Understanding Gas Laws and Applications

So there are gas laws that relate the different properties of gases. So the four important properties of gases are pressure, represented with a capital P, volume, represented by capital P, capital V, temperature with a capital T, and the amount of gas in moles, and that's given the symbol N. So the amount in moles. These properties are interrelated. If you choose change one, the others will change as well. And so the simple gas laws are just looking at pairs of properties. Might look at the pressure and the volume, or the pressure and the temperature, or the volume and the amount of moles. Not looking at all four of them at once, because that's kind of complicated. We'll keep it simple and just look at two. And these laws were deduced from observations where they held two of these properties constantly. changed one and looked at how the other one changed. And these laws date back a long time ago. So here's Boyle's Law, relating volume and pressure. Robert Boyle lived in the 1600s, right? And so... This is a representation of the type of experiment he did. He used a J-tube, a J-shaped tube with mercury in it. And in the closed end, he put a gas sample. And then you can measure the volume of the gas. You could calculate that based on the diameter of the tube and how tall it was. And you can look at the pressure between the outside and the inside, the pressure difference. is the height of the mercury columns, right? Just like with a manometer. So he found that if he poured more mercury in here, then that would cause the gas to squish together and the volume of the gas got smaller. The pressure of the gas increases. We see that the difference in heights here is larger. So the pressure goes up and the volume goes down. A lot of these relationships, if we think about balloons or other real-life experiences with gases, they make sense, right? So if you imagine taking a balloon and squishing it on all sides, putting pressure on it, it's going to get smaller, right? So this is an inverse relationship. When one thing gets smaller, the other thing gets bigger. This is an example of Boyle's Law graft. So here's the volume of the gas and the pressure. So as the pressure increases, the volume of the gas decreases. So that's, you know, kind of a pretty graph, but what's a lot more useful is if we can have a graph that gives us a straight line instead of curves. Because, you know, this was before they discovered electricity, so. They didn't have computers that did linear regression for them. So drawing straight lines is just so much easier. Now we can have a computer that will take this data and fit a quadratic or an exponential function to it and tell us what the equation is. But back in the 1600s, they didn't have stuff like that. So Boyle's Law says that the volume of a gas is proportional to 1 over the pressure. And this is when we keep the temperature and the moles of gas constant. We're keeping those the same because if we change those, it's going to mess everything up. So we say the pressure of a gas is inversely proportional to its volume. We saw that the graph of volume versus pressure is curved. If we graph volume versus 1 over pressure, the inverse of pressure, we get a straight line. So this is volume versus pressure. This is volume versus one over pressure, and now we get a beautiful straight line. This shows me that volume is directly proportional to one over the pressure. That can be described with a line equation. So from this proportionality, we realize that pressure times volume is... equal to a constant. Now we don't care what that number is, but it's going to be constant at a given temperature and amount of moles. And so when we change the pressure and the volume changes, or vice versa, we can say that this relationship is true. And this is... how we generally express Boyle's Law, the pressure 1 times the volume 1 is equal to pressure 2 times volume 2. Now, I am not going to quiz you on whether you remember that that is called Boyle's Law. As I've mentioned before, I'm not teaching chemical history. I think it's good to include that. on the side, but I'm not going to require you to pick Boyle's law out of a list of equations. I personally have a hard time remembering which one belonged to Boyle's and which one belonged to Charles, and does it actually matter? No, it doesn't. So if we look at what gas molecules are doing, so here we have a container where we can change the volume. It's a bit like a syringe or a piston. The lid here can slide up and down. So we can change the volume of the container, and we'll see that we talk about this a lot with gases. So here the lid has a one kilogram weight on it, and this lid does not just fall to the bottom because there's a gas in here, and the gas is holding it up. But there's a pressure. So the pressure here is measured with a more modern pressure gauge, and we're told it's one atmosphere. The volume of this gas is one liter. Over here now we've increased the mass sitting on the lid, which causes the lid to move down. So now the volume here is less than it was. It's half a liter. So we've gone from one liter to half a liter. So we've divided the volume by two. And what happens to the pressure? It increases by a factor of two. P1V1 equals P2V2. Pressure 1 was 1 atmosphere. Volume 1 was 1 liter. Pressure 2 is 2 atmospheres. And volume 2 is half of a liter. And this is equal. It does not matter what kind of gas it is. Is that the same as the one we're using? Yep. Yeah, we'll talk about that later, but one of the strange things about gases is you can, the behavior of gases is almost identical regardless of what the gas is. What the particles are doesn't matter, it's just how many particles there are. So Boyle's Law has implications in scuba diving. So if you've ever dived down to the deep end of a swimming pool, you know that you start to feel some pressure in your head when you get way down to the bottom, right? Because as you go below the surface of the water, the water Water is putting pressure on you. Water is more dense than air, obviously, and so the pressure will increase much more rapidly underwater than it does with an elevation change above water. So this is a general... relationship for every 10 meters of depth the pressure increases about one atmosphere due to the weight of the water. So here we have a diver up here at the surface depth of zero meters the pressure experienced is one atmosphere which is just normal atmospheric pressure. Come down here to a depth of 20 meters you go down 10 meters you add one atmosphere of pressure. you go down another 10 meters you've added two atmospheres of pressure so starting with one up here adding two we get a pressure of three atmospheres that's a lot of pressure So if you are breathing 20 meters below, your lungs are expanding against three atmospheres of pressure. The only way you can breathe down there is with pressurized gas. So the pressurized gas needs to match the outside pressure fairly closely so that you can expand your lungs. So down there you take a breath and... You, you know, big lung full of air, and you hold it. The pressure down there is three meters, and then you come up to the surface very quickly, holding your breath. Now the external pressure is one atmosphere. The internal pressure in your lungs was three atmospheres, and so it's going to make your lungs expand a lot. But I don't think you'd actually be able to hold your breath. It would hurt too much, and you'd let it go. before your lungs exploded, kind of a self-preservation thing. So the volume of the air, Boyle's Law predicts, will expand. So the pressure decreases by a factor of 3. The volume would expand by a factor of 3. That would not be good for your general health. Always exhale when rising from a dive. I remember thinking, you know, scuba diving, I don't know if you know this, in order to go scuba diving, you have to take the lessons and get certified and stuff. And I remember thinking, well, you go swimming, you don't need to figure it out, right? Why is it so different to go scuba diving? Well, there's a lot of things like this that are complicated and potentially fatal. And so there's a lot of stuff about gases that's actually really chemistry and physics that you have to understand in order to scuba dive safely. Personally, I'm not interested in going that far under the surface of the water. I don't like caves either. So let's do this problem. A snorkeler takes a syringe filled with 16 milliliters of air from the surface, where the pressure is 1 atmosphere, to an unknown depth. The volume of the air in the syringe at this depth is 7.5 milliliters. What's the pressure at this depth? If the pressure increases by 1 atmosphere for every additional 10 meters of depth, how deep is the snorkeler? I'm not sure why the author gives this as the first simple gas law problem. This one's challenging. But we can figure this out. So, this is a problem that we have not seen anything like before. And it seems a little, um, just unusual. I mean, who takes syringes underwater, right? Syringes of air. That's very strange. strange. So what we have to do is we have to take these words and get comprehension out of them, which can be challenging sometimes. So I find it very helpful to draw pictures. My pictures