In this lecture, we're going to wrap up our study of the physical properties of gasses. This will conclude the material from Concept Development Study number 14. So you may want to review that reading as well. In this lecture, we're going to discuss a few applications of the Ideal Gas Law which we developed in the previous two lectures. And we're also going to extend the Ideal Gas Law to a discussion of mixtures of gasses. To review very quickly, you'll recall that the Ideal Gas Law tells us that for any gas regardless of it's type, type of molecules, that the pressure times the volume is equal to the number of moles of the gas multiplied by the temperature multiplied by a constant R. That constant is given here on the screen. It is in fact just experimentally measured to be 0.08206 liter atmospheres per mole Kelvin. The Ideal Gas Law is a summary of the three experimental laws that we've produced. Boyle's Law, relating pressure and volume. Charles's Law, relating volume and temperature. And Avogardo's Law, relating the volume and the numbers of moles. Remember that, ideal means that it is independent of the type of gas molecule. What that means is, we can apply it to just about any kind of gas that we are interested in. Here is a particular example. What volume of gas is occupied by one mole of the gas if we are at room temperature and one atmosphere pressure? Again the constant is given here on the screen. Well, let's write this out. We know that PV is equal to nRT and we are now asking for what the volume of the gas is, so we can solve this equation for the volume. It's just the number of moles times RT divided by the pressure. And we have given here in this equation, or in this problem, the temperature which is in degrees centigrade 25. the pressure on atmosphere and the number of moles here. So you can tell we have n and temperature and pressure. And it's just a matter of then plugging those numbers in to get the calculation correct. So what do we have? 1.00 moles multiplied by R which is 0.08206 liter atmospheres per mole Kelvin multiplied by the temperature which is 25 degrees centigrade which is 298.15 Kelvin. We always have to use absolute temperatures there. And we divide that by the pressure which is given to be 1.00 atmospheres and this now just becomes a calculator problem. You can just plug these in. We'll check the units here. Here is moles per mole. Here is Kelvin per Kelvin. Here is atmosphere per atmosphere. The only unit left is liters. So in fact our answer's going to be in liters. Plugging the numbers in and calculating them out. We wind up with a value 24.5 litres is the volume of a single mole of gas. And it is interesting that because this is the Ideal Gas Law, whatever gas this is, regardless of the identity of that gas, a single mole of gas under normal conditions, 25 degrees centigrade and one atmosphere of pressure will occupy 24.5 liters. Let's consider then, another application sort of stepping it up one level farther. In this case, we're actually going to use the Ideal Gas Law to determine the molar mass of a gas. So we're going to study now a hydrocarbon gas, that'll be a gas consisting entirely of hydrogen and carbon or molecules. And consists entirely of hydrogen and carbon. And what we are given here is the density of the gas at 25 degrees centigrade and one atmosphere. But that density is not moles per litre but rather grams per litre. But from the Ideal Gas Law, we can write that the number of moles divided by the volume is equal to the pressure divided by R times T. That's just a rewriting of PV is equal to nRT. If we then calculate the pressure divided by the temperature and with R, you'll notice back over here that we do in fact have the temperature. And we have the pressure. So we have everything that we need to know to be able to plug in this equation and find the number of moles per volume. Let's do that. The pressure is, what did we say, 1.00 atmospheres. R is 0.8206 liter atmosphere per mole Kelvin. And the temperature is, let's see, 298.15K. And if we plug all those numbers in, let's check the units first. Atmospheres per atmospheres, Kelvin per Kelvin and the denominator, we still have units of litre per mole which will ultimately then give us units of moles per litre. And if we just plug the numbers into the calculator, we discover that the number of moles per litre is 0.0409 moles per litre. But of course that's not the question we were asked. We were asked what is the molar mass of this hydrocarbon? And right now what we have is, instead the number of moles per litre of the hydrocarbon. But what we also know of course is that the [SOUND] [COUGH] number of moles per litre or the number of moles is equal to the mass divided by the molar mass. So, if we simply write this as the mass per volume times molar mass. Just substituting n into the equation up above, is equal to 0.0409 moles per litre. Then let's see, we have mass per volume is here. That's the number that we've been given. That's 1.80 grams per mole. So, we get 1.80 grams per litre times one over the molar mass, is equal to 0.0409 moles per liter. If we solve now for the molar mass, that's just