eq 1........ V= nRT/P
Since the problem here is that as T --> zero or as P --> infinity, the volume goes to zero. So let's just add a size correction to the equation:
eq 2 ........V = nRT/P + b
Now our gas does not disappear as T --> zero or as P --> infinity. Now as those variables go toward those limits, V goes to b. So b is the size effect of the gas and would have a different value for every gas. That was easy!
Is there another problem with the ideal gas equation of state? Yes. Let's look at solving for P:
eq 3 ........P = nRT/V
Now remember what the pressure is for gases -- it is the force of the molecules or atoms colliding with an area of the container. If the molecules don't have any attraction or repulsion for each other then this prediction for P will be ok. However there is a potential energy of interaction between the molecules or atoms of the gas. (By the way, the Kinetic Molecular Model assumes that there is only kinetic energy which is one of its failings.) So let's try to very simply take into account an attractive interaction between the molecules or atoms of the gas. We know that these attractive interactions will reduce the pressure. Do an experiment where you are now sitting to understand why there is a reduction in pressure: Strike out with your right hand at a pretend wall in front of you as a simulation of a gas molecule hitting the wall and causing a pressure effect. Now take a rubber band and wrap it around your left wrist and your right wrist. Now keep your left hand still and strike out again at the same imaginary wall the same distance in front of your right hand. Can you feel the resistance of the rubber band which reduces the force with which your hand strikes toward the wall? The rubber band represents the attractive forces between the atoms or molecules which reduces the pressure.
These forces between different molecules or atoms (represented by your rubber band) are called intermolecular forces. The forces which hold a molecule together are called intramolecular forces.
So correcting the pressure for the intermolecular forces, we have
eq 4 ........P = nRT/V - something
It turns out that this something is a/(V/n)2 -- the derivation is not too complicated, but you can see it in another course later in your career.
Combining equation 2 and equation 4 gives us
eq 5 ........P = nRT/(V - b) - a/(V/n)2
and this is called the van der Waals equation which is the simplest equation of state that we can develop which corrects for both size and intermolecular forces. The b and a constants are different for every gas which is a bit of a disadvantage because we would really like to have an EOS which does not contain variables which are different for every gas -- however that is another story for a more advanced course. But no matter how complicated the EOS is, the corrections are still just for those two items: size and intermolecular forces.
For a more detailed discussion of intermolecular forces go to this site. But take a deep breath before you click on that for it is a bit deep there.
Here is a help site more on the level of this course which includes all of the topics we have covered in these four sections.
|Now take a practice quiz to help you understand if you understand the basic concepts.|
|You must use your real name when it asks for a name.|
|The test will only submit when you have answers all of the questions correctly.|
|If you are not taking this course for credit please do not answer all the questions correctly for I don't want to be flooded with email answers to the tests.|
Web Author: Dr. Leon L. Combs
Copyright �1999 by Dr. Leon L. Combs - ALL RIGHTS RESERVED