are really ugly, but they still help. And then when you get better at solving problems like this, you don't actually have to draw the picture anymore. You can picture it in your mind. You've got a whiteboard or a three-dimensional simulation thing in your head. It's so much better than drawing it. But we'll just settle for drawing it. So the snorkeler, I'm not even going to attempt to draw the snorkeler, but here's the surface of the water, right? A syringe filled with 16 milliliters of air from the surface. Okay, so here we are at the surface. Here is the syringe filled with air, and this must be a syringe that's closed on the end. So there's the plunger. So this is the volume in here, that's 16. milliliters of air. And the diver goes down. So here's our Our syringe again, and it says now the volume is only 7.5 milliliters. So the plunger comes in, and this volume is 7.5 milliliters. Why did the gas get smaller? Because as you go down in the water, the pressure gets bigger, right? What's the pressure up here? The surface, one atmosphere. So one atmosphere up here. What's the pressure down here? Well, that's the question. What is the pressure at this depth, right? So that's what we're trying to figure out. So I have two different pressures. I have two different volumes. This should make me think of Boyle's Law, which relates pressure to volume. Now, it really doesn't matter whether you call this pressure 1 and volume 1, or if you call that 2. It doesn't matter. Generally, people like to call this one 1, because that one came first. So we'll just do that. So I'm going to call this, this is pressure 1, and the other pressure is pressure 2. This is volume 1. Because this is the volume at this pressure, these two go together. They happen at the same time. And this is volume 2. Okay? So we have P1, V1 is equal to P2, V2. And we are trying to find pressure 2. Yes? So we should rearrange the equation before we put the numbers in. So I want to get P2 all by itself on one side. I don't care which side of the equal sign, but on one side. And here it's multiplied by V2, so if I divide by V2 on this side, V2 over V2 cancels out. And then to keep it equal, I need to do the same thing to the other side. So P2 is equal to P1 times B1 divided by V2. This sort of algebra is really common and important in Chem 1A, and so if you are algebra challenged and have a hard time rearranging equations like this, please come and talk to me. can explain it much better one-on-one and I can go into all sorts of crazy analogies and I'll keep at it as long as you will until it clicks and you go oh I see now it's so easy so once you see it it is but until then not so much So I rearrange the equation to solve for the thing I want. And then I'm going to take these other pieces and put them into the equation. You need to be careful here because if you get your ones and twos, if you don't copy it down correctly, now your equation is wrong and your answer is wrong. So some people have harder time tracking and copying things like that. Just be careful. So pressure 1 is 1 atmosphere, and volume 1 is 16. milliliters. Always, always, always put the units in there. And this is divided by volume 2, which is 7.5 milliliters. What happens with the units? They cancel. Milliliters divided by milliliters cancels. Unit cancels out. I'm left with atmosphere, which is a pressure unit, and I'm calculating pressure, so that's a good thing. So now I have 1 times 16 divided by 7.5. Pressure 2 is equal to 2.133 repeating atmospheres. How many sig figs would that number have? This one atmosphere can be a little confusing. One atmosphere is defined as the pressure at sea level. And so we can consider that to be an exact number. That's a little iffy, but that's what we do. This has two sig figs, this has two sig figs, so I'm just going to keep two sig figs here. 2.1 So the answer to the first question, what's the pressure? The pressure at that depth is 2.1 atmospheres. Any questions yet? That's this question. If the pressure increases by one atmosphere for every additional 10 meters of depth, how deep is the snorkeler? This one also requires a little bit of problem solving. So this is a relationship, one atmosphere for every additional 10 meters. So one atmosphere for each 10 meters. That's the increase. We aren't starting at zero, we're starting at one atmosphere. How much did the pressure increase? I'm going to fill this in now because we know what it is. Yeah, it increased by 1.1. So the difference between these two is 1.1 atmospheres. That's how much it increased. This is telling me that it increases one atmosphere for every 10 meters. I'm going to change colors here so it stands out a little better. So we had an increase of 1.1 atmosphere, and that relationship tells me that I'm going down 10 meters for every one atmosphere. And so the snorkeler must be 11 meters under the surface of the water. Any questions? Why wouldn't you use the 2.1? That's an excellent question. Why wouldn't I use the 2.1? If I was starting at zero pressure here, I would use the 2.1. But this is a one atmosphere increase in pressure for every 10 meters. And so I have to consider where I'm starting from. Does that make sense? Anybody else? Yes, so I had a mental debate while I was talking to you and I decided to do it this way, but thank you for bringing that up. So here I had my result and I wrote down extra digits like I always do. always tell you to do, and then I rounded my answer. Over here, we really should use the unrounded value when we do the calculation, and so the difference in pressure is 2.133 repeating minus one atmosphere, which is 1.133 repeating atmospheres. And then when I put that in here, my answer comes out to be 11.33 meters. But when I think about the significant figures, going with two significant figures, it ends up getting rounded to the same place. There are times where if you round your intermediate values, your final answer is going to be different. Here they end up the same. Any other questions? Jay, is he Charles? 1700s, early 1800s, looked at the relationship between the volume and the temperature of a gas and found that the volume of a gas is directly proportional to its absolute temperature. So Charles' Law can be stated as the volume is proportional to the temperature at constant pressure and constant amount of gas. You can say that the volume divided by the temperature is equal to a constant. And so Charles' Law is often expressed like this, V1 over T1 equals V2 over T2. The temperature must always be expressed in kelvins. The volume could be expressed in any volume unit as long as the two volumes are the same. The problem with temperature is that all the temperature scales are related with adding or subtracting. It's not just multiplying or dividing. And so the conversions with temperatures do not cancel out. The other problem that arises is on the Fahrenheit and the Celsius scale, you can have a temperature of zero or you can have a negative temperature. If you try to plug zero into this equation, what happens? It's undefined. It's not going to work, right? So the Kelvin scale, we don't ever have zero. If you use negatives here, you can also end up with negative volumes, which are ridiculous. So here's a Charles Law graph. So here's the volume of a gas at different temperatures. This is one mole of gas, this is half a mole of gas, and this is a quarter of a mole of gas. And we see that for each of them, the volume increases linearly with the temperature. So the data is going down here to about zero Celsius. We don't have data down here, but we can extrapolate. This is a nice line, and so we can draw the line past the data, extrapolating, and we find that all of these lines intercept at the same place, and that's called absolute zero. It happens to be minus 273.15 degrees Celsius, which should be familiar from when we learned about converting temperatures. The idea here is as you decrease the temperature of the gas, its volume gets smaller and smaller and smaller. If you keep decreasing it, eventually... Maybe you get to zero. Could you go past zero to negative volume? No. So if we got to zero volume, that would have to stop. All of these things intercept here at the same point, and that's not a coincidence. So this is the absolute theoretical lowest possible temperature. And negative temperatures are not possible. We need to use kelvins because on the Kelvin scale, the intercept of the line is zero, whereas for the other ones it will not be zero. One of my less good explanations. Sorry about that. So here's something you could try at home, demonstrating Charles' Law. Take a balloon, don't fill it up super much, but you know, kind of fill it up. If you put the balloon into boiling water, heat it up, the balloon will grow in size. it'll get bigger. And if you take that same balloon and put it in ice water and make it cold, it'll shrink and shrivel down. And you can go back and forth, back and forth, until the balloon breaks or you get bored. All right? So what's happening? Well, when we increase the temperature of the gas, we're increasing the kinetic energy of the gas particles. The gas particles are going to be moving faster. When they move faster, they're going to bump into the surface with more force and more frequently. The trains aren't bad enough. Is