So we learned about these simple gas laws which relate two different properties of gases while keeping the other two constant. We can combine those relationships into a single law that encompasses all of them. And the way this is done is we can say, well, volume is proportional to 1 over p, volume is proportional to t, and volume is proportional to n.
So then volume must be proportional to nT over P, and in order to make this proportionality into an equation, we just need to add a constant, proportionality constant. So we get that, which is often expressed this way, PV equals nRT. That's the ideal gas law, where R is the ideal gas constant.
So this constant can be expressed with different units, so the value of it depends on what units are being used. What we're going to use almost all the time is 0.08206 liter atmospheres per mole Kelvin, and that's using atmospheres for pressure and liters for volume. And these units here look really messy, but these are the units of the other things.
N is moles, T is in Kelvin, volume is in liters, P is in atmospheres. And so units like this on the ideal gas constant cause all the other units to cancel out. We can get all of the other gas laws. from the ideal gas law. This is one way to think about it.
Another way to think about it is, so if PV equals NRT under one set of conditions, and I divide this by something that's equal PV equals nRT under a second set of conditions. It's still equal. Because I'm doing the same thing to both sides, I'm dividing by the same quantity, it looks different.
So I'm taking the ideal gas law and dividing it by itself, essentially. And to get any of the other gas laws, all we have to do is cross out the things that are not changing. So R never changes. So we can cross out R.
If we want to get Boyle's Law, where N and T are constant, Then we can cross out n because n1 is equal to n2. And we can cross out t because t1 is equal to t2. Well, what do we get on the right side then? We canceled everything out. Yeah, so that's equal to 1. So this side now is equal to 1. Well, if I rearrange that, I get P1V1 equals P2V2.
Boyle's Law. Okay? I could do, which one should I do, Charles Law or Avogadro's? Avogadro's. Okay, so P1V1 equals N1RT1 divided by the same thing with twos.
Avogadro's Law, we've got constant pressure and constant temperature. So the pressures go away and the temperatures go away. And of course R doesn't change, so R goes away. Those cancel out. Now this looks a little bit like what we learned as Avogadro's Law.
What we looked at was a rearrangement, v1 over n1 equals v2 over n2, Avogadro's Law. So, all you really have to memorize is the Ideal Gas Law, and you can quickly derive the other ones in this manner. Now, I'm not going to require that you... do that.
I believe on the exam I will give you all of the individual simple gas laws as well. But this can be a really handy tool to have because occasionally it'll give you a problem where we didn't learn a specific gas law. These are the three we looked at. What if we had constant N and constant V, and we just had pressure and temperature changing? Well, we don't have a specific gas law for that.
How could we figure it out? Well, we could derive one like this. Any questions?
So the ideal gas law, PV equals NRT, is used when we just have one set of conditions. So we'll have problems like this. An 8.5 liter tire contains 0.552 moles of a gas at a temperature of 305 Kelvin.
What's the pressure in atmospheres and PSI of the gas in the tire? Now, sometimes we recognize, oh, this is an ideal gas law problem, which, you know, given the title up here, we should be able to figure that out. But what if this is on an exam jumbled up with all kinds of other questions?
How do we recognize that we use the ideal gas law instead of one of the other gas laws? Well. The simplest thing to do is just to start making a table like we did for the simple gas loss. Just identify what the numbers are.
So the pattern we used before is we made a table because we were expecting two conditions. And so we had condition 1 and condition 2, and then we were just going to label everything as we came to it, re-reading the problem. So 8.50 liters, that's the first thing I come to. And what is that? Pressure, volume, temperature, amount of gas.
That's the volume, right? The volume is in liters, that tells us it's volume. Okay, and then we've got 0.552 moles. And what letter do we use for that? N, amount of gas in moles.
And then we're given a temperature, 305 Kelvin. And it says temperature, so that's temperature. Do I need to convert the units on that? No, because Kelvin is what I need. I went through the problem and I used all the numbers.
There's nothing in the second line. Because nothing was changing. So I'm not going to have ones and twos.
This was not a waste of time, though, because I now have my variables all nicely labeled. So I've got volume, amount of gas in moles, and temperature, and what are they asking for? Pressure.
So the ideal gas law, PV equals nRT, and we want to find out what P is. So we're going to take the equation and rearrange, divide by V on both sides, so these V's cancel out. So pressure is equal to nRT over V. And R is a constant, 0.08206 liters times atmospheres divided by moles times Kelvin.
The units on R tell us what units we have to use for the other quantities. So here are the volumes in liters. That matches. Amount of gas in moles.
That matches. Temperature in Kelvin. That matches.
So I don't have to do any unit conversions before I can use this equation. So N is 0.552. R is 0.08206. I know it's a pain, but write out the units times the temperature divided by the volume.
So this part here. is the minimum of what you should write in showing your work. Write the equation in its rearranged form, bless you, and then write it with all of the values plugged in with their units. That's the bare minimum.
