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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.

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  • starky ultimate style avatar for user zihad maruf
    Wonderful lecture but sir,I can't understand how you put n/v^2
    (43 votes)
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    • purple pi purple style avatar for user Normunds Palabinskis
      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)
  • winston baby style avatar for user AJ
    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)
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  • duskpin ultimate style avatar for user Kalp Aghada
    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)
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    • piceratops ultimate style avatar for user Darmon
      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)
  • blobby green style avatar for user Colin Westphal
    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)
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  • spunky sam blue style avatar for user tanvi.khairkar22
    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)
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  • leafers seedling style avatar for user Mintie
    What is the difference between the two (n/v)s?
    (3 votes)
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    • male robot donald style avatar for user Irfaan
      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)
  • blobby green style avatar for user Prabhakar Pandey
    Why you take n/v whole square
    (3 votes)
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  • blobby green style avatar for user Hailey Milakovich
    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)
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  • blobby green style avatar for user gurmatthecommando
    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)
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    • aqualine sapling style avatar for user AB
      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)
  • blobby green style avatar for user Colorado
    What are the values for a and b?
    (2 votes)
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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.