that a power drill or something? It's an electric blower. He's cleaning this outside sidewalk with the blower. He's blowing all the dust down there. So now that I know what it is, I can be okay with it. Before it was... I mean it's still a little distracting, but what was I talking about? Gas particles, when they're moving faster, in order for them to stay at the same pressure, they're going to cause the volume of the gas to expand. When we cool them down, they slow down, and the balloon is, because of the external pressure, is going to be at the same pressure, and so the balloon is going to shrink. When the temperature increases, the gas particles move faster. The collisions are more frequent. The force exerted with each collision is greater. The only way for the pressure to be constant is for the gas to occupy a larger volume, so the collisions become less frequent and occur over a larger area. This is a Dewar flask with liquid nitrogen in it, and if you put that balloon into liquid nitrogen, it's much, much colder and it shrinks and gets really, really small. This is a hot air balloon. I remember being fascinated by the... these as a child. I mean, balloons are just fun, right? And you can blow a balloon up by just blowing into it. And that balloon is, you can bat it around, but it's just going to fall to the ground, right? But then like, if you go to the fair and you buy a balloon, right? Right? It's got helium in it and it floats. It's like, oh, this is really cool. And then you see hot air balloons. And not only is this balloon full of air, which regular balloons, that sinks, but this balloon has a ginormous hole in the bottom. If you have a hole in your regular balloon, is it going to stay inflated? No, it goes flat. So what's going on here? And how does this thing rise off the ground? Well, the secret is this ginormous burner here. So they light the burner, which heats up the air. Right? So the air starts to explode. expand and starts to fill up the balloon and it's going to bring more air in because the pressure is getting less and so the balloon will expand. Now as we keep heating that air, The volume expands more and more, and the extra air that doesn't fit in the balloon will come out, but the air inside now is much less dense than the air outside. And so just like a balloon... full of air will float in water, a balloon full of this hot air, less dense air, will float in regular air. And so you can control up and down by how hot you make the air. It doesn't have any directional steering, so you're kind of at the mercy of the winds, but you can make it go up and down. Anybody have any questions about that? It's the same idea with the little paper lanterns with a candle inside. How do those things float? Well, the candle warms up the air enough to cause it to expand, and the density of it then is reduced, and they will float, lift up into the air. So let's do a Charles Law problem. A gas in a cylinder with a movable piston has an initial volume of 88.2 milliliters. If we heat the gas from 35 degrees Celsius to 155 degrees Celsius, what is its final volume in milliliters? So we end up with a lot of numbers in these problems. Typically we're given at least three numbers and asked to find a fourth. So I find it useful to make a little table to keep track of all of these things because we're going to end up with a condition one and a condition two. So here it says it has an initial volume of 88.2 milliliters. So I'm going to make my table, and I'm going to put that one down first because it came first in the problem. So that's initial. That's the first one. And this is a measurement of what? Volume, right? So I'm going to call this column V for volume. And then it says if we heat the gas from 35 to 155. So this is the initial volume. This is the initial temperature. So 35 degrees Celsius. This is the temperature. But what did I say about the temperature? It has to be in kelvins. So before we go any further and get caught up in our calculations, let's convert it to kelvins right away. So I convert to kelvins by adding 273, and I get 308. It goes from 35 to 155, so my second temperature is 155 degrees Celsius. 155 plus 273. This is the sort of math that, you know, you certainly could do in your... ...head, but then sometimes we make silly mistakes and then get the whole problem wrong. So just use the calculator. So here I've got my little table and I have one blank. And this is asking what's the final volume and that matches up with my blank. I'm going to call this V2. So this is v1 and t1. v2 is unknown, and here's t2. Any questions about that? So I'm going to use my Charles Law equation. v1 over t1 is equal to v2 over t2. There's a reason that it's usually given like this, but it's kind of a pain because this equation has fractions in it. And my experience is that students end up doing stupid things when they rearrange equations with fractions. So let's get rid of the fractions and then rearrange it so we can avoid doing something dumb. So hopefully you remember the idea of cross multiplying. So cross multiplication says I can take the top of one side and multiply it by the bottom of the other side. And then I can do the opposite. I'm not sure why that ended up in an arrow word, but whatever. These two multiply, and those will be equal to each other. So I can cross multiply, and get v1 times t2 is equal to t1 times v2. v1 times t2, over here, equals t1 times v2. I got rid of the fractions, and so now I'm more likely to rearrange correctly. I want to find v2. So this is v2. I want to rearrange my equation so that's all by itself. So I'm going to divide by T1, so T1's cancel out. Whatever I do to the right side of the equation, I have to do the same thing to the left side. So volume 2 is equal to volume 1 times temperature 2, divided by temperature 1. I have these values all labeled over here, and so I'm going to put them in. Be careful because it's easy to get them mixed up. Volume 1 is 88.2 milliliters. Temperature 2 is down here is 428 Kelvin. And then this is divided by temperature 1 which is 308 Kelvin. The Kelvin units cancel out, and I'm going to be left with a unit of milliliters. I use my calculator, 88.2 times 428, divided by 308. 122.56, and it's going on and on and on and on, and I don't want to do that, and so I ask myself, well, how many sig figs should this answer have? It should have three. This temperature only has two significant figures. But when I convert to Kelvin, I'm adding. This was to the nearest one's place. This is still to the nearest one's place. And so now it also has three significant figures. So they all have three sig figs. So this should have three sig figs. And so we're going to call this 123 milliliters. With the simple gas laws, you can think about balloons and ask yourself if your answer makes sense. I took this gas and I heated it up. Do I expect it to get bigger or smaller? I expect it to get larger, right? And this number is larger than the initial volume, so that's encouraging. It doesn't prove anything, but it's a good sign. Any questions? So, conceptual question. The pressure exerted on a sample of a fixed amount of gas is doubled at constant temperature, and then the temperature of the gas in kelvins is doubled at constant pressure. What's the final volume of the gas? Students often have a hard time with questions like this because we develop skills of being able to calculate stuff, but then this is not giving us actual numbers to do a calculation with, it's making us think about it. So let's look at the two things that are happening here, one after the other. So we've got a fixed amount of gas, so the N, the moles of gas, is the same. The pressure is doubled. If I double the pressure, what does that do to the volume? Causes it to be cut in half, right? This is asking about the volume. So the initial volume, when I double the pressure, pressure times 2, now I've got half the volume. Because when the pressure goes up, the volume goes down. Then it says the temperature in kelvins is doubled. Well, when I increase the temperature, does the gas expand or contract? It expands. So here, if I've got the temperature times 2, that will multiply the volume times 2. And so I've got 2 times that 1 half volume, and I'm back to my original volume. Doubling the pressure made the volume decrease by half. Doubling the temperature made it increase by half again back to the original. So what's the correct answer? E. The final volume of the gas is the same as the initial volume. Any questions? Another simple gas law, Avogadro's Law. The name might be familiar, because of Avogadro's number. Amadeo Avogadro, 17 and 1800s. He was the first one to formally state the relationship between the volume of a gas and the amount of gas in moles. So the volume of a gas is proportional to the number of gas molecules. It's not proportional to their mass. It's proportional to how many there are. So Avogadro's law is that the volume is proportional to the amount of gas in moles at constant temperature and pressure. And so that can be expressed like this, V1 over N1 equals V2 over N2, which looks a little bit like Charles' law with the temperature. Instead of counting individual gas molecules, we count them using moles, the chemist's dozen of molecules. So, equal volumes of gases contain equal numbers