That way I can look and double check, did I get everything in the correct place? I can check my units. I see that the moles, oops.
The moles cancel, and the kelvins cancel, and the liters cancel. And so then when I do the math, I'm going to end up with the pressure in atmospheres. 0.552 times 0.08206 times 305 divided by 8.5. How many significant figures should this answer have?
Three. So as I'm writing down digits from my calculator, 1.62, that's the third sig fig, and so I'm going to write down two extras, not worrying about rounding or anything yet, and my unit is atmospheres. Now it asks for the pressure in atmospheres, so I'm going to report the pressure in atmospheres by rounding this appropriately 1.63 atmospheres. That's one answer.
It also asks for the pressure in psi. And I want to use the unrounded version here, the unrounded value, when I convert to avoid rounding errors. Pressure conversions can always be done in one step. I want psi, so I'll divide by atmospheres.
I go to the table of pressure relationships and I see that 14.7 psi is equal to 1 atmosphere. So this answer should also have 3 sig figs, 23.892 psi. And that will round to 23.9. Yes, sir. Yes.
Which one? Like, for example, are we supposed to memorize, like, the 14th? 14.7? Yes. No.
Okay. No. You should memorize, just to save yourself a lot of time, that one atmosphere is 760 millimeters of mercury or 760 torr. We use that one a lot.
But it will be given on the exam. Oh, okay. Any other pressure conversions will be given on the exam. Yeah.
Any other questions? There's the answers. Let's do another one. What volume does 0.556 moles of gas occupy at a pressure of 715 millimeters of mercury at a temperature of 58 degrees Celsius? So, aside from the title, Ideal Gas Law 2, as I'm reading this, I'm seeing there's one amount of gas, there's one pressure, there's one temperature, there's no mention of things heating up or cooling or changing.
changing in any way. That's a big clue that this is an ideal gas law problem. So another way to do these problems without making a table is to just take the values and list them and label them.
So the first number that I come to is 0.556 moles. So I have the variables P, V, N, and T, what belongs with moles? N. So this is N.
Pressure of, so this is the pressure, 715 millimeters of mercury. And temperature of T equals 58 degrees Celsius. And they are asking what volume. So nothing's changing here.
I'm going to be using the ideal gas law, PV equals NR. This time I'm trying to find the volume, so I'm going to rearrange my equation to solve for volume. So I need to divide both sides by the pressure so the pressures cancel out.
Instead of rewriting this, I'm just going to clean it up. If you're still having trouble rearranging equations, please come and talk to me. I can explain it to you. And then you'll go, oh, that's so easy. Why did I have so much trouble with it?
Seriously, we can do that. Okay, so I've got n, I've got t, I've got p. I need r.
Well, r is a constant. And I'll give that to you on an exam also. So before I go plugging my numbers into the equation, I'm going to check my units.
It'll save me some time later. The units on these other terms need to match the units in the ideal gas law, and gas constant. So pressure should be in atmospheres. Here it's in millimeters of mercury. So I can save some time by converting that before I start doing any calculations.
Again, pressure conversions, one step. I want atmospheres, so I'm going to multiply by atmospheres. I'm going to divide by the unit that I've given, millimeters of mercury. I go look up the relationship.
One atmosphere is equal to 760 millimeters of mercury. So I get 715 divided by 760. And that is 0.940. So my 715 here is 3 sig figs. So my converted pressure also has 3 sig figs.
The 760 to 1 atmosphere, that's exact because that's how it was defined. So 940, and I'm just going to write down 2 extra digits. That's atmospheres.
I settled on writing two extra digits because by writing two digits instead of just one, I don't even have to think about rounding this one. It doesn't matter. I'll be fine.
And I don't have to write down a whole bunch of extra stuff, but this is enough to avoid rounding errors. So that's my pressure. Is 58 degrees Celsius good?
What unit do I have? I need Kelvin. So, and you might wonder, well, should I use 273 or 273.15? Well, 273.15 is always better.
Here, this temperature has its uncertain digit in the ones place, so using 273 would be fine. But 273.15 is better. So my calculator gives me 331.15 Kelvin. This is adding. Here I had two sig figs ending in the ones place, and so my temperature is also going to end in the ones place.
Notice I started with two sig figs in my temperature and now I have three. That happens a lot converting to Kelvin. So now I can take my equation that's up here.
I don't have space right under it. I can write in here the volume is equal to n, .556 moles. times R, 0.08206 liter atmospheres per mole Kelvin, times the temperature, 331.15 Kelvin.
Divided by the pressure, 0.94078 atmospheres. If I did everything correctly, my units should cancel out. Moles and moles, atmospheres and atmospheres, kelvins and kelvins. As I'm writing my volume down, I'm looking at significant figures and predicting I need three significant figures. In my answer, 1, 6, 0, there's three.
