 I just wanted to remind everybody, of course, that we're still having the lab practical midterm this week. The exam two is on Friday. We'll cover up only till Monday's stuff, OK? So everything today won't be on the exam. There's practice exams and solutions are posted. Inez is having her exam review 3 to 4.50 PM on Thursday, right? And what's the rate that is? 118. 118. I have two reviews already. I posted the videos for both of those reviews. The second review, as probably many of you know, doesn't have volume, so I apologize for that. But I've also posted the page of problems that I got, all of the questions that I went over in the review on that same kind of folder in the folder where it says exam reviews on the Blackboard site. So if you want to look for that stuff, you can find the videos of the two reviews I think they're on both like about an hour and a half long, and the problems are there, too. So the last thing we talked about last time was gay loosex law. Let's go over this problem. I don't think we've got a chance to do this last time. So if you recall, gay loosex law is written there. It relates pressure and temperature. So PI over TI equals PF over TF. And if you remember, I means initial, and F means final. So if we look at this, the way I usually do these problems is go ahead and write all of them out. TI equals TI equals PF equals and TF equals. Because this way, you can really kind of figure out which variable you're looking for, and then you can manipulate your equation. So if you look, it says the pressure of a gas is 1 ATM at 273 Kelvin. So 1 ATM must be the initial pressure, 1.001. And it says that it's at 273 Kelvin. So that must be the initial pressure. Remember, if this temperature was in degrees Celsius, you would have to convert it to Kelvin, OK? If you use these temperature measurements in degrees Celsius, you'll get the incorrect answer, OK? And then it says, while you double your temperature to 546, what's the final pressure? So 546 Kelvin is the final temperature. And what are we looking for? The final pressure. Then you can come over here and say, OK, this is the variable we want to isolate. So we just, so we want to get this part out of this side, get the TF off of the right side of this equation. So in order to do that, since we're dividing by TF, we'll multiply by TF. That'll cancel that out. And remember what you do on one side of the equation you have to do to the other side of the equation. So we'll just multiply this side by TF as well. And when we do that, our new equation is TF equals TF times PI over PI. And that's cool because the only thing we're looking for is TF, and we have all of the other variables there. So just come over here and plug in your numbers. So TF is 546 Kelvin PI, which is PI. Notice Kelvin cancels there. And our units are going to be units of ATM. What you have to do is take 546, multiply it by 1, and then divide it by 273, and then write it to the correct number of significant figures of which, in this case, would be 3.0018. And then, as we're on the test, you want to box your answer to show me that you got it right. So that's the last thing we'll be looking at for Friday's test. So everything subsequent to that will be on ATM 3. OK, so let's look at the combined gas law. So the combined gas law is very similar to Boyle-Charles and Gay-Lusak's law. It's actually just squishing all of those laws together. So you can see, hopefully, Boyle's law in it. Here, let's write this down. Maybe if I have enough color to chop it off. Let's circle them all. OK, so I'm going to read. If you remember Boyle's law, that's P i times V i equals P f times V f. So that's the header in it. Gay-Lusak's law, the one we just did, pressure and temperature relation. And Charles law, you can see all three of them are in there. OK, so you just squish them all together and you get the combined gas law. So the combined gas law is used when you're having pressure changes, volume changes, and temperature changes. So not just pressure and volume or pressure and temperature or pressure or volume and temperature. So when you've got more than just the two variables changing, you're going to have to use the combined gas law. So here's an example of using the combined gas law. It says a gas sample has a volume of 2.50 liters when it's at a temperature of 30 degrees Celsius and a pressure of 1.80 ATM. So let's write down all that stuff. So it says the volume, so it's going to be V i, equals 250 liters when it has a temperature of 30 degrees Celsius, the pressure of 1.80. If you saw this, you should be like, OK, we're going to be using the combined gas law because we've got these three variables. More than likely to be using the combined gas law. We've got to see if two of them are going to change and then we'll definitely be. So let's see if we can find V f. Well, no, it looks like that's what we're looking for. We don't know that one. T f is going to be, so since you've