Greetings, in this lecture I will continue to discuss the concept of pressure and show a way that pressure can be measured. Then, the concepts we've already discussed related to Charles' Law and Boyle's Law, will be combined with Avogadro's Law to make the combined Ideal Gas Law. It's important to realize that on earth, we have quite a large volume of air pressing down upon us. For example, if one considers one square meter of area at sea level, there is a mass of about 10,000 kilograms of air that exerts a pressure of 101 kilopascals onto that square meter of air. That pressure of 101 kilopascals is equal to 760 millimeters of mercury. The unit of measure called millimeters of mercury comes from this measuring device called a mercury manometer. In this case, the air is surrounding an open container containing the metal mercury, which is a liquid metal. Put down into that mercury is a glass tube which is open on one end, the end that is submerged in the mercury, and it's closed on the top end. Inside that tube there's a vacuum. And then there are millimeter hash marks along the edge of the tube. Normal sea level pressure gives us a column height of 700mL of mercury. [BLANK_AUDIO] Under normal atmospheric pressure, the air molecules are coming down and colliding with the surface of the mercury and bouncing off. And that's happening millions of times. And that exerts enough pressure to push the mercury up the column of the tube. At sea level, the normal atmospheric pressure is about 760 millimeters of column height which we call millimeters of mercury. When you watch the weather man you might hear he or she talk about either a high pressure system or a low pressure system coming through. And that's what's happening when the atmospheric pressure is either slightly above 760 millimeters of mercury or slightly below 760 millimeters of mercury. So this unit represents a very literal measure of the pressure. What if instead of measuring the atmospheric pressure, so just the air around us, we want to measure the pressure of another gas? This type of pressure measuring device is called a barometer. But what if we'd like to measure the pressure of a different type of gas? Not the air that's around us, but some other gas sample? In that case we can use a device called the manometer. In a manometer we have a gas of unknown pressure. [SOUND]. Here it's in this bulb and the gas is shown as yellow. In this case the atmospheric pressure, which is sometimes called the barometric pressure, was a little bit higher than normal. It's a high pressure day, 766 millimeters of mercury. In a manometer, in this particular type of manometer, there is mercury inside a tube that's in a U shape. So I still have mercury. Here, the mercury is shown as dark black. The air here is just shown as light blue. So this is just air all around us. And the unknown gas is yellow. Is the pressure of the unknown gas in this picture greater than or less than the barometric pressure of 766 millimeters of mercury? Go ahead and submit your answer now. [BLANK_AUDIO] Thank you for submitting your answer. In this case, we see that the height of the mercury in this U tube is lower on the side of the unknown gas. So, the mercury atoms are being hit harder on the side where the unknown gas is. Or more frequently, than they are on the side where the air is hitting the mercury. So, there's a surface of our mercury on both sides. Therefore, the unknown gas must, must have higher pressure than the mercury. We can calculate that if we know the height of this difference. Here are some pressure conversion factors just to remind you. Here we're measuring millimeters of mercury, that can be converted to atmospheres, which can also be converted to kilopascals. In order to calculate the pressure of the unknown gas we can use this equation. Pressure of the unknown gas is the barometric pressure. In this case, we're adding the height and we're adding the height because the unknown gas pressure is higher than barometric pressure. So, for example, if the height was 18 millimeters, then we would take the 766 barometric pressure. And add to it the 18 millimeters of height difference and that gives us 784 millimeters of mercury for the pressure of the unknown gas. Now, if you don't like the unit of millimeters of mercury for some reason, I don't particularly have a bias against it. But if you don't like it, you can then convert that into bar or pascal or whichever other pressure unit you'd like to use. Psi if you want, whatever pressure unit makes you comfortable. But this is how it's measured, traditionally. Let's do another example. This is another scenario. In this case, the trapped gas pressure is less than the barometric pressure. Can you see how I determined that? I'm looking at the relative heights of the mercury column on either side of this U tube. And you see that the mercury column is lower on the side where the air is pushing down. The barometric pressure is the air And the mercury is up higher on the side when there's the unknown gas. So that must mean that the air is pushing harder on the mercury. It's colliding, those gas