exam questions =(

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hey i'm going through an exam paper and i'm having a difficult time working them out. any help will be greatly appreciated. thanks!

the first question is for the size of an atom..
The second virial coefficient, B, at high temperatures equals half the excluded volume due to the repulsive part of the intermolecular forces. Use B=11.9 cm3 mol-1 for Argon to estimate the range (in nm) of the repulsive part of the interatomic forces between argon atoms. (Hint: Imagine the Argon gas to be like a collection of billiard balls in 3D with the excluded volume being the volume occupied by the balls themselves.) :uhh:

i have gone through that question many times and havent been able to find an equation for it.. all i could think of was avragardo's no. NA/B.. what else can i do..?

my second question is to..
determine the ratio of the mean speeds of oxygen and nitrogen molecules in air at 28°C and at a pressure of 1 bar
do i just find the mean speed of each molecule and find the ratio? using the question (8RT/(pi*M)^0.5 = (3kT/m)^0.5 where M=m*NA=mass of one mole. would M be in g or Kg..? (i'm too use to chemistry, using grams) :confused:

and my final question i am having trouble with uses a big formulae i think..
it asks me to find the pressure (in Pa) at which the mean free path of argon at 23°C becomes comparable to the diameter of a 1L spherical vessel that contains it (cross section of argon = 0.36nm^2
i'm using the equation where
λ =
c/((N/V)*σ*crel = kT/(sqrt2)σp and rearanging it in terms of P to find the pressure :cry: so much hassle

sorry for writing so much but i find these questions really tough. am i going about these questions the wrong way.. or are these tougher than usual :bugeye:
pls help mee...=(
thanks!
 

Answers and Replies

  • #2
OlderDan
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koomanchoo said:
the first question is for the size of an atom..
The second virial coefficient, B, at high temperatures equals half the excluded volume due to the repulsive part of the intermolecular forces. Use B=11.9 cm3 mol-1 for Argon to estimate the range (in nm) of the repulsive part of the interatomic forces between argon atoms. (Hint: Imagine the Argon gas to be like a collection of billiard balls in 3D with the excluded volume being the volume occupied by the balls themselves.) :uhh:

I will pass along what comes to mind on this one. As I interpret the question, one mole of gas atoms (not the space between them) occupies a space of 2*11.9 cm^3. From that you can figure out the space occupied by a single atom. If you assume each atom is a sphere, you can calculate the radius of the sphere. I assume the range is twice the radius, since that is the separation of the centers of two colliding atoms.
 
  • #3
Gokul43201
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OlderDan said:
As I interpret the question, one mole of gas atoms (not the space between them) occupies a space of 2*11.9 cm^3.
Why the 2 ? Ar is a monoatomic gas. Or am I missing something here ? Also, you shouldn't have to double the radius to get the range.
 
  • #4
OlderDan
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koomanchoo said:
my second question is to..
determine the ratio of the mean speeds of oxygen and nitrogen molecules in air at 28°C and at a pressure of 1 bar
do i just find the mean speed of each molecule and find the ratio? using the question (8RT/(pi*M)^0.5 = (3kT/m)^0.5 where M=m*NA=mass of one mole. would M be in g or Kg..? (i'm too use to chemistry, using grams) :confused:

(8RT/(pi*M)^0.5 = (3kT/m)^0.5

This appears to me to be a confusion of two different kinds of averages. I think the R is supposed to be a k.

Edit: I see now why you have R instead of k. Your M is Avagadro's number times m. [tex] R/N_A = k[/tex]. This does not affect anything I said that follows

The left side of your equation would be the "normal" average velocity, and the right side would be the root-mean-square average velocity. You want to be using the same average for the two different kinds of molecules. You also need to look at where these equations come from. For a monatomic gas, kinetic theory says that

[tex] KE_{ave} = \frac{3}{2}kT[/tex]

so

[tex] \frac{1}{2}m(v^2)_{ave} = \frac{3}{2}kT[/tex]

[tex] (v^2)_{ave} = \frac{3kT}{m}[/tex]

The rms velocity average comes from taking the square root of this equation. I have seen the average on the left side of your equation described as the median velocity, where half the atoms have lower speed and half the atoms have greater speed, but I think it is more likely the average obtained by taking the square root of the distribution of the velocity squared and then taking the true average of that distribution. I don't think it matters because you don't have to use it in my opinion.

I think you probably want to use the rms average for both your gasses. However, your gases are not monatomic. Kinetic theory extended to diatomic molecules suggest that for rigid molecules

[tex] KE_{ave} = \frac{5}{2}kT[/tex]

and for vibrating diatomic molecules

[tex] KE_{ave} = \frac{7}{2}kT[/tex]

One of these last two is appropriate for your gasses, so you need to figure out the correct expression for the average velocity from these.
 
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  • #5
OlderDan
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Gokul43201 said:
Why the 2 ? Ar is a monoatomic gas. Or am I missing something here ? Also, you shouldn't have to double the radius to get the range.

The two comes from the problem statement

The second virial coefficient, B, at high temperatures equals half the excluded volume due to the repulsive part of the intermolecular forces.

So, 2B is the excluded volume.

You might be right about the range, but I am inclined to interpret it as I have done for this collection of atoms and this model. If the range is one radius, then the force of one atom would only act on part of the other atom, and part of one atom would start acting on part of the other atom before they came into contact, which of course is more realistic, but doesn't fit the rigid ball model. I think a reasonable way to look at the model of the atoms is as if the stuff the atoms are made of is concentrated at the atomic centers with some rigid force field at a distance that no other atom can penetrate. From that point of view, the range is twice the radius. It is problematic no matter how you do it. I like my way, but I have to concede there are other points of view.

Edit

For what it's worth, I was just starting to look at the last question and wanted to verify the mean free path formula; I came across this. This is equivalent to what I am saying about the range.

http://hyperphysics.phy-astr.gsu.edu/hbase/kinetic/menfre.html
 
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  • #6
Gokul43201
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The second virial coefficient, B, at high temperatures equals half the excluded volume due to the repulsive part of the intermolecular forces.
Didn't see that, sorry.

As for the range, I'm trying to think of this as merely an order of magnitude estimate, so I don't really see the need to double the radius...especially considering that neither 2R nor R is going to be the actual range of the repulsion (likely, some number in between), even within the set-up of this system. That said, I don't think doubling would hurt, if adequately explained.
 
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  • #7
Gokul43201
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koomanchoo said:
λ = c/((N/V)*σ*crel = kT/(sqrt2)σp and rearanging it in terms of P to find the pressure :cry: so much hassle
This is just laziness. We can't help you here unless you show your working. Calling it a hassle and expecting others to do it for you is not in keeping within the guidelines.

If you show your steps, we can point out possible errors.
 
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  • #8
OlderDan
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koomanchoo said:
and my final question i am having trouble with uses a big formulae i think..
it asks me to find the pressure (in Pa) at which the mean free path of argon at 23°C becomes comparable to the diameter of a 1L spherical vessel that contains it (cross section of argon = 0.36nm^2
i'm using the equation where
λ =
c/((N/V)*σ*crel = kT/(sqrt2)σp and rearanging it in terms of P to find the pressure :cry: so much hassle

This one looks pretty straightforward. You are given the cross section, σ, and the temperature, T. You just need to find the diameter of a 1L sphere and set it equal to λ, then solve for P in the appropriate units.
 

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