Van der Waals Equation

the Vander Waals equation is an equation used for real gas is it combines the effects of attractive forces and gas volume to describe non ideal behavior. So the behavior of real gasses, we're gonna say deviations from this ideal gas behavior happens at high pressures and low temperatures. Remember, an ideal gas is imaginary, and ideal gasses behave as though they are alone. This is not possible if the pressure is incredibly high inside the container. At high pressures, it forces gas molecules to come closer together so they can't be alone and at lower temperatures that also causes gas is to start to condense downward. This also forces them to becoming closer contact with one another. Now, with the Vander Waals equation, we have two coefficients, which we call variables or Vander Waals constants. The polarity coefficient is the Vandals constant with the letter A and it corrects for the attractive forces felt between gas molecules, the size coefficient, it is the Vander Wal's constant be that corrects the volume of gas molecules. Now with the Vander Wal's constant be. What we need to realize is, as we increase the molecular weight of a gas, then this causes an increase for this Vander Wal's constant. So the greater the molecular weight of a gas, the greater its Vander Wal constant be now that we've seen the whole idea of these coefficients, and we know that the Vander Waals equation is used for real gasses, click on the next video and let's take a look at the formula involved.

now remember that the Vander Waals equation is used for real gasses. You'll notice that there's no purple box because you're not expected to memorize it. If you do have to use it on an exam, it's usually embedded within the question or on a formula shoot. So if we take a look here, we say that for the Vander Waals equation, it is pressure, plus your moles squared times your vander wal's constant a divided by your volume squared because this portion contains Are Vandals constant? A. It is a correction for attractive forces or just the polarity. General polarity of gas molecules. Then that's gonna be times volume minus your moles and times your Vandals constant be because B is involved here. This is a correction for the volume off gasses, which could be tied to their size. Now this equation may not look like it, but it's connected to your ideal gas law. The Van Waals equation is used for real gas is the ideal gas law is used for ideal gas is but ideal. Gasses are imaginary. If ideal gas has existed popular, one says that their volume is so small that it's insignificant and not important. It's negligible, so sense they're so small. That means their size would be equal to zero. Let's look if if event, if an ideal gas is involved, be a zero so end times zero is zero. This would mean that this entire part here is just V. And then if we're dealing with an ideal gas and ideal gas has completely elastic collisions, it has no attractive forces or repulsive forces, so it's polarity would be equal to zero. So take that zero and plug it in. So n squared times zero divided by the square, all that just become zero. So all you have left is P. So if we're dealing with an ideal gas, the Vander Waals equation becomes simplifies into the ideal gas law. So that's the connection. They have to one another. Now, if we take a look here at the columns we have, all these gas is the second column. Are are a values and they're in units of atmospheres. Times leader squared over mole squared. And then here we have beat, hear their units are leaders over moles with polarity, attractive forces. It's harder to see a pattern. But remember be is our size coefficient, we said, as the molecular weight increases your value for B increases. So if you were to look and look at all the weights of each of these gasses, it would make sense in terms of the numbers, as as you see them more or less. There's a few deviations, of course, because it's chemistry. But the general trend is, as you increase the weight of a gas, you should see your B value increase as well. All right, so that's how our Vander Waals equation is connected to the ideal gas law. And this long equation is the Vander Waals equation. Again. Don't worry too much about memorizing it. It's more important to understand the ideas of the coefficient A and B.

here. The example Question states using the Vander Waals equation determine the pressure of g oxygen gas in 250 ml graduated flask when the temperature is 50 degrees Celsius. Alright, So we're going to say the Vander Waals equation is pressure plus n squared times, a over V squared times volume minus end times B equals and are cheap. Now all we gotta do is plug in the values that we have. So we need to have the moles off oxygen. Remember, oxygen is diatonic. So one mole of 02 ways 32 g, though, too. So when we do that, we're gonna get our moles of 02 comes out as 20.6 to 5 moles of 02 volume has to be in leader. So that's 0.250 leaders and then temperature needs to be in Kelvin. So at 2 73 15, that gives us 3 23 points. 15 Kelvin. Since we're dealing with 02 in the charts up above, we would see that the a constant it comes out to 1. The B constant comes out 2. with this information, we plug it into the formula. All right, so let's see, we're looking for pressure, so we don't know it. So this is gonna be 0.6 to 5 squared times 1.360 divided by volume squared, which is 0.250 squared. And this is gonna be times volume, which is 0.250 minus moles, which is 0.6 to 5 times 0.318 And this is gonna equal. My moles are and t so, moles we wanna actually move all this out of the way. So we have space to write this out. So here are moles again. Come out to 0.6 to 5. So the Vander Waals equation related to the ideal gas law so are again is 0.8 to 06 and then temperatures 23 15 Kelvin, When I multiply these three together on this side, it comes out to be see, we're gonna multiply those out together. That comes out to be 16. 356 When I when I do this value minus these two multiplying, what I get is I'm going to get 0. Then when I work all of this out in here, this comes out to be 8. plus p. So now I need to isolate my P. So I'm gonna divide both sides here by 0. So plea P plus eight five equals 72.0 Subtract 85 from both sides. So P equals 63. atmospheres. So that would be the pressure of 02 when utilizing the Vander Wal's equation.