of molecules. The identity of the gas doesn't matter. This is weird. This is not true with liquids or solids. Here's a graph of Avogadro's law, the volume of a gas, as we increase the amount of gas in moles. And this makes sense. If I add more gas to the container, the volume should get larger, right? I'm keeping the pressure and the temperature the same. The volume of the gas increases in directly proportional to the number of gas particles. And if we take this line and extrapolate it to zero particles, zero moles, we get zero volume, which makes sense. If there's no gas, it would occupy zero volume. Another conceptual problem. Which action causes the volume of a sample to increase? Decreasing the pressure. Is that going to cause the volume to increase or decrease? That's going to cause the volume to increase. So that sounds like a possible answer here. Decreasing the temperature. Is that going to make the volume increase? No. That's going to make the volume decrease. Decrease. Decreasing the number of moles of gas. So, think about a balloon. If I take air out, does that make the balloon bigger? No, it makes it smaller, right? Well, we found that A was true, so none of the above isn't good. It must be decreasing the pressure. Any questions? Avogadro's Law. A chemical reaction occurring in a cylinder equipped with a movable piston produces 0.621 moles of a gaseous product. If the cylinder contained 0.120 moles of gas before the reaction and had an initial volume of 2.18 liters, what was its volume after the reaction? Assume constant pressure and temperature and that the initial amount of gas completely reacts. So, this problem is nicely labeled, Avogadro's Law, but the problems on the exam won't be labeled like that for you. We should always read through the whole question, and then I'm seeing that I have something that's changing, right? And I've got these numbers, I've got moles and moles and a volume, and they're asking me about another volume. I'm going to make a table and organize these numbers to decide what I should do with them. So I'll have row 1 and row 2. So this produces 0.621 moles of gas. Well, I'm just going to write it here, 0.621 moles of gas. And what is the variable we use to represent moles of gas? If the cylinder contained 0.12 moles before the reaction, so this is another amount in moles, that needs to go in this same column. Now, is anybody bothered by this being the first thing and labeled 2? Somebody has to be bothered by that. Well, how about if I just relabel my table? Let's call this 2 and call that 1, and now we feel a little better. This is the initial thing, and that's the second thing. We maybe didn't anticipate that when we first started writing things down, but we can kind of fix it. It had an initial volume of 2.18 liters. So this is the initial volume, goes with the initial moles of gas. 2.18 liters. And that's volume. And then the question here, what was its volume after the reaction? So that's V2. Everybody okay with that? We have to identify what the numbers are and label them. Exactly what you label them doesn't matter, but it's important to get them paired up correctly, right? So I have n's and v's and 1's and 2's, and so I need the equation v1 over n1 equals v2 over n2. And again, we've got an equation here with a fraction. So I'm going to cross multiply. And get rid of the fraction before I rearrange the equation. V1, N2 equals N1, V2. I know some of you can just look at that and rearrange it. Good for you. I'm talking to... the rest of the class, which is most of you. So I got rid of my fractions. Here's my equation. I'm trying to solve for v2. So I'm trying to get v2 by itself. So v2 is multiplied by n1, so I need to divide by n1 to make it go away. And I need to divide the other side by n1. So V2 equals V1 times N2 divided by N1. So you should always show your work when doing problems like this. Do you have to show how you rearrange the equation? No, you don't. If you can do that in your head, you don't need to show that. You don't necessarily have to write out this table. either, but I do want you to write down the rearranged equation, and then write it down with the numbers in it. So volume 1 is 2.18 liters with their units, and that's multiplied by N2, which is 0.621 moles. Divided by N1, which is 0.120 moles. The moles cancel out. I have 2.18 times 0.621 divided by 0.12. And how many sig figs should this answer have? Three. 11.281 liters, so I'm going to round this to 11.3 liters of gas. So that part right here, that's showing your work. You need to learn how to show your work. I know that some of you can do this and you like to do it. You put the numbers in and you get the right answer. Why do I have to show my work? Well, because sometimes we need to write stuff down. so that someone else later can look at what we did and understand, even though we're not there to explain it to them. The other benefit of doing this is if your answer is wrong, you can go back and very clearly see what you did. You can find the mistake and fix it without starting over. If you're doing an exam, you're very concerned about the answer being correct, but you don't know if it's correct, you can look back over this, and you can, you know, work through it in your head again and confirm. Does that work out? Does that look correct? Or you may realize, oh shoot, I, you know, this is N1, and I actually wrote it in for N2. And you can fix your mistakes. Any questions? 11.3 liters. Here's another example. The scale model of a blimp rises when it's filled with helium to a volume of 55 cubic decimeters. When 1.1 moles of helium is added to the blimp, the volume is 26.2 cubic decimeters. How many more grams of helium must be added to make it rise? Assume constant temperature and pressure. So this, like that first one we did, is a Chem 1A problem. The other couple that we've done, the simpler ones, are Chem 3A problems because you covered these things in your prerequisite class. So hopefully most of this is review. But then applying it to a problem like this, this requires a little bit of critical thinking. And again, we might want to draw pictures here. So a scale model of a blimp. I'm going to make my blimp red. They're pretty, right? Terrible. It's going to float when it has a volume of 55 cubic decimeters. That's a weird unit. It's okay. So that's floating. It says when 1.1 mole of helium is added to the lamp, the volume is 26.2. This is 55. If the volume is 26.2, is it going to be floating? No. It's going to be laying on the floor, kind of all squishy, right? So here its volume is 26.2 cubic decimeters. We're starting to fill it up, but we don't have enough helium. helium in it yet. Down here, the amount of helium in there is 1.10 moles. It's asking how many more grams of helium must be added to make it rise. So what we're looking for is from here to here, how many grams of helium do we have to put in? It's kind of funny to think about. The balloon is on the ground, and by putting more stuff into it, you can make it go up. Yeah, that doesn't happen with liquids and solids. This is a volume, right? And that's a volume. And this is an amount of gas in moles. Which one should we call 1? This down here, the beginning. So this is volume 1. And this is the amount of gas in moles at that condition. When we get it to float... This is volume two. Now the gas law is not going to tell us directly how much gas we need to add in grams. What can we calculate using Avogadro's law? The number of moles, right? We have V1 over N1 equals V2 over N2. We have V1 and V2 and we have N1. We can calculate what N2 is. So let's cross multiply this to get rid of the fractions. V1 N2 is equal to N1 V2. I'm solving for N2. So I'm going to divide by V1. Divide by V1. So there's my rearranged equation, and I'm going to put my numbers in here. N1 is 1.10 moles. Yeah. Moles. Volume 2 is 55.0 decimeters cubed. I screwed up, huh? I go to fill this in and I'm like, V2, wait a minute, I've got V2 down here. What happened? I copied it down wrong. It's supposed to be V1 in the bottom. So let's fix that. V1 is this guy, 26. 0.2 cubic decimeters. I'd like to be able to say I did that on purpose to illustrate something, but it was just screwed up. The cubic decimeters, which we were a little uncomfortable with, cancel out and go away. Yay. So if we calculate here, we get 1.1 times 55 divided by 26.2. I did that wrong. I ended up with negative 0.999, which is just not even possible. I pressed minus instead of times. 2.309. That's better. So N2 equals 2.309 moles. That would have three significant figures. 3091. Let me keep two extras. So if I start with 1.1 moles, and I need to have 2.3091 moles, how many moles do I have to add? The difference, right? I want to end up with 2.3091 moles. I already have 1.10 moles. Need to add 1.2091 moles of helium. This is an H. Looked like an N there for a while. Ooh, worked that time. Can I figure out how many grams that is? Sure. 1.2091 moles of helium times grams divided by moles is 4.003 grams of helium per mole. The moles will cancel out. And so my answer should be 4.84 grams of helium is how much I need to add. Any questions? This is kind of a tricky problem because it's not just looking at initial amount and final amount, it's looking at the difference. This is a very common tricky problem.