molecules in the air are colliding with the mercury surface more often on the side of the air. And that height difference we then say is minus 25 millimeters. You can use this same equation you did last time. You can say Pbar plus height and you can say the height here is negative or you can just remember you need to subtract because you're using common sense. So here's the equation. And I'm actually going to fix it, because if we're going to call that height minus 25, that needs to say plus h, there. And so, in this case the pressure of the unknown gas is 741 millimeters of mercury. [BLANK_AUDIO]. In the last video, we learned about Charles' Law and Boyle's Law, and that told us how temperature, volume and pressure were related to each other. We saw some wonderful demos of that. We also talked about how volume and temperature are linear related at all amounts of gas. Remember, we had that graph? Well, it turns out we can combine all of these small gas relationships into one large Ideal Gas Law that has this equation. Pressure times volume equals the number of moles of gas, times the ideal gas constant R, times the temperature. So, let's go through these variables one at a time. First is the pressure. Now when I use the Ideal Gas Law, I always convert the pressure to atmospheres. I do this for a personal reason. I have the ideal gas constant memorized with units of atmosphere, but that was a personal preference. So, I chose to always do it in atmospheres because of a number that I have memorized, but you might prefer to use bar or pascal and that's fine. The volume here, I always make it expressed in liters. So if I'm given the volume in some other unit, milliliters or gallons, I convert it to liters, again, because of the value I have memorized for R. N is the number of moles of gas, which tends to only have units of moles but sometimes you'll be given the mass of the gas. If you know how many grams of gas you have, can you convert that to moles? Sure you can. Just use the periodic table and look up the masses of the elements that you need and do some calculations. So the number of moles of gas is fairly easy to determine if you're given the mass. The volume of R that I have memorized for the gas law constant is 0.0821 liter atmospheres per mole Kelvin. That's why I convert my pressure to atmosphere in my volume to liters. Probably can deduce then that I always express my temperature in Kelvin. So, it doesn't matter if I'm giving it in Celsius, I convert it to Kelvin before I do any calculations related to the Ideal Gas Law. So, I just want to point out that pressure, volume, and temperature can all be expressed in other units. But if we use other units for those variables, then we need to be sure that we're using the correct value for the ideal gas constant R. Because we need to have all of our units cancel out properly when we're doing these calculations. So be very careful about that. [BLANK_AUDIO] One of the laws I've alluded to that we haven't really explored yet is Avogadro's Law. So, you probably have observed in your life that when you blow a balloon up it gets larger and larger. And you probably could explain to someone that the reason that the balloon is getting larger is that you are putting more air into the balloon. Just like this young man is doing in this picture. If we look at how that can be described by the Ideal Gas Law, what we would have to do then is, if we want to have the volume, right? So why is the balloon getting bigger and bigger? That's a question about volume. We want to have the volume related somehow to the amount of gas that we've added. Because that's our explanation; well the volume's getting bigger because I'm adding more gas to it. So, let's divide the left side and the right side both by the pressure times the number of moles. And that simplifies the Ideal Gas Law just through algebra. If I do some cancellations, right, pressure cancels on this side. Over on the right side, number of moles cancels. And that leaves me with this expression. The volume divided by the number of moles equals the ideal gas constant times the temperature divided by the pressure. Well, normally we're at constant temperature and pressure. We're not in a room where the temperature is fluctuating wildly, wildly. We're not changing elevations, so that our barometric pressure's changing. So normally, this part of the equation, R times T divided by P, is being kept constant when we're just blowing up a balloon. Because that's constant we can write Avogadro's Law. If volume divided by the number of moles equals a constant. Then the volume at condition one divided by the number of moles at condition one is equal to the volume at a different condition, condition two. Divided by the number of moles at condition two. What the heck does that mean? That sounds more complicated than it is. All this means, is that the volume and the number of moles are directly proportional. Well, I'm pretty sure anybody could explain that. The volume gets bigger when you add more gas to the inside of the balloon. Let's do an example. Let's suppose that this young man really has a balloon