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Topic: Van Der Waals constants  (Read 7042 times)

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Offline Disturbia

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Van Der Waals constants
« on: October 10, 2007, 01:52:23 AM »
Hello,
I have a little problem interpreting the Van Der Waals' constants 'a' and 'b' in his equation:

I know that the constant a  depends upon the type of gas and it has something to do with the attractive force while the b constant represents the repulsive force between the molecules. My question is why the constants a and b of H2 (respectively ~0.244 L2atm/mol and ~0,0266 L/mol) is lower than Cl2's (respectively ~6.49 L2atm/mol and ~0,0562 L/mol) ?

Offline lemonoman

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Re: Van Der Waals constants
« Reply #1 on: October 10, 2007, 08:21:03 AM »
My question is why the constants a and b of H2 (respectively ~0.244 L2atm/mol and ~0,0266 L/mol) is lower than Cl2's (respectively ~6.49 L2atm/mol and ~0,0562 L/mol) ?

So if a has to do with the attractive forces between molecules, then it will be related to the intermolecular forces.  The Cl2 molecules are more attracted to each other than the H2 molecules are attracted to other H2 molecules ... this is because Cl2 has a big, mushy electron cloud that can be pushed around ... which leads to temporary dipole on the molecule.  Follow?

b, most simply, is related to the size of the molecules....and Cl2 is way bigger than H2.

This may be interesting:

if a and b were both 0, then this equation would simplify to P = RT/Vm --> PV=nRT --> the ideal gas law!


Offline Disturbia

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Re: Van Der Waals constants
« Reply #2 on: October 10, 2007, 11:43:12 AM »
My question is why the constants a and b of H2 (respectively ~0.244 L2atm/mol and ~0,0266 L/mol) is lower than Cl2's (respectively ~6.49 L2atm/mol and ~0,0562 L/mol) ?

So if a has to do with the attractive forces between molecules, then it will be related to the intermolecular forces.  The Cl2 molecules are more attracted to each other than the H2 molecules are attracted to other H2 molecules ... this is because Cl2 has a big, mushy electron cloud that can be pushed around ... which leads to temporary dipole on the molecule.  Follow?

b, most simply, is related to the size of the molecules....and Cl2 is way bigger than H2.

This may be interesting:

if a and b were both 0, then this equation would simplify to P = RT/Vm --> PV=nRT --> the ideal gas law!



Thank you.

So in other words, the factors that explain the values of a of different gases should be somehow similar to its boiling temperature right? Also, lets say fictionally that the atoms occupy the same volume how will this affect the VDW equation ? do you consider removing both a and b or just b ?

Offline lemonoman

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Re: Van Der Waals constants
« Reply #3 on: October 10, 2007, 12:16:07 PM »
So in other words, the factors that explain the values of a of different gases should be somehow similar to its boiling temperature right?

There is probably a good relationship between them....but I'm not sure if it's 100% true all of the time.  Generally a good rule though because they're both related to intermolecular forces.  Good call on that one, Disturbia.

Also, lets say fictionally that the atoms occupy the same volume how will this affect the VDW equation ? do you consider removing both a and b or just b ?

Not sure I follow you.  If the molecules took up the exact same volume and packed the exact same way, then their b values would be the same.  So you'd just plug the same value in for b in the Van der Waals equation.

All I was really saying with that last bit, is that the ideal gas law makes two assumptions:

1. Molecules take up no space.  This means b = 0
2. Molecules don't interact.  This means a = 0

And when you make these two assumptions in the Van der Waals equation, the ideal gas law comes out.  It's kinda cool.  And so, just as you were probably told in class, the Van der Waals equation (and constants) are really just there to adjust for non-ideality.

:)

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