These properties are interrelated. If you choose change one, the others will change as well. And so the simple gas laws are just looking at pairs of properties.

Might look at the pressure and the volume, or the pressure and the temperature, or the volume and the amount of moles. Not looking at all four of them at once, because that's kind of complicated. We'll keep it simple and just look at two. And these laws were deduced from observations where they held two of these properties constantly. changed one and looked at how the other one changed.

And these laws date back a long time ago. So here's Boyle's Law, relating volume and pressure. Robert Boyle lived in the 1600s, right? And so...

This is a representation of the type of experiment he did. He used a J-tube, a J-shaped tube with mercury in it. And in the closed end, he put a gas sample. And then you can measure the volume of the gas. You could calculate that based on the diameter of the tube and how tall it was.

And you can look at the pressure between the outside and the inside, the pressure difference. is the height of the mercury columns, right? Just like with a manometer. So he found that if he poured more mercury in here, then that would cause the gas to squish together and the volume of the gas got smaller. The pressure of the gas increases.

We see that the difference in heights here is larger. So the pressure goes up and the volume goes down. A lot of these relationships, if we think about balloons or other real-life experiences with gases, they make sense, right? So if you imagine taking a balloon and squishing it on all sides, putting pressure on it, it's going to get smaller, right?

So this is an inverse relationship. When one thing gets smaller, the other thing gets bigger. This is an example of Boyle's Law graft. So here's the volume of the gas and the pressure. So as the pressure increases, the volume of the gas decreases.

So that's, you know, kind of a pretty graph, but what's a lot more useful is if we can have a graph that gives us a straight line instead of curves. Because, you know, this was before they discovered electricity, so. They didn't have computers that did linear regression for them.

So drawing straight lines is just so much easier. Now we can have a computer that will take this data and fit a quadratic or an exponential function to it and tell us what the equation is. But back in the 1600s, they didn't have stuff like that. So Boyle's Law says that the volume of a gas is proportional to 1 over the pressure.

And this is when we keep the temperature and the moles of gas constant. We're keeping those the same because if we change those, it's going to mess everything up. So we say the pressure of a gas is inversely proportional to its volume. We saw that the graph of volume versus pressure is curved. If we graph volume versus 1 over pressure, the inverse of pressure, we get a straight line.

So this is volume versus pressure. This is volume versus one over pressure, and now we get a beautiful straight line. This shows me that volume is directly proportional to one over the pressure. That can be described with a line equation.

So from this proportionality, we realize that pressure times volume is... equal to a constant. Now we don't care what that number is, but it's going to be constant at a given temperature and amount of moles. And so when we change the pressure and the volume changes, or vice versa, we can say that this relationship is true. And this is...

how we generally express Boyle's Law, the pressure 1 times the volume 1 is equal to pressure 2 times volume 2. Now, I am not going to quiz you on whether you remember that that is called Boyle's Law. As I've mentioned before, I'm not teaching chemical history. I think it's good to include that. on the side, but I'm not going to require you to pick Boyle's law out of a list of equations. I personally have a hard time remembering which one belonged to Boyle's and which one belonged to Charles, and does it actually matter?

No, it doesn't. So if we look at what gas molecules are doing, so here we have a container where we can change the volume. It's a bit like a syringe or a piston.

The lid here can slide up and down. So we can change the volume of the container, and we'll see that we talk about this a lot with gases. So here the lid has a one kilogram weight on it, and this lid does not just fall to the bottom because there's a gas in here, and the gas is holding it up.

But there's a pressure. So the pressure here is measured with a more modern pressure gauge, and we're told it's one atmosphere. The volume of this gas is one liter. Over here now we've increased the mass sitting on the lid, which causes the lid to move down. So now the volume here is less than it was.

It's half a liter. So we've gone from one liter to half a liter. So we've divided the volume by two. And what happens to the pressure? It increases by a factor of two.

P1V1 equals P2V2. Pressure 1 was 1 atmosphere. Volume 1 was 1 liter. Pressure 2 is 2 atmospheres. And volume 2 is half of a liter.

And this is equal. It does not matter what kind of gas it is. Is that the same as the one we're using? Yep. Yeah, we'll talk about that later, but one of the strange things about gases is you can, the behavior of gases is almost identical regardless of what the gas is.

What the particles are doesn't matter, it's just how many particles there are. So Boyle's Law has implications in scuba diving. So if you've ever dived down to the deep end of a swimming pool, you know that you start to feel some pressure in your head when you get way down to the bottom, right?

Because as you go below the surface of the water, the water Water is putting pressure on you. Water is more dense than air, obviously, and so the pressure will increase much more rapidly underwater than it does with an elevation change above water. So this is a general... relationship for every 10 meters of depth the pressure increases about one atmosphere due to the weight of the water.

So here we have a diver up here at the surface depth of zero meters the pressure experienced is one atmosphere which is just normal atmospheric pressure. Come down here to a depth of 20 meters you go down 10 meters you add one atmosphere of pressure. you go down another 10 meters you've added two atmospheres of pressure so starting with one up here adding two we get a pressure of three atmospheres that's a lot of pressure So if you are breathing 20 meters below, your lungs are expanding against three atmospheres of pressure. The only way you can breathe down there is with pressurized gas. So the pressurized gas needs to match the outside pressure fairly closely so that you can expand your lungs.

So down there you take a breath and... You, you know, big lung full of air, and you hold it. The pressure down there is three meters, and then you come up to the surface very quickly, holding your breath.

Now the external pressure is one atmosphere. The internal pressure in your lungs was three atmospheres, and so it's going to make your lungs expand a lot. But I don't think you'd actually be able to hold your breath.

It would hurt too much, and you'd let it go. before your lungs exploded, kind of a self-preservation thing. So the volume of the air, Boyle's Law predicts, will expand.

So the pressure decreases by a factor of 3. The volume would expand by a factor of 3. That would not be good for your general health. Always exhale when rising from a dive. I remember thinking, you know, scuba diving, I don't know if you know this, in order to go scuba diving, you have to take the lessons and get certified and stuff.

And I remember thinking, well, you go swimming, you don't need to figure it out, right? Why is it so different to go scuba diving? Well, there's a lot of things like this that are complicated and potentially fatal. And so there's a lot of stuff about gases that's actually really chemistry and physics that you have to understand in order to scuba dive safely. Personally, I'm not interested in going that far under the surface of the water.

I don't like caves either. So let's do this problem. A snorkeler takes a syringe filled with 16 milliliters of air from the surface, where the pressure is 1 atmosphere, to an unknown depth. The volume of the air in the syringe at this depth is 7.5 milliliters. What's the pressure at this depth?

If the pressure increases by 1 atmosphere for every additional 10 meters of depth, how deep is the snorkeler? I'm not sure why the author gives this as the first simple gas law problem. This one's challenging.

But we can figure this out. So, this is a problem that we have not seen anything like before. And it seems a little, um, just unusual.

I mean, who takes syringes underwater, right? Syringes of air. That's very strange. strange. So what we have to do is we have to take these words and get comprehension out of them, which can be challenging sometimes.

So I find it very helpful to draw pictures. My pictures are really ugly, but they still help. And then when you get better at solving problems like this, you don't actually have to draw the picture anymore.

You can picture it in your mind. You've got a whiteboard or a three-dimensional simulation thing in your head. It's so much better than drawing it. But we'll just settle for drawing it. So the snorkeler, I'm not even going to attempt to draw the snorkeler, but here's the surface of the water, right?

A syringe filled with 16 milliliters of air from the surface. Okay, so here we are at the surface. Here is the syringe filled with air, and this must be a syringe that's closed on the end. So there's the plunger. So this is the volume in here, that's 16. milliliters of air.

