Main content
Course: MCAT > Unit 9
Lesson 8: Gas phase- Gas phase questions
- Absolute temperature and the kelvin scale
- Pressure and the simple mercury barometer
- Definition of an ideal gas, ideal gas law
- Derivation of gas constants using molar volume and STP
- Boyle's law
- Charles's law
- Avogadro's law
- The van der Waals equation
- Partial pressure
© 2024 Khan AcademyTerms of usePrivacy PolicyCookie Notice
The van der Waals equation
By adding corrections for interparticle attractions and particle volumes to the ideal gas law, we can derive a new equation that more accurately describes real gas behavior. This equation, known as the van der Waals equation, can be used to calculate the properties of a gas under non-ideal conditions. Created by Sal Khan.
Want to join the conversation?
- Wonderful lecture but sir,I can't understand how you put n/v^2(43 votes)
- Interesting question!
Let's examine proportional factors that we have.
1.) The attractive force is proportional to the density of molecules (the more molecules we have, the stronger the force), hence proportional to n/V
2.) On the same time not all molecules experience the same attractive forces. As it was said in the video, the attractive forces between the molecules in the center of the container cancel out, as they are getting "pulled" in many directions simultaneously.
Hence, the attractive forces are experienced only by those molecules which are closer to the wall of the container. The number of molecules which are "on the edge" of the container is also proportional to the density (the more molecules there are, the more of them will be located near the "edge"), hence we again have n/V
Therefore we have two equal proportionalities. After we multiply them (n/V)*(n/V) we get (n/V)2
I hope it helped!
P.S. If you wanna dig a little bit deeper, here's the link: https://chemistry.stackexchange.com/questions/70616/why-is-the-pressure-correction-in-the-van-der-waals-equation-proportional-to-n(19 votes)
- why did they need n^2 and v^2? couldn't they just use a single n and v or perhaps n^3 and v^3? Why did they choose the squared versions?(15 votes)
- It means that for a single molecule, its inward force is proportional to the concentration of other molecules. And the total inward force of all the molecules is also proportional to the concentration. So there is (n/v)^2.(11 votes)
- I don't understand how the real volume will be greater than the ideal volume. Suppose n ideal molecules occupy say "x" space, since they have no volume, the volume they occupy will be "x". But if n real molecules occupy say "x" space, the actual space available to the molecules to move will be less than x and so the volume they occupy will be less than x, won't it?(7 votes)
- You are on the right track. If n real gas molecules occupy a container of the same volume as one occupied by n ideal gas molecules, then the real gas molecules will occupy more space in that container; this means the real gas molecules will have less room to move around. Now what if we want to keep the pressure in both containers constant? This means we would have to give the real gas molecules the same amount of space to move around, thus we would need to make their container larger. So in essence, this whole question boils down to deciding on which value you want to keep constant; if you want the pressure of a real gas to be equal to that of an ideal gas, then you have to increase the volume of the real gas's container, but if you want to keep the volume constant, then the real gas will exert a higher pressure because it takes up more space in the container. :)(22 votes)
- Despite the previous questions and answers, I still do not understand why one is supposed to square n/v, I understand that there is a difference in density between the edge and the center of the container, but why would one not simply find the average density rather than square the value for density?(6 votes)
- what are the values and the units of 'a' and 'b'? i mean, since they are constants, they are bound to have some unit and value , right? help me if i'm wrong :)(3 votes)
- Yes, "a" and "b" do have values, but the values depend on what the gas is. There is a list of some of the values here - https://en.wikipedia.org/wiki/Van_der_Waals_constants_(data_page).(6 votes)
- What is the difference between the two (n/v)s?(3 votes)
- As there is differece between (pressure real and pressure ideal) as Sal stated as Pressure real< Pressure ideal so following this this creates a variation between areas of their particular container in which they have been placed, this follows to density (n/v) and experimantally it is found to be some constant times (n/v) i.e. a(n/v)(3 votes)
- Why you take n/v whole square(3 votes)
- In the final "Van Der Waals Real Gas Equation," wouldn't the P and V values be the "ideal" values that you are adding adjustments to?
Example: [Vr - bn] ... wouldn't that be [Vi - bn] because you are adjusting the ideal?(3 votes)- I think they are explaining wrong here. We for certainty know the equation is correct. But why one term(a) is added and the other term(b) is subtracted is not explained precisely. I too came here looking for this, but I guess I have to search more.(1 vote)
- Density is mass/volume not (moles/volume) and what is the sense of considering it density and applying to derive the formula
explain please..................(2 votes)- I think that the density should be considered as number density and not as mass density because in case of gases what matters more is how many 'number' of molecules we have. So, that's the first point.
