Interpretation of the Van der Waals Equation

AI Thread Summary
The discussion centers on the interpretation of the Van der Waals (VDW) equation, specifically why pressure is adjusted differently than volume. The equation modifies the ideal gas law by decreasing volume to account for the finite size of molecules and adjusting pressure to reflect attractive forces between them. The key point is that the volume "V" in the VDW equation represents the ideal gas volume, while pressure "P" reflects real gas pressure. The adjustment for attractive forces is subtracted from the pressure, leading to the final form of the VDW equation. This clarification helps reconcile the seemingly contradictory adjustments made to pressure and volume in the equation.
nezahualcoyot
Messages
5
Reaction score
1
There is a silly detail about the interpretation of the Van der Waals (VDW) equation that I cannot fully understand. Say we have the Van der Waals equation for one mole:

(P + a / V^2 ) (V - b) = RT

The usual interpretation is that if you start from the ideal gas law PV=RT, you have to "decrease" the volume to take into account the finite size of molecules, so you replace "V" by "V-b". The attractive forces also reduce the pressure, so you... replace "p" by "p+a/V^2 " ? Why not "p-a/V^2 " ? Why if both pressure and volume are reduced, you subtract a quantity to volume but add a quantity to pressure? I know the equation is correct as it reproduces experimental results within its domain of applicability, but I would like an intuitive explanation for this. Thanks!
 
Physics news on Phys.org
There is a formal way to see that the "a" term does indeed represent an attractive force, by using the virial expansion. However, I don't know an intuitive explanation to supplement the formal way. Kardar's notes http://ocw.mit.edu/courses/physics/8-333-statistical-mechanics-i-statistical-mechanics-of-particles-fall-2007/lecture-notes/lec17.pdf give the formal way and some explanations which are supposed to be intuitive, but I don't understand the latter.

(Kardar's full set of notes is at http://ocw.mit.edu/courses/physics/8-333-statistical-mechanics-i-statistical-mechanics-of-particles-fall-2007/lecture-notes/ .)
 
Last edited by a moderator:
Thank you atyy. I got an answer for my question elsewhere. Its basically that, in the VDW equation, "V" stands for the ideal gas volume, as this is the quantity you measure experimentally (the volume of the container), but "P" stands for the real gas pressure, as this is what you measure.

To derive VDW from the ideal gas law, P = RT/V, the real gas pressure "P" will depend on the real gas volume, which is the ideal gas volume "V" minus a factor, so you have

P = RT/(V-b)

Finally you must subtract a factor from the real pressure to account for the attractive forces, so you get

P = RT/(V-b) - a/V^2

which is the VDW equation (P + a / V^2 ) (V - b) = RT.
 
  • Like
Likes 1 person
Thanks!
 
Thread 'Question about pressure of a liquid'
I am looking at pressure in liquids and I am testing my idea. The vertical tube is 100m, the contraption is filled with water. The vertical tube is very thin(maybe 1mm^2 cross section). The area of the base is ~100m^2. Will he top half be launched in the air if suddenly it cracked?- assuming its light enough. I want to test my idea that if I had a thin long ruber tube that I lifted up, then the pressure at "red lines" will be high and that the $force = pressure * area$ would be massive...
Back
Top