And the diver goes down. So here's our Our syringe again, and it says now the volume is only 7.5 milliliters. So the plunger comes in, and this volume is 7.5 milliliters.

Why did the gas get smaller? Because as you go down in the water, the pressure gets bigger, right? What's the pressure up here?

The surface, one atmosphere. So one atmosphere up here. What's the pressure down here? Well, that's the question.

What is the pressure at this depth, right? So that's what we're trying to figure out. So I have two different pressures. I have two different volumes. This should make me think of Boyle's Law, which relates pressure to volume.

Now, it really doesn't matter whether you call this pressure 1 and volume 1, or if you call that 2. It doesn't matter. Generally, people like to call this one 1, because that one came first. So we'll just do that. So I'm going to call this, this is pressure 1, and the other pressure is pressure 2. This is volume 1. Because this is the volume at this pressure, these two go together.

They happen at the same time. And this is volume 2. Okay? So we have P1, V1 is equal to P2, V2.

And we are trying to find pressure 2. Yes? So we should rearrange the equation before we put the numbers in. So I want to get P2 all by itself on one side. I don't care which side of the equal sign, but on one side.

And here it's multiplied by V2, so if I divide by V2 on this side, V2 over V2 cancels out. And then to keep it equal, I need to do the same thing to the other side. So P2 is equal to P1 times B1 divided by V2.

This sort of algebra is really common and important in Chem 1A, and so if you are algebra challenged and have a hard time rearranging equations like this, please come and talk to me. can explain it much better one-on-one and I can go into all sorts of crazy analogies and I'll keep at it as long as you will until it clicks and you go oh I see now it's so easy so once you see it it is but until then not so much So I rearrange the equation to solve for the thing I want. And then I'm going to take these other pieces and put them into the equation. You need to be careful here because if you get your ones and twos, if you don't copy it down correctly, now your equation is wrong and your answer is wrong.

So some people have harder time tracking and copying things like that. Just be careful. So pressure 1 is 1 atmosphere, and volume 1 is 16. milliliters. Always, always, always put the units in there. And this is divided by volume 2, which is 7.5 milliliters.

What happens with the units? They cancel. Milliliters divided by milliliters cancels. Unit cancels out.

I'm left with atmosphere, which is a pressure unit, and I'm calculating pressure, so that's a good thing. So now I have 1 times 16 divided by 7.5. Pressure 2 is equal to 2.133 repeating atmospheres.

How many sig figs would that number have? This one atmosphere can be a little confusing. One atmosphere is defined as the pressure at sea level. And so we can consider that to be an exact number.

That's a little iffy, but that's what we do. This has two sig figs, this has two sig figs, so I'm just going to keep two sig figs here. 2.1 So the answer to the first question, what's the pressure?

The pressure at that depth is 2.1 atmospheres. Any questions yet? That's this question.

If the pressure increases by one atmosphere for every additional 10 meters of depth, how deep is the snorkeler? This one also requires a little bit of problem solving. So this is a relationship, one atmosphere for every additional 10 meters. So one atmosphere for each 10 meters.

That's the increase. We aren't starting at zero, we're starting at one atmosphere. How much did the pressure increase?

I'm going to fill this in now because we know what it is. Yeah, it increased by 1.1. So the difference between these two is 1.1 atmospheres. That's how much it increased. This is telling me that it increases one atmosphere for every 10 meters.

I'm going to change colors here so it stands out a little better. So we had an increase of 1.1 atmosphere, and that relationship tells me that I'm going down 10 meters for every one atmosphere. And so the snorkeler must be 11 meters under the surface of the water. Any questions?

Why wouldn't you use the 2.1? That's an excellent question. Why wouldn't I use the 2.1? If I was starting at zero pressure here, I would use the 2.1.

But this is a one atmosphere increase in pressure for every 10 meters. And so I have to consider where I'm starting from. Does that make sense?

Anybody else? Yes, so I had a mental debate while I was talking to you and I decided to do it this way, but thank you for bringing that up. So here I had my result and I wrote down extra digits like I always do. always tell you to do, and then I rounded my answer.

Over here, we really should use the unrounded value when we do the calculation, and so the difference in pressure is 2.133 repeating minus one atmosphere, which is 1.133 repeating atmospheres. And then when I put that in here, my answer comes out to be 11.33 meters. But when I think about the significant figures, going with two significant figures, it ends up getting rounded to the same place. There are times where if you round your intermediate values, your final answer is going to be different.

Here they end up the same. Any other questions? Jay, is he Charles?

1700s, early 1800s, looked at the relationship between the volume and the temperature of a gas and found that the volume of a gas is directly proportional to its absolute temperature. So Charles' Law can be stated as the volume is proportional to the temperature at constant pressure and constant amount of gas. You can say that the volume divided by the temperature is equal to a constant.

And so Charles' Law is often expressed like this, V1 over T1 equals V2 over T2. The temperature must always be expressed in kelvins. The volume could be expressed in any volume unit as long as the two volumes are the same. The problem with temperature is that all the temperature scales are related with adding or subtracting.

It's not just multiplying or dividing. And so the conversions with temperatures do not cancel out. The other problem that arises is on the Fahrenheit and the Celsius scale, you can have a temperature of zero or you can have a negative temperature. If you try to plug zero into this equation, what happens? It's undefined.

It's not going to work, right? So the Kelvin scale, we don't ever have zero. If you use negatives here, you can also end up with negative volumes, which are ridiculous. So here's a Charles Law graph.

So here's the volume of a gas at different temperatures. This is one mole of gas, this is half a mole of gas, and this is a quarter of a mole of gas. And we see that for each of them, the volume increases linearly with the temperature. So the data is going down here to about zero Celsius. We don't have data down here, but we can extrapolate.

This is a nice line, and so we can draw the line past the data, extrapolating, and we find that all of these lines intercept at the same place, and that's called absolute zero. It happens to be minus 273.15 degrees Celsius, which should be familiar from when we learned about converting temperatures. The idea here is as you decrease the temperature of the gas, its volume gets smaller and smaller and smaller. If you keep decreasing it, eventually...

Maybe you get to zero. Could you go past zero to negative volume? No.

So if we got to zero volume, that would have to stop. All of these things intercept here at the same point, and that's not a coincidence. So this is the absolute theoretical lowest possible temperature. And negative temperatures are not possible.

We need to use kelvins because on the Kelvin scale, the intercept of the line is zero, whereas for the other ones it will not be zero. One of my less good explanations. Sorry about that. So here's something you could try at home, demonstrating Charles' Law.

Take a balloon, don't fill it up super much, but you know, kind of fill it up. If you put the balloon into boiling water, heat it up, the balloon will grow in size. it'll get bigger.

And if you take that same balloon and put it in ice water and make it cold, it'll shrink and shrivel down. And you can go back and forth, back and forth, until the balloon breaks or you get bored. All right?

So what's happening? Well, when we increase the temperature of the gas, we're increasing the kinetic energy of the gas particles. The gas particles are going to be moving faster.

When they move faster, they're going to bump into the surface with more force and more frequently. The trains aren't bad enough. Is that a power drill or something? It's an electric blower. He's cleaning this outside sidewalk with the blower.

He's blowing all the dust down there. So now that I know what it is, I can be okay with it. Before it was...

I mean it's still a little distracting, but what was I talking about? Gas particles, when they're moving faster, in order for them to stay at the same pressure, they're going to cause the volume of the gas to expand. When we cool them down, they slow down, and the balloon is, because of the external pressure, is going to be at the same pressure, and so the balloon is going to shrink.

