# Van der Waals equation for arbitrary gas - need HELP!

• April 5th 2010, 03:00 PM
Intrusion
Van der Waals equation for arbitrary gas - need HELP!
Hi guys, I've been working on this problem for so long but I cannot figure out how to approach it...what am I supposed to do? Here is the question, and what I have tried...

(Background info is included so you know what's going on...)

http://imgur.com/juykx.jpg

http://imgur.com/XT2oh.jpg

What I've done: b) I've taken the derivative of the van der Waals gas equation to get 2a/v^3 - RT/(v-b)^2

Thanks so much to anyone that can help!
• April 5th 2010, 03:15 PM
skeeter
$p = \frac{RT}{v-b} - \frac{a}{v^2}$

$\frac{dp}{dv} = -\frac{RT}{(v-b)^2} + \frac{2a}{v^3}$

$\frac{d^2p}{dv^2} = \frac{2RT}{(v-b)^3} - \frac{6a}{v^4}$

use the chemists' definition of a critical point.
• April 5th 2010, 03:25 PM
Intrusion
Wait a minute, wouldn't 2RT/(v-b)^3 - 6a/v^4 would be the inflection point for the van der Waals gas? That's actually the next question and what I had done. However, when I get to the last question in that series I have no idea what to do, which is why I was questioning my first two results.

http://imgur.com/GyX34.jpg
• April 5th 2010, 03:31 PM
skeeter
Quote:

Originally Posted by Intrusion
Wait a minute, wouldn't 2RT/(v-b)^3 - 6a/v^4 would be the inflection point for the van der Waals gas? That's actually the next question and what I had done. However, when I get to the last question in that series I have no idea what to do, which is why I was questioning my first two results.

http://imgur.com/GyX34.jpg

the second derivative expression is not an inflection point ... an inflection point occurs at a defined point on a curve where the second derivative changes sign.
• April 5th 2010, 03:38 PM
Intrusion
Right, where f'' = 0, but I still don't see how that can be used in the last problem....
• April 5th 2010, 03:44 PM
skeeter
Quote:

Originally Posted by Intrusion
Right, where f'' = 0, but I still don't see how that can be used in the last problem....

you are looking for points where f' = 0 and f'' = 0 simultaneously.
• April 5th 2010, 05:46 PM
Intrusion
How would I find that? I don't understand what it means by "solve the equations"....
• April 5th 2010, 08:51 PM
Intrusion
I don't mean to "bump" this just for the hell of it because I know how annoying that is to mods, but I am still having trouble with this.

Can anyone help me out? I didn't pick up on the clues skeeter provided.