Large amounts of nitrogen gas are used in the manufacture of ammonia, principally for use in fertilizers. Suppose 120.00 kg of N₂(g) is stored in a 1100.0-L metal cylinder at 280 °C. (c) Under the conditions of this problem, which correction dominates, the one for finite volume of gas molecules or the one for attractive interactions?

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Welcome back everybody. Our next problem says use the Van der Waals equation and the ideal gas equation to calculate the volume of 1.00 moles of neon at a pressure of 300.0 atmospheres and a temperature of 275.0. Kelvin explain why the two values are different. Hint. One way to solve the Van der Waals equation for V is to use successive approximations, use the ideal gas law to get a preliminary estimate for B. So here we need to get at what is the difference between the Vanderwaal Equation and the ideal gas equation? If we're solving this problem to obtain the volume of a certain amount of gas at specific pressure and temperature. Well, let's recall that in the ideal gas law, using the ideal gas equation, we make an assumption that so we assume that the gas particles. So the actual molecules or atoms of gas have no volume and no intermolecular interactions. Normally, this is pretty safe gas particles are usually quite far apart in the typical sample of gas at for instance, standard temperature and pressure. So the tiny little volume of a molecule of gas is so insignificant compared to the total volume of the gas. And then of course, the gas particles are so far apart that any intermolecular interactions, whether dipole dipole interactions or London dispersion forces don't really have an effect. However, look at the conditions of our problem, we have a pressure of 300.0 atmospheres, that's really high pressure. And when you have a gas under that much pressure, your particles are being forced much closer together. And the, the volume of the particles themselves becomes more significant as you've compressed the gas and the intermolecular interactions start having more effect. So our problem said, explain why the two values are different. They are different. Our equations are going to give different values for B because the very high pressure in this particular situation means that our values for volume and pressure need correction. We can't use, if we use the ideal gas law, we'll get an inaccurate result due to the high pressure. We need those correction factors that are provided in the vander walls equation. So let's look at our first, we need to calculate the volume based on the ideal gas equation. So we recall our ideal gas equation so that PV equals N RT. So if we solve that per volume, we have V equals N RT over P. So we'll go ahead and plug in our values here that we're given in our problem. So we have volume equals our number of moles, we know we have 1.000 moles of neon multiplied by R are gas constant. Now, in this case, we have gasses, we're dealing with volumes and pressures. So we want to use the value for R that has the units liters atmospheres per mole Kelvin. So it is 0.08206 leaders atmosphere per mole. Kelvin. Now, one quick note, we're dealing with neon. It can be easy to mix up the idea of noble gasses and ideal gasses. Neon's a noble gas with its full octet. It doesn't interact much with other elements. It's not an ideal gas because even though it's non reactive, it's non polar, it doesn't form dipole interactions just like any element out there. It has Vanderwaal forces. Thus, our Vander Wal's equation, it has electrons in a cloud, they'll have momentary areas or intermediate or temporary areas of positive or negative charge and can still have those intermolecular forces. They're very weak, but they're there and under high pressure have a little more impact. So we've said 1.00 moles, 0.08206 leaders atmosphere per mole. Kelvin. And then our temperature, we've been given that in Kelvin. So 275.0 Kelvin, all that divided by our pressure of 300.0 atmospheres. Let's check that our units cancel properly. That also always gives us a tip of whether we've set up our equation properly are moles canceled our atmospheres cancel. We've got them in the numerator and the denominator of the whole fraction. Kelvin cancels. We're left with leaders, which is what we want. So we solve this equation and we end up with our volume equaling according to the ideal gas law, 0.075 2 2 L, double check our significant figures, our values and our initial problem all have four significant figures. Let's put a little note four sig figs. So that's the format of this volume with four significant pictures. So this answer is solving for volume using the ideal gas equation. So we've answered why they're different. We have our ideal gas volume. Now we hit the bigge the van der Waals equation. So we're going to be correcting those values of pressure and volume with factors to account for volume of particles intermolecular interactions. And we've been told to use the method of successive approximation. So we'll talk about that and the volume we're going to use for initial approximation is the one we just calculated, which is why we did that first. So our Van der Waals equation, we can see that the setup is very similar to the ideal gas equation. I'm going to go ahead and highlight in blue, the two equations just so we can easily compare them. So that equals in parentheses P plus and then A and squared divided by V squared and then in parentheses V minus NB equals N RT. So you see we have that corrective factor on the pressure and the corrective factor on the volume. So we have this A value and A B value. These are factors based on individual gasses. So you look them up in a reference book, our gas is neon. So we're going to look up those values for neon. So we have a, our, a value for neon is 0.211 liter squared atmospheres per mole squared. And our B value for neon is zero point 0 7 1 L per mole. With that a factor being a corrective for inter macular attraction. You can think a attraction and the B factor being a corrective for the volume of the particles. So we need to solve this for V. And of course, we know right away that that's not going to be a simple answer. So we go ahead and just start by isolating that V minus NB. So then we have V minus NB equals N RT divided by the expression P plus and in parenthesis A N squared over V squared. And that means that we can add and B to both sides. And we'll have an equation in the form of V equals N RT divided by P plus A N squared, divided by V squared plus NB. Of course, we can see it right away. The issue is that we've got a V squared in the denominator of the denominator So we can't very simply solve our V. But that's where our successive approximations come in. And we're going to use that ideal gas calculated value to plug in for V squared. Then successive approximations means we'll come up with the V value. After doing that, we take that V value, use it as our next approximation. And then we keep going plugging the new V value in until that V value is no longer changing excessive approximations. So let's do out the first one. So I have filled in the values. Quick note, I didn't include all the significant figures. We're going to keep in mind that our values are provided in the initial problem with more significant figures. But things like 1.00 moles or excuse me, 1.000 moles take up a lot of space. I just wrote one mole in here, but we'll just keep in the back of our mind and I'll make a note. So we don't forget that we started with poor sig figs for everything. So let's look at what we plugged in, make sure our units cancel. This is a really complicated equation, make sure we didn't forget anything. So at the top of V equals at the top, we've got our N one mole times R 0.08206 L atmospheres per mole. Kelvin times our temperature. 