When the temperature increases, the gas particles move faster. The collisions are more frequent. The force exerted with each collision is greater. The only way for the pressure to be constant is for the gas to occupy a larger volume, so the collisions become less frequent and occur over a larger area.

This is a Dewar flask with liquid nitrogen in it, and if you put that balloon into liquid nitrogen, it's much, much colder and it shrinks and gets really, really small. This is a hot air balloon. I remember being fascinated by the...

these as a child. I mean, balloons are just fun, right? And you can blow a balloon up by just blowing into it.

And that balloon is, you can bat it around, but it's just going to fall to the ground, right? But then like, if you go to the fair and you buy a balloon, right? Right?

It's got helium in it and it floats. It's like, oh, this is really cool. And then you see hot air balloons. And not only is this balloon full of air, which regular balloons, that sinks, but this balloon has a ginormous hole in the bottom.

If you have a hole in your regular balloon, is it going to stay inflated? No, it goes flat. So what's going on here?

And how does this thing rise off the ground? Well, the secret is this ginormous burner here. So they light the burner, which heats up the air. Right? So the air starts to explode.

expand and starts to fill up the balloon and it's going to bring more air in because the pressure is getting less and so the balloon will expand. Now as we keep heating that air, The volume expands more and more, and the extra air that doesn't fit in the balloon will come out, but the air inside now is much less dense than the air outside. And so just like a balloon... full of air will float in water, a balloon full of this hot air, less dense air, will float in regular air. And so you can control up and down by how hot you make the air.

It doesn't have any directional steering, so you're kind of at the mercy of the winds, but you can make it go up and down. Anybody have any questions about that? It's the same idea with the little paper lanterns with a candle inside. How do those things float? Well, the candle warms up the air enough to cause it to expand, and the density of it then is reduced, and they will float, lift up into the air.

So let's do a Charles Law problem. A gas in a cylinder with a movable piston has an initial volume of 88.2 milliliters. If we heat the gas from 35 degrees Celsius to 155 degrees Celsius, what is its final volume in milliliters?

So we end up with a lot of numbers in these problems. Typically we're given at least three numbers and asked to find a fourth. So I find it useful to make a little table to keep track of all of these things because we're going to end up with a condition one and a condition two. So here it says it has an initial volume of 88.2 milliliters.

So I'm going to make my table, and I'm going to put that one down first because it came first in the problem. So that's initial. That's the first one. And this is a measurement of what?

Volume, right? So I'm going to call this column V for volume. And then it says if we heat the gas from 35 to 155. So this is the initial volume.

This is the initial temperature. So 35 degrees Celsius. This is the temperature. But what did I say about the temperature? It has to be in kelvins.

So before we go any further and get caught up in our calculations, let's convert it to kelvins right away. So I convert to kelvins by adding 273, and I get 308. It goes from 35 to 155, so my second temperature is 155 degrees Celsius. 155 plus 273. This is the sort of math that, you know, you certainly could do in your... ...head, but then sometimes we make silly mistakes and then get the whole problem wrong.

So just use the calculator. So here I've got my little table and I have one blank. And this is asking what's the final volume and that matches up with my blank. I'm going to call this V2.

So this is v1 and t1. v2 is unknown, and here's t2. Any questions about that?

So I'm going to use my Charles Law equation. v1 over t1 is equal to v2 over t2. There's a reason that it's usually given like this, but it's kind of a pain because this equation has fractions in it.

And my experience is that students end up doing stupid things when they rearrange equations with fractions. So let's get rid of the fractions and then rearrange it so we can avoid doing something dumb. So hopefully you remember the idea of cross multiplying. So cross multiplication says I can take the top of one side and multiply it by the bottom of the other side.

And then I can do the opposite. I'm not sure why that ended up in an arrow word, but whatever. These two multiply, and those will be equal to each other. So I can cross multiply, and get v1 times t2 is equal to t1 times v2.

v1 times t2, over here, equals t1 times v2. I got rid of the fractions, and so now I'm more likely to rearrange correctly. I want to find v2. So this is v2.

I want to rearrange my equation so that's all by itself. So I'm going to divide by T1, so T1's cancel out. Whatever I do to the right side of the equation, I have to do the same thing to the left side.

So volume 2 is equal to volume 1 times temperature 2, divided by temperature 1. I have these values all labeled over here, and so I'm going to put them in. Be careful because it's easy to get them mixed up. Volume 1 is 88.2 milliliters. Temperature 2 is down here is 428 Kelvin.

And then this is divided by temperature 1 which is 308 Kelvin. The Kelvin units cancel out, and I'm going to be left with a unit of milliliters. I use my calculator, 88.2 times 428, divided by 308. 122.56, and it's going on and on and on and on, and I don't want to do that, and so I ask myself, well, how many sig figs should this answer have?

It should have three. This temperature only has two significant figures. But when I convert to Kelvin, I'm adding. This was to the nearest one's place. This is still to the nearest one's place.

And so now it also has three significant figures. So they all have three sig figs. So this should have three sig figs. And so we're going to call this 123 milliliters.

With the simple gas laws, you can think about balloons and ask yourself if your answer makes sense. I took this gas and I heated it up. Do I expect it to get bigger or smaller? I expect it to get larger, right?

And this number is larger than the initial volume, so that's encouraging. It doesn't prove anything, but it's a good sign. Any questions?

So, conceptual question. The pressure exerted on a sample of a fixed amount of gas is doubled at constant temperature, and then the temperature of the gas in kelvins is doubled at constant pressure. What's the final volume of the gas?

Students often have a hard time with questions like this because we develop skills of being able to calculate stuff, but then this is not giving us actual numbers to do a calculation with, it's making us think about it. So let's look at the two things that are happening here, one after the other. So we've got a fixed amount of gas, so the N, the moles of gas, is the same. The pressure is doubled. If I double the pressure, what does that do to the volume?

Causes it to be cut in half, right? This is asking about the volume. So the initial volume, when I double the pressure, pressure times 2, now I've got half the volume. Because when the pressure goes up, the volume goes down. Then it says the temperature in kelvins is doubled.

Well, when I increase the temperature, does the gas expand or contract? It expands. So here, if I've got the temperature times 2, that will multiply the volume times 2. And so I've got 2 times that 1 half volume, and I'm back to my original volume. Doubling the pressure made the volume decrease by half.

Doubling the temperature made it increase by half again back to the original. So what's the correct answer? E.

The final volume of the gas is the same as the initial volume. Any questions? Another simple gas law, Avogadro's Law. The name might be familiar, because of Avogadro's number. Amadeo Avogadro, 17 and 1800s.

He was the first one to formally state the relationship between the volume of a gas and the amount of gas in moles. So the volume of a gas is proportional to the number of gas molecules. It's not proportional to their mass.

It's proportional to how many there are. So Avogadro's law is that the volume is proportional to the amount of gas in moles at constant temperature and pressure. And so that can be expressed like this, V1 over N1 equals V2 over N2, which looks a little bit like Charles' law with the temperature. Instead of counting individual gas molecules, we count them using moles, the chemist's dozen of molecules. So, equal volumes of gases contain equal numbers of molecules.

The identity of the gas doesn't matter. This is weird. This is not true with liquids or solids. Here's a graph of Avogadro's law, the volume of a gas, as we increase the amount of gas in moles. And this makes sense.

If I add more gas to the container, the volume should get larger, right? I'm keeping the pressure and the temperature the same. The volume of the gas increases in directly proportional to the number of gas particles.

And if we take this line and extrapolate it to zero particles, zero moles, we get zero volume, which makes sense. If there's no gas, it would occupy zero volume. Another conceptual problem. Which action causes the volume of a sample to increase? Decreasing the pressure.