taking let's see, we'll move mass to, molar mass to the other side. Divide both sides of the equation by 0.0409. If we do that calculation we will wind up with the molar mass being 44.0 grams per mole. The units of litres cancel out. The last part of this question then, asks us to identify what is this hydrocarbon's molecular formula? Well that's a trickier problem. But, remember, this hydrocarbon consists only of hydrogen and carbon. So we have to have a comb, combination of the mass of carbon and the mass of hydrogen which adds to 44. And with a little bit of work, we can see that, that is three times 12 plus eight times one would give us in fact the gas which is C3H8 or propane. So in fact just by measuring the density of the gas under specified conditions, temperature and pressure, we can figure out what the molar mass of the gas is. Let's try another example. Here is actually applying still at the Ideal Gas Law to do some Stoichiometry. In this particular case, we are going to react calcium hydride, which is a solid, with water. When calcium hydride reacts with water, it produces hydrogen gas as well as calcium hydroxide. In this case, we're going to say that an experiment has been one in which we have a five gram solid sample that contains calcium carbon calcium hydride. But is not exclusively calcium hydride. It contains other impurities that don't react with water. The question then is, if we produce a specified amount of hydrogen gas as the product. Here we have under given conditions. How much calcium hydride was there in the original sample? To do the stoichiometry, we have to know what the balanced equation is. But we're given the information we need. Calcium hydride reacts with water to produce calcium hydroxide and hydrogen gas. Let's be careful to make sure that this equation is balanced. We have one calcium on each side of the equation. We have an one oxygen on the left side but two on the right side. So we'd better put a two in front of the water so now we have two oxygens on each side. How about hydrogen? There are six hydrogens in the reactants. And only two hydrogens in the products plus two more here. We need to double the amount of hydrogen gas there. So one calcium hydride produces two hydrogen molecules. That's important because we can calculate the number of moles of hydrogen which are produced from the Ideal Gas Law. The Ideal Gas Law tells us that the number of moles is equal to the pressure times the volume divided by R times T. Notice if I multiplied both sides of the equation by RT, I would have PV as equal to NRT again. And this is useful because, let's see, we have been given a pressure although it's in torr. And we're given a temperature although it's in degrees centigrade. And we are given a volume 3.90 litres. So we have the pressure, we have the volume and we have the temperature. Let's write it all in here then. The pressure is 750 torr, but we know R in values of atmosphere. So we need to convert this to atmospheres. And there are 760 torr per atmosphere. That's the pressures in atmospheres. The volume is 3.90 litres. R is 0.08206 litre atmosphere per mole Kelvin. And the temperature is let's see, 32 degrees centigrade which is 305 Kelvin. We can check all the units here K K's cancel out. Torrs cancel out. Atmosphere cancels out. Liters cancel out. The only unit left is one over one over moles which is units of moles. And if we simply plug all these numbers into our calculator and do the calculation, we discovered that the number of moles of the hydrogen is 0.154 I, I'm sorry 0.154 moles. That's the number of moles of hydrogen gas which are produced by the reaction of the calcium hydride with water. But it is not the number of moles of calcium hydride. The number of moles of calcium hydride is going to be half that amount because one mole of calcium hydride produces two moles of hydrogen. That means that the number of moles of calcium hydride is 0.077. actually zeros, yeah, 0.0770 moles. And now we know the number of moles of calcium hydride. But the question asks us for what mass percentage was the sample of the calcium hydride? So we need to know the mass of the calcium hydride. And the mass of the calcium hydride is then given by the number of moles of calcium hydrate multiplied by the molar mass of the calcium hydride using the stoichiometry we learned in the first semester. The molar mass of the calcium hydride is easily calculated to be 42.1 grams per mole. Multiplying the two numbers together then, we wind up with the mass of calcium hydride is 3.24 grams in the original sample. But the original sample, you recall, was five grams large. It was not 3.24 grams. That's because apparently there was some impurity present. What fraction of the calcium hydride was present? The percent of the calcium hydride will be equal to 3.24 grams divided by five grams total. And that as a percentage works out to then be 64.7%. So we can actually use the Ideal Gas Law to assay samples of solid samples by reacting those solid samples out and getting a volume of a gas produced. That's three examples now of how we can use the Ideal Gas Law. We'll continue in the next lecture.