I'm going to write down the next two digits. The unit is liters, and so this is going to round to 161 liters. Any questions? In some ways, ideal gas problems are more student-friendly because you don't have to identify which things go together, but you do have to make sure all the units match up. Question?
Hmm. Well, thank you. I obviously did something wrong. So looking back in the history of my calculator, I see that what I did is I left out the .08. Yes, so thank you, thank you for pointing that out.
And we can remedy that. Well, I'm going to have to move some things over a little bit. There we go. Fixed. Any other questions?
There's proof. It'll be on YouTube later. I make mistakes too. It's okay. We all make mistakes.
No questions? 16.1 liters. Another one. Determine the pressure in millimeters of mercury of a 0.133 gram sample of helium gas in a 648 milliliter container at 32 degrees Celsius. So I recognize this is an ideal gas problem.
If I didn't recognize that, I could start making a table like I did for the first example today. There's no harm in that. It's not a waste of time because you're still writing stuff down and getting it labeled. Determine the pressure. So that's what I'm trying to find.
I've got 0.133 grams. Not sure what to do with that, so I'm going to leave it alone. 648 milliliters.
What is that? Volume. That's a volume.
And then I've got 32 degrees Celsius. And that's a temperature. And I'm trying to find the pressure. So ideal gas law, that's kind of messy, PV equals nRT.
If I want to solve for pressure, I'm going to divide both sides by volume. Volumes cancel out. So pressure is equal to nRT over V. I've got V and T. Was I given the amount of gas in moles? No, I was given the amount of gas in grams.
Can I convert grams to moles? Sure I can. So this is going to be n.
So this is helium gas. So I need to use the molar mass. Multiply by moles, divide by grams. I look up the mass of helium.
One mole weighs 4.03 grams. 1.03 divided by 4. I'm going to keep track of sig figs as they go along. This should have three significant figures.
3, 3, 2. Write down two extras. Remember, leading zeros don't count. That's probably...
One of the most common serious mistakes students make with sig figs is they count that leading zero, and that can end up in you rounding a number off to literally nothing, which should never happen. So that's moles. I don't like where this is, so I'm going to move it because I can.
Well, we also need R. I find things like the ideal gas constant to be somewhat comforting because when everything else is changing, that's the same, right? It's like I don't have to worry about that one. It's the same. The units on the ideal gas constant tell me what units I need for the rest of my things.
What happened to my... Oh, I know what happened. This got messed up here. So, is milliliters going to work?
No. So we can multiply by 10 to the minus 3 liters over milliliters and cancel the units out. Always works. Or I can say, well, instead of writing this prefix, I'm just going to...
write what it means in there and I'm not going to worry about the rest of the math. So this is the same as 648. The m means times 10 to the minus 3. So I'm going to write times 10 to the minus 3 liters. Millie is an abbreviation for times 10 to the minus 3. So is this improper scientific notation? No.
Does your calculator care? No. This is fast.
If you want to do it the other way, go right ahead. Celsius never works with gases, so we need to convert this. That's going to be 305.15. That's a dangerous thing to do that in your head.
It's a good idea to double check. So now I've got all my variables. I've got my rearranged equation.
So I can put the numbers in. Pressure is going to equal the amount of gas in moles, 0.033225 moles, times R. And I can tell I'm not going to have enough space here. I'm going to move it over where there's a little more room. You guys can use the whole piece of paper.
I've just got this one screen. Ideal gas constant, 0.08206 liter atmospheres per mole kelvin times the temperature 305.15 kelvin. Whole business divided by the volume. I check my units. Malls cancel.
Leaders cancel. Whoops. That was scary, huh?
And Kelvin's canceled. 0.033225 times 0.08206 times 305.15 divided by 648EE-3 equals. Anybody else get that number?
Thank you. Is that my answer? No, it's asking for pressure.
Millimeters of mercury. So I just need to convert that. This time I've got atmospheres on the bottom and millimeters of mercury on the top. One atmosphere is 760 millimeters of mercury. This is still going to have three significant figures.
And so I will ground that to 976 millimeters of mercury. Yes? We don't want two sig figs because they're too many.
That's an excellent question. Don't we have two sig figs because of the 32? Temperature conversions are weird because they involve adding.
So when you change temperature to a different unit, you can gain or lose significant figures. What does it know that it's 32.0? Well, it doesn't matter because when we're adding, we're not counting the number of sig figs. We're counting the number of decimal places. So here, this has been measured to the nearest one.
degree Celsius. So when we convert it to Kelvin, it should be measured to the nearest 1 Kelvin. 1 Kelvin is exactly the same as 1 degree Celsius.
So by converting to Kelvin, we've gained a significant figure. Yeah, that can be a little confusing. And, you know, if you are one of the multitude of students who just really hates sig figs, three sig figs is often a good guess, right?
Keep some extras and round it to three at the end, and most of the time that works. Any other questions?