with a large capacity and he has blown it up to a volume of 11.2 liters, that's a pretty big balloon. In that experiment we determined that he had to add a half of a mole of gas to the balloon to get it to inflate to that volume, 11.2 liters. At atmospheric pressure, so 101 kilopascals, or one atmosphere and a temperature of zero degrees C. So he's outside on a cold day. Now what happens if he doubles the amount of gas inside the balloon? So his balloon was 11.2 liters. The young man's going to continue blowing into the balloon and double the amount of gas. Well, of course, the volume gets larger. The volume doubles because we know we have this relationship between the volume and the number of moles of gas. So, at atmospheric pressure under standard conditions at zero degrees C, one mole of gas occupies a volume of 22.4 liters. Let's do another example of applying the Ideal Gas Law in a calculation. Here's kind of a, a bland example. But we have 1.8 grams of helium at 330 degrees C. So this is a hot helium balloon, not a cold balloon like we had in the last example. What is the pressure of 280 milliliter container that contains this helium at 330 degrees C? Well, I told you that I have memorized the ideal gas constant R as a value of 0.0821 liter atmosphere per mole Kelvin. So, because I have that in my head and I don't want to have to look anything up, because it takes more time. I'm going to convert the pressure to atmospheres. Well, I wasn't given a pressure, so I'm going to find the pressure in atmospheres. I'm going to convert the volume to liters. And I'm going to convert the number, the mass of gas, the amount of gas into number of moles. And finally, I'm going to convert the temperature to Kelvin. So to find the number of moles, I can look up the molar mass of helium on the periodic table. And I see that helium has a molar mass of about four grams per mole. So I can take the mass that I was given divide by the molar mass and see that I have 0.45 moles of helium. To convert to Kelvin from Celsius, remember that we're going to add 273.15. In this case we should only have two sig figs, so I gave too many sig figs. I really should have 6.0e2Kelvin. But I'm not going to intermediately round anyway, so it's okay that I have 603 here because I'm still going to plug it into an equation. I'll round at the end. So I'll plug the values that I know into PV equals nRT. But I need to do a little bit of algebra, don't I? Because I really want to solve for P. So I can bu-, divide both sides of the expression by V. If I divide both sides of the expression by V, I get this expression. The pressure equals nRT divided by V. And then I can do the calculation and see that my pressure is about 80 atmospheres. going to assume that's supposed two sig figs and put a period there. I want to caution you about this ideal gas constant R because sometimes I see students plug in a value for R and they haven't thought about the units. One value that you will, frequently see sited in reference literature for R is this value R equals 3.154 Jules per Kelvin mole. Now that is the same constant, it's still R but it's in different units. So if I plug that into my PV equals nRT expression and I didn't worry about units canceling out, it would give me a, a incorrect value for my answer. So, just be careful that your units are matching up. As I said before, these are the units on the value of R that I use. You can use a different value if you want, but be sure that your volume is in the same units of volume that your ideal gas constant has. Here's one for you to try. What is the temperature of 0.43 moles of gas at a pressure of 2.7 atmospheres and a volume of 10.9 liters? Please go ahead and try to answer that now. [BLANK_AUDIO] Thank you for your submission. Well, remember because I have R memorized with a certain units R equals 0.0821 liter atmosphere per mole Kelvin. I'm going to convert my pressure to atmospheres, my volume to liters, and my temperature to Kelvin to figure things out. So in this case, I have PV equals nRT. I need to solve for unknown, which is T. So the temperature, just doing a little quick algebra is going to be PV divided by nR and I need to plug in what I was given. So, the pressure I was given is 2.7 atmospheres, the volume was 10.9 liters. Well, that's handy, that already lines up. [SOUND]. The number of moles of gas is 0.43 moles. And then I can plug in the value that I have memorized of, for R, 0.0821 liter atmosphere, mole Kelvin. And if I cancel my units, atmospheres cancels. Liters cancels, right? Moles cancel and I'm left with Kelvin down here. But it, we have one over one over X, so Kelvin becomes the units and my final answer for this, is that the temperature equals 830 Kelvin. [BLANK_AUDIO] So that was simply an application of the Ideal Gas Law. PV equals nRT. [BLANK_AUDIO] This concludes our introduction to the Ideal Gas Law. In this lecture we've learned how pressure can be measured from a practical stand point. We've also learned about Avogadro's Law, and done a little bit of practice of using algebra to apply the Ideal Gas Law to real problems. [BLANK_AUDIO]