Is that going to cause the volume to increase or decrease? That's going to cause the volume to increase. So that sounds like a possible answer here. Decreasing the temperature.

Is that going to make the volume increase? No. That's going to make the volume decrease. Decrease. Decreasing the number of moles of gas.

So, think about a balloon. If I take air out, does that make the balloon bigger? No, it makes it smaller, right?

Well, we found that A was true, so none of the above isn't good. It must be decreasing the pressure. Any questions? Avogadro's Law. A chemical reaction occurring in a cylinder equipped with a movable piston produces 0.621 moles of a gaseous product.

If the cylinder contained 0.120 moles of gas before the reaction and had an initial volume of 2.18 liters, what was its volume after the reaction? Assume constant pressure and temperature and that the initial amount of gas completely reacts. So, this problem is nicely labeled, Avogadro's Law, but the problems on the exam won't be labeled like that for you.

We should always read through the whole question, and then I'm seeing that I have something that's changing, right? And I've got these numbers, I've got moles and moles and a volume, and they're asking me about another volume. I'm going to make a table and organize these numbers to decide what I should do with them. So I'll have row 1 and row 2. So this produces 0.621 moles of gas.

Well, I'm just going to write it here, 0.621 moles of gas. And what is the variable we use to represent moles of gas? If the cylinder contained 0.12 moles before the reaction, so this is another amount in moles, that needs to go in this same column.

Now, is anybody bothered by this being the first thing and labeled 2? Somebody has to be bothered by that. Well, how about if I just relabel my table?

Let's call this 2 and call that 1, and now we feel a little better. This is the initial thing, and that's the second thing. We maybe didn't anticipate that when we first started writing things down, but we can kind of fix it.

It had an initial volume of 2.18 liters. So this is the initial volume, goes with the initial moles of gas. 2.18 liters.

And that's volume. And then the question here, what was its volume after the reaction? So that's V2. Everybody okay with that? We have to identify what the numbers are and label them.

Exactly what you label them doesn't matter, but it's important to get them paired up correctly, right? So I have n's and v's and 1's and 2's, and so I need the equation v1 over n1 equals v2 over n2. And again, we've got an equation here with a fraction. So I'm going to cross multiply. And get rid of the fraction before I rearrange the equation.

V1, N2 equals N1, V2. I know some of you can just look at that and rearrange it. Good for you.

I'm talking to... the rest of the class, which is most of you. So I got rid of my fractions. Here's my equation.

I'm trying to solve for v2. So I'm trying to get v2 by itself. So v2 is multiplied by n1, so I need to divide by n1 to make it go away.

And I need to divide the other side by n1. So V2 equals V1 times N2 divided by N1. So you should always show your work when doing problems like this.

Do you have to show how you rearrange the equation? No, you don't. If you can do that in your head, you don't need to show that.

You don't necessarily have to write out this table. either, but I do want you to write down the rearranged equation, and then write it down with the numbers in it. So volume 1 is 2.18 liters with their units, and that's multiplied by N2, which is 0.621 moles. Divided by N1, which is 0.120 moles. The moles cancel out.

I have 2.18 times 0.621 divided by 0.12. And how many sig figs should this answer have? Three.

11.281 liters, so I'm going to round this to 11.3 liters of gas. So that part right here, that's showing your work. You need to learn how to show your work. I know that some of you can do this and you like to do it. You put the numbers in and you get the right answer.

Why do I have to show my work? Well, because sometimes we need to write stuff down. so that someone else later can look at what we did and understand, even though we're not there to explain it to them. The other benefit of doing this is if your answer is wrong, you can go back and very clearly see what you did.

You can find the mistake and fix it without starting over. If you're doing an exam, you're very concerned about the answer being correct, but you don't know if it's correct, you can look back over this, and you can, you know, work through it in your head again and confirm. Does that work out?

Does that look correct? Or you may realize, oh shoot, I, you know, this is N1, and I actually wrote it in for N2. And you can fix your mistakes. Any questions?

11.3 liters. Here's another example. The scale model of a blimp rises when it's filled with helium to a volume of 55 cubic decimeters.

When 1.1 moles of helium is added to the blimp, the volume is 26.2 cubic decimeters. How many more grams of helium must be added to make it rise? Assume constant temperature and pressure. So this, like that first one we did, is a Chem 1A problem.

The other couple that we've done, the simpler ones, are Chem 3A problems because you covered these things in your prerequisite class. So hopefully most of this is review. But then applying it to a problem like this, this requires a little bit of critical thinking.

And again, we might want to draw pictures here. So a scale model of a blimp. I'm going to make my blimp red.

They're pretty, right? Terrible. It's going to float when it has a volume of 55 cubic decimeters.

That's a weird unit. It's okay. So that's floating. It says when 1.1 mole of helium is added to the lamp, the volume is 26.2. This is 55. If the volume is 26.2, is it going to be floating?

No. It's going to be laying on the floor, kind of all squishy, right? So here its volume is 26.2 cubic decimeters.

We're starting to fill it up, but we don't have enough helium. helium in it yet. Down here, the amount of helium in there is 1.10 moles. It's asking how many more grams of helium must be added to make it rise. So what we're looking for is from here to here, how many grams of helium do we have to put in?

It's kind of funny to think about. The balloon is on the ground, and by putting more stuff into it, you can make it go up. Yeah, that doesn't happen with liquids and solids.

This is a volume, right? And that's a volume. And this is an amount of gas in moles. Which one should we call 1?

This down here, the beginning. So this is volume 1. And this is the amount of gas in moles at that condition. When we get it to float...

This is volume two. Now the gas law is not going to tell us directly how much gas we need to add in grams. What can we calculate using Avogadro's law?

The number of moles, right? We have V1 over N1 equals V2 over N2. We have V1 and V2 and we have N1.

We can calculate what N2 is. So let's cross multiply this to get rid of the fractions. V1 N2 is equal to N1 V2. I'm solving for N2. So I'm going to divide by V1.

Divide by V1. So there's my rearranged equation, and I'm going to put my numbers in here. N1 is 1.10 moles. Yeah.

Moles. Volume 2 is 55.0 decimeters cubed. I screwed up, huh? I go to fill this in and I'm like, V2, wait a minute, I've got V2 down here. What happened?

I copied it down wrong. It's supposed to be V1 in the bottom. So let's fix that.

V1 is this guy, 26. 0.2 cubic decimeters. I'd like to be able to say I did that on purpose to illustrate something, but it was just screwed up. The cubic decimeters, which we were a little uncomfortable with, cancel out and go away. Yay. So if we calculate here, we get 1.1 times 55 divided by 26.2.

I did that wrong. I ended up with negative 0.999, which is just not even possible. I pressed minus instead of times. 2.309. That's better.

So N2 equals 2.309 moles. That would have three significant figures. 3091. Let me keep two extras.

So if I start with 1.1 moles, and I need to have 2.3091 moles, how many moles do I have to add? The difference, right? I want to end up with 2.3091 moles.

I already have 1.10 moles. Need to add 1.2091 moles of helium. This is an H.

Looked like an N there for a while. Ooh, worked that time. Can I figure out how many grams that is? Sure. 1.2091 moles of helium times grams divided by moles is 4.003 grams of helium per mole.

The moles will cancel out. And so my answer should be 4.84 grams of helium is how much I need to add. Any questions? This is kind of a tricky problem because it's not just looking at initial amount and final amount, it's looking at the difference.

This is a very common tricky problem.

Transcript for:Understanding Gas Laws and Applications

Transcript for:
Understanding Gas Laws and Applications