Now the reason why we consider number density here:
First we are looking at just one gas molecule and it is experiencing a net inward (in this case) from not just one but many molecules. so we multiply 'a' by the number of molecules that we have per unit volume (you could also imagine a 1 cubic unit container). But we don't have just one molecule in the container. So, we multiply this by the number of molecules we have per unit volume to take into account all the molecules.
Hope it helps.(2 votes)
- What are the values for a and b?(2 votes)
- It depends on what molecules you are dealing with.(2 votes)
Video transcript
- [Instructor] We have
so far spent many videos talking about the ideal gas law, that pressure times volume is equal to the number of moles
times the ideal gas constant times temperature measured in kelvin. What we're going to do in this video is attempt to modify the ideal gas law to try to take into account when we're dealing with real gases, gases where the volume
of the actual particles are worth considering,
that we don't just say they're negligible compared to
the volume of the container. And intermolecular forces are something that we would like to
take into consideration. So let's think about how
we could modify this. And to help us a little bit, I'm just gonna solve for P. So I'm gonna divide both sides by volume, so we can say pressure is
equal to number of moles times ideal gas constant times temperature measured
in kelvin divided by volume. So first, how would we adjust this if we want to take into
account the actual volume in which the molecules can move around? Well, if we wanted to do that, we would replace this
volume right over here with this volume minus the
volume of the actual particles. So what's the volume of the
actual particles going to be? Well, it's going to be
the number of particles times some constant, based on how large each
of those particles are, maybe on average. And let's just call that b. So we could view this as a
modified ideal gas law equation, we're now all of a sudden,
we are taking into account the fact that these particles
have some real volume to them, but of course we also
know it's not just about the volume of the particles,
we also need to adjust for the intermolecular
forces between the particles. And we know that in many cases
those intermolecular forces are attractive forces, and so they would take
away from the pressure. And so we need some term
that accounts for that, a term that accounts for
taking away the pressure due to intermolecular forces. So term for intermolecular forces. Now I know what some
of y'all are thinking. "Do we always subtract? Might not there be some situations in which we actually have
repulsive forces between particles and it would actually
add to the pressure?" And there could be scenarios like that. You could imagine if they all
have a strong negative charge, they wanna get away from each
other as far as they can. And that could actually
add to the pressure, but in that situation, we
could subtract a negative and then that would be additive. Now, how could we take
this into consideration? So we know from Coulomb's
law that the force between two particles,
two charged particles is going to be proportional to the charge on one
particle times the charge on the other particle divided
by the distance squared. Now, obviously if we're
dealing with a lot of particles in a container, we're not
gonna be able to think about the forces for
between any two particles. But one way to think about it is in terms of how concentrated
are the particles generally. So when we're trying to think of a term that takes into account
the intermolecular forces or how much we're reducing the pressure because of those intermolecular forces, maybe that can be proportional
to not just the concentration of the particles, and that'd
be the number of particles divided by the volume,
but that times itself, because we're talking
about the interaction between two particles at a time, very similar to what we
see in Coulomb's law, because the end of the day these really are just Coulomb forces. So this thing right over
here is gonna be proportional to the concentration times itself. Or we could maybe call this some constant, for the proportionality,
times n over v squared, where a would depend on
the attractive forces between gas particles. And what we have just constructed, and let me rewrite it again,
this ideal gas equation, and actually let me put this orange term back on the left hand side. So if I write it this
way, that pressure plus a times n over v squared is equal to n R T over the volume of our container minus the number of molecules we have times some constant b, based
on how large on average those molecules or those particles are. This right over here is a
pretty good competence equation for when we're dealing
with more real gases, ones that have intermolecular forces, and one where the actual
particles have volume. And this actually does a pretty good job, and there's a name for it, it's called the Van der Waals equation. And there's many different
ways you might see it, you could see it written like this, or we could try to take this blue part and get it on the left hand side so it really looks like
what we saw at the top. Where there it would be
written as, and I'll write it, actually write it this way. Pressure plus some constant
times the density squared, let me close that parentheses, times the volume minus
the number of molecules times some constant is
going to be equal n R T, is going to be equal to n R T. And all of this looks really complicated, but the end of the day it
is just our ideal gas law modified for intermolecular forces and the actual volume of the particles.