got all of these variables, you know that you're going to be using the combined gas law. A lot of times it'll say the pressure is 1.80 ATM and then it's constant or something like that. In those cases, you won't be using the combined gas law. They're just giving you a little extra information to kind of make it a little more confusing. But when two of them change and you've got all three of them and you're looking for the change in the last one, you're always going to be using this combined gas law here. So let's go ahead and manipulate this. So remember the combined gas law, P i, V i. So that's oil's locked over T i. We're looking for V f, which is up here. So what do we want to do? Well, first, we can multiply both sides by T f to cancel out T f, like that. And now we have this equals to P f V f. This V f is still not isolated, so we need to get rid of P f. How do we do that? Divide by P f, like that. And what we do to one side, we have to do to the other side. So now what we need to do is look and make sure we've got our units the same for pressure. So they're both ATM, so that's cool. For volume, we'll just get this one in liters. But unfortunately, our temperatures aren't in kelvin. So we've got to do something to each one of these temperatures. So what is that that we have to do? Add 273. So for that one, it's going to be 373 kelvin. Our values in kelvin, P i, so kelvin's out of there. Cancel, cancel, ATN's out of there. And we get liters here. And that's cool because volumes are liters. So all we've got to do now is over there, just take 373, multiply that by 1 point, multiply that by 250, and then divide that by 303, and then divide that by 3. So unfortunately, because this one only has two sick things. Oh, it's 2.5. Well, let's just do this, this is 250. And we'll put a decimal point there. So that'll give us three sick things. So I got under, hopefully, everybody else got that same answer if you didn't go through it a couple more times. And you guys can do a different one up there that's 2.5. Second problem that you might like to try on your own. So Apogadro, there's Apogadro. Apogadro's law is very similar to Boyle, Carlson, and Jay-Lucak's law, except it's the equal volume, or the volume of a gas is proportional to the number of moles. So v related to v and n. And you can see it's very similar, especially to Charles and Jay-Lucak's law, where you've got, or Carl's law, especially, because you've got volume, but instead of super clear, you've got number of moles. And number n is the number of moles. What you'll find is if you do this calculation at standard temperature and pressure, volume per number of moles, you'll get the volume always to be 22.4 liters. And so we consider that the molar volume. That's the volume occupied by one mole of any gas. Apogadro's law, it's more of the same sort of stuff. I'll let you go through this one on your own and see if you can get some sort of answer for that. The density, let's try this one out together. Calculate the density of 4 grams of helium at STP. This one actually is a little more involved than what you would first be thinking, probably by the density. What's the formula for density, does anybody remember? Do we know the mass of this? 4.0 grams. 22.4 liters. How did you figure that out? The molar volume, right? So how do we get, but this isn't moles, right? This isn't moles, this is grams, right? So how did you get that the volume was this? So you converted it to moles, right? So this should be, if it's, the volume is this, this should be one mole of helium, right? That makes sense because remember at STP, one mole of any gas, right? So in this case, one mole of helium is going to equal 22.4 liters, right? So this is the conversion factor that you know. But we don't have, or it doesn't tell us that this is one mole. How did you figure out that was one mole? So how would you figure out? How many moles of helium do we have? So yeah, use the mole, or the atomic mass, right? So the atomic mass of helium, right? Or the molar mass of helium, 0.0 grams per mole. So we've got to figure out the number of moles of helium, 4 divided by 4 is 1.00 mole. If we know that one mole of helium, okay, so this is what you would have had to do. So you don't just say, just because you got some mass here, but that's one mole. You've got to figure out each, you know, each calculation. So if I were to give you, you know, a number that's like 6.32 or something like that, you know, this wouldn't be this volume here, okay? So you've got to watch that. You can't just say, just because it says STP, that is one mole. But let's figure this out, okay? So the density, remember, is the mass of the volume. So we've got the mass, so we've got to use that mass. So we're not going back to this moles over here. We just use that moles to figure out the number of liters, right? So we've got that mass divided by 22.4 liters. So it's very important that, you know, you've got to