275 Kelvin again left off some zig figs there divided by in the denominator, we have a pressure 300 atmospheres plus. And then in brackets we have that A N squared. So A is 0.11 L squared atmosphere per mole squared. And then our N squared and as one mole that quantity squared divided by volume squared. So using our ideal gas volume, 0.07522 L that quantity squared. Now note I said I started with four sigs but my A value and my B value have and excuse me, I just realized my B value here is written in incorrectly. So I apologize for that. Go back up to correct this B value. I'm missing a digit here. So erase that my B value should be 0.0171 m per mole. We haven't entered it into our equation yet. So I haven't ruined anything yet. But as I was saying, our A and RB value now that I have the correct version have just three significant figures, three SS. So our answer at this stage will drop to three. So just to keep in mind when we get to the very, very, very end, but of course, we don't use them until we get to the end. The only ones I've dropped are ones that don't matter to the volume. So like 0.0. Now notice that this is just the first part of the equation. Do not forget plus and B which um I almost did here. So we still need to add it. So little here might be good to put brackets around this whole phrase so that you don't lose it. I always like adding brackets or parentheses when I'm worried about losing a negative sign or a term. So plus and B we've got and being one more and our B factor being 0.0171 L per mole. So now I'm just going to check that all my units cancel to make sure I have put everything in correctly. The simple expression is NB because I have moles canceling out and I end up with leaders. So I should have leaders plus leaders in this equation. Let's look at that big fraction. Here in the numerator, I have moles canceling out. I have Kelvin's canceling out. So I'm left with leaders and atmospheres in the numerator. So I should end up with atmospheres in the denominator to end up with leaders. So I have of course an addition in the denominator, the 1st 300 atmospheres, I've got the atmospheres there. I should end up with atmospheres of the units of that fraction. This one's a little trickier because of all the squaring here. Be careful. I'm just going to cross, I have one mole. I'm just going to cross out mole here. It's mole squared. It cancels out the mole squared in the denominator of my A factor. But if I cross out mole squared there, which happened the first time I solved this problem. Not an issue with the one mole. It is an issue with the volume. I forgot to square the volume. So be real careful when canceling out units. If you then forget, you need to square the actual numerical value there. So the mole squared cancel out and then liter squared. And then again, same thing, my volume in the denominator is squared. I'm just going to cross out the L not the squared. I know that those units cancel, but I don't want to forget to square that value. So units all cancel, I end up with, of course, then my atmospheres are going to cancel after I add those. And on the top, I have liters plus liters. So that's all good. So I'm not going to work out how to simplify each term here. This video will take forever. I'm going to let you plug this all into your calculator, tell you what you should end up with here. But I'll just take a little pause. You can work on this and make sure you end up with the same value I do. So you should have ended up with 0.084005 L. We're not going to round off into significant figures because we have to use this value again in our successive approximations. We'll wait till the very end. But we recall that we have three significant figures to work with when we get to our final answer. Now, just note, be careful if you came up with a different answer, make sure you squared those values in there. So let's scroll down and look at how this is our first approximation. So now we're going to take that value of volume and plug that in to the denominator of the squared factor as our second approximation. And just keep going until we get two V values that don't change. So let's set up the second equation. So you can see that we took our value for our first approximation and put that in for the volume to get our second approximation. When you plug that in, you'll end up with an answer of 0.085504 L. So see that now we differ at the 1000th place in our answer. That's compared to the di difference from the first approximation to the original ideal gas law value which differed in the 100th place and we're getting closer. So now we just repeat this again and again. So I'm going to go ahead and put up the next couple of approximations. We can look at them, you know how to set up the equation. I won't take your time running through the whole thing. But let's see what happens as you keep plugging this in. So you can look over this equation. See it's the same equation over and over again, we have uh the number for volume from our first approximation plugged into that spot for V squared and solved for a new second approximation 0.085504 L. Notice that this only differs by that third decimal point. Then we plug in that value into the volume squared to get our third approximation 0.085720 L. So we're down another decimal place to differ. Plug that in to get our fourth approximation value of 0.085751 L. Keep on going to 1/5 approximation. I end up with 0.085755 L. Plug that in sixth approximation. And what do we have here? 0.085755 L. And we've got two values for volume that are the same. So we've reached our endpoint finally, last step as we recall, we have three significant figures to work with. So the answer that we will have here, we'll round that off 0.0858 L. So that would be our volume as determined by the Van Der Waal equation. So I'm going to bring that value up so we can look at it side by side so that we don't, we can see that comparison, 0.0858 scroll on back up, so we can see all of our answers. So again, that Vander Bals equation gives us a value for volume of 0.0858 L where the ideal gas equation gave us 0.0570 0.07522 L. So we see the volume is actually greater when calculated by our Vander Waals equation. And again, the reason those two differ in this case where usually the ideal gas would be a good um approximation is because of the very high pressure in this situation where we have 300 atmospheres of pressure. So hopefully after this, you have a new appreciation for why the ideal gas law is such a nice way to simplify this when it applies when pressure isn't so high. See you in the next video.

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Key Concepts

Ideal Gas Law

Real Gas Behavior

Van der Waals Equation