convert mass to moles, okay? You should be so familiar with converting, doing that conversion by now. That shouldn't be too much of a problem when you get to it. Okay, so we talked about the combined gas law. We talked about the individual gas laws. And we talked about molar volume, okay? Let's take all of that stuff and combine it into one more thing, okay? This is called the ideal gas law. And probably if you remember anything from general chemistry or a lot of people what they remember from general chemistry is the ideal gas law. Because it's very easy to get it in your head when you start saying it, I guess. But the ideal gas law is Nr. So if you recall, T is pressure, V is volume, N is number of moles, T is temperature in Kelvin, okay? So for the ideal gas law, the temperature has to be in Kelvin. Of course, it has to be in moles. The volume, it's got to be in liters. The pressure has to be in ATM. So if they're not in these units, you've got to convert them to those units. And that's because this thing here are, okay, this is a constant. What does that mean? It's kind of like a fudge factor, okay? So you've got to multiply these numbers by this constant in order to get these things equal to each other, okay? And so R is always going to be the same. It's written up there, 0.0821 liter, 18. So you'll see it written like this a lot of times. I like to write it like, oh, I'm going to say 1 mole Kelvin. So you guys remember that it's this divided by this. But that's what we call the ideal gas constant. So the thing is, if your units aren't in ATM liter moles in Kelvin, if you notice the ideal gas constant has these units in it, okay? And your problem won't come out to the right answer if you can't cancel these things out. So that's why you've got to have them in this unit. Okay, so PV equals NRT. And if you look at NRT even a little bit more, what you find is all of the gas laws come from that. Okay? So PV equals NRT. So if we put initial, initial, initial, initial, initial, well, R is the same. Divided that by cancel out like that. So we would get this constant. So if you said number of moles in temperature are constant, you would get PI. So if they were all constant, this would equal 1, right? This side would equal 1. And then you would get PI, VI equals PF, PF, and that's boils law. Okay? You could do that for everything. If you want to hold, I don't know, pressure in number of moles constant, you get Charles law. Okay? If you want to hold temperature and pressure constant, you get Apogadro's law. Okay? So holding two of those constant will give you one of those four laws, Charles, Apogadro's, Gay-Luzak's, or Boll's law. Okay? So it all comes from this equation here. Okay? This equation. And you get that equation from this equation here. Okay? So you can get all of those from... You could do any gas problem using this equation and this equation here. This is if they changed and this is if they stayed the same. Okay? So let's talk about them staying the same, which we haven't done yet. We've just been analyzing saying, okay, if the pressure changes from this to this, what happens to the volume? Okay? Now we're going to look at, well, what if we have the number of moles, temperature, pressure, and the gas constant? Well, what would be the volume of that thing? Okay? So we're not changing it. We're just looking at the state it is now. Let's do this one. What's the volume of gas occupied by five grams of methane, blah, blah, blah? RT, so the first thing we want to do, or the first thing I always do, is write down P, V, and T here, instead of equal to something. Okay, number one, what's pressure? How'd you know that? Kuba says ATM, right? That's pressure. What about me? What's that? What we're looking for, right? That's what we're looking for, is T. Okay? So we've got to do the ideal gas law to figure that thing out. Okay? What about N? What is N? Moles. Yeah, you've got to convert the grams to moles. So we've got to have a different one, too. So we've got M over here. M is 5.0 grams, right? So let's hold off on that thing right now. So it's grams of methane. We've got to do. And temperature, what is that? 25 Celsius. 25 Celsius. But we've got to convert that, too, right? So how do we do that? That's a little easier. 25 plus 273. We still need N before we consult for B, right? So how do we get N from this? We multiply this by 1. So what do we want down here? Grams. Grams. We've got to have that because we don't want grams, right? So this has got to be grams of methane. And what's got to be up here? Moles. Moles. And in fact, it's one mole of methane. So what is this conversion factor? One mole of methane equals how many grams, right? How many grams does one mole of methane equal? Yeah, 16.0. What is it? Something like that. 16.0. How did you get that number? You took C, which equals 12.01, right? Grams per mole. And multiply that, or add that to 4 H's, which equals 1.008 grams per mole. This grams of methane cancel out. Now we get moles of methane. So 5 divided by 16.04. And with this calculation, I'm just going to take it out to a d digits because my final answer is here, okay? For volume here, that's when I'm going to condense it down to the right number of significant digits, okay? So we've got now number of moles, right? So let's put that over here. And we've got moles, temperatures, pressure, and volume. And we're looking for volume. And the ideal gas constant's right in there. So let's erase this and figure out how we solve for volume. Okay? So we've got Pv equals Mrt. But we want V by itself. So what do we have to do? We've got to get rid of P, right? So divide by P. That'll cancel P out. And what you do to one side, you've got to do to the other side. Remember your algebra rules. So P cancels. And what do we get? We get volume equals number of moles times R times T over P, okay? So now let's just plug these numbers in. N is 0.3117 moles. I usually do like that, a big one for R, because I put that line like that. And I say 0.0821 liter ATM. Close brackets. Over one mole. Okay, you might not see this written like this very often. In the books or anything. I like to do it just so we can emphasize what we're canceling out. Okay? So since we've got it written there, I'm going to erase it. So we can kind of put stuff in the place of it. Okay? And then we've got to put RT to 90K. Okay? And then we're dividing the whole thing by pressure, which is 1.0 ATM. So what's the significant digits of the volume measurement going to be 2? How many significant digits? Are you going to have 2? 2, where does that come from? ATM. ATM down there, okay? So let's go ahead and cancel these units out. So look, mole cancels with mole there. Cancels with kelp in there. And ATM, cancels with ATM there. So you've got to think of this as being over one. Right? So that would be the same thing. Without cancels with the kelp. Okay? So when we cancel all those out, all we've got left is this unit there, which is liters, which is cool because volume should be in liters, right? So we've got V equals something liters. And all we've got to do is take that number, multiply it by 0.0821. And then divide that by 1 onto 2, 7.7. Methane, it's 5 grams at 25 degrees Celsius in 1 ATM. This is like Pete's methane bubble. It's probably about the size of one of those things. Well, a big one, right? Because at 7.6 liters, it's kind of big, right? But you might get a good flame out of that thing. Notice the next problem here. I'll let you do that on your own. But it asks for the mass of nitrogen. So you're going to have to convert that from moles back to mass, okay? But it also gives you the temperature and degree Celsius. So you've got to convert that to Kelvin. But it gives you the pressure in millimeters of mercury. You've got to convert that to ATM, okay? So if you remember, do you guys remember how to convert millimeters of mercury to ATM? Yeah, what is this? 1 ATM equals what? 760 millimeters of mercury. Well, 760 millimeters of mercury. So if we said that we got 700 millimeters of mercury. The other thing is we could have gone so far. Remember, if we wanted to ask, well, what's the density of methane gas for the first one? We could have said, well, remember, density equals mass divided by volume. We just learned the volume from doing all that ideal gas stuff, right? And we have the mass given to us 5 grams that we've already figured out in your notebook earlier. Notice these gases are pretty dilute, if you will, not very dense, right? A liter is very big and one gram is not a lot of weight, okay? So if you imagine two grams of this is already bigger than a liter, right? Like 132, 1.32 liters. So that's like that or something. This piece of chalk is probably about two grams, you know? So you can see the difference between the density of solids and the density of, you know, a gas. Pretty dramatic, you know? Okay, so let's talk about Dolman's Law now. Dolman's Law is a mixture of gases exerts a pressure that's the sum of the pressures that each of the gases would exert if they were present alone under the same conditions, okay? So what does this mean? We have a mixture of gases. The total pressure equals the pressure of one of the gases plus pressure of the other gases plus the pressure of the other gases, okay? So if we look, we want to know, well, what's the pressure of air? Well, if it's at sea level, it's 1 atm. That equals the partial pressure of nitrogen plus the partial pressure of oxygen because those are the two major components of it. Okay, so that's it for the gaseous phase, okay? So the cool thing about that that we're moving on to the liquid phase and then from there to the solid phase is that the gas phase has really been, you know, analyzed to death, if you will, especially in general chemistry courses. Liquid phase and solid phase you don't have to do as much stuff but it's like with these ideal gas. Okay, so that's cool. Okay, so let's look at this liquid state. What I'd like you to see here is this is the same volume of liquid, same solution, but in different containers, okay? So you're probably very familiar with this. The liquids could take on different shapes depending on the container that they're in but notice again they're the same volume. So that's kind of revealing, if you will. So the liquid state is characterized by high density. Remember relative to gases. Remember we were talking about, you know, the methane gas, the density of it is very low, okay? We measure density of gases in grams per liter, right? With volumes or with densities of liquids and solids but with liquids we measure them in grams per milliliter, okay, because they're so much more dense. Solids we measure essentially grams per milliliter, grams per cubic centimeter, okay? But anyway, so high density, indefinite shape that depends on the shape of its container just like what's picture here, small compressibility says very difficult to smash them together, okay? Because the molecules are already, remember, rolling on top of each other. And very small thermal expansion. In fact if you heat liquids up too much, right, they won't expand in the liquid state. What they'll do is go to the gaseous state and expand that way. So practically incompressible. So this enables things like brake fluid to work on your car. If the brake fluid was all squeezed out, you know, your car would kind of seize up. Viscosity is a measure of the liquid's resistance to flow. So if you have a very viscous liquid, it doesn't flow very much. You have a very non-viscous one. It flows quite a bit. It's a function of both the attractive forces between molecules and molecular geometry, okay? So molecules are just like anything else. The shape, their properties depend on their shape. So if they're very, very long and stringy, and they're going to be trying to roll all over each other, it takes them a long time to do so, okay? Or if they're very sticky, they'll stick together and they can't flow very well, okay? So this gives them increased viscosity. So if we wanted to compare something like... So first, stickiness, right? So if we go this way and they're very sticky to each other, right? They'll stick to each other and they won't flow very well, right? Compare that to ones that aren't sticky, right? They'll flow quite easily, okay? So this would be a more viscous, this would be less viscous, okay? So this would be sticky, and this would be not. Different structure. What is the structure of the molecule look like this? It would be harder just for these things to flow on top of each other, because they're just rolling around on top of each other. They would be long and kind of peeling on each other, right? So this also gives more viscosity to liquid. So it's a function of the attractive forces and molecular geometry. Flow occurs because the molecules can fly easily past each other. Glycerol is an example of a very viscous liquid. It's got these kind of attractive forces that make it sticky to other... glycerol molecules. Viscosity decreases with temperature. You've probably seen this on your own. If you heat up like some sort of like honey or syrup or something like that, it flows much easier than it does. So like syrup, I mean you could feel how sticky it is when you get it all over you, right? That's one of the reasons. It's also kind of, you know, it's like a sugar molecule that's a couple of sugar stuff together, so it's kind of got a weird geometry and it's got a lot of stickiness. That's why it's viscous, but decreases with increased temperature. Surface tension. I know it's about time to go, but let's just go over this couple of slides. Surface tension is a measure of the attractive forces exerted at the surface of the liquid. So what happens is molecules like to stick together, okay? So when you're at the surface of the liquid, the attractive forces are very great, okay? So if you ever see like water bugs walking on pools of water, it's because they're taking advantage of this inherent stickiness of the molecules at the surface layer sticking to each other and it allows them to put their weight on the water even though they're heavier and more dense than the water. They should be falling through. So the net attractive force is pull the molecules toward each other and that's why raindrops are like spherical in shape or you see beads on like your shower curtain because they want to kind of be next to each other and the easiest way to do that is to form a sphere. A surfactant is a substance that decreases the surface tension. For example, soap. So any sort of soaps will decrease the surface tension and you'll see instead of like beading, you'll see streaking when you put soap into it, okay? So if you have soapy water, those little water bugs, they'll fall through the surface. They won't be able to stand on. And we'll talk about paper pressure next time, okay?