 Alright, so we've seen the Boltzmann distribution, which is an equation that can be used for lots of different things. As we go forward in this course, we're going to use it many, many different times. One additional example that we can use it for in addition to the ones we've seen so far is understanding the pressure of gases at altitude, just to show you another example. So the key here is that this expression, which describes the probability of finding some molecule, some system, some object in some state is related to both the energy of that object and the temperature. And that energy may be the conformational energy of the molecule, as we discussed in the example of the butane molecule, but it could be any type of energy at all. So including, for example, gravitational energy. So if I have an object that I've had to lift some distance against gravity or that finds itself at some height in a gravitational well, its energy, as you might remember from a physics class, is mass times a gravitational constant times its height. So if I use that as the description of the energy that we're talking about, the Boltzmann expression then becomes the probability looks like 1 over q e to the minus energy. Gravity is m times g times h if we're talking about gravitational energy and I need to divide that by k t. So that's the probability of the system in this state, the probability of the system having this particular height in a gravitational well. So the probability of being at that height is given by this expression. That's useful enough, it's a little bit inconvenient because it's kind of challenging to figure out the value of this constant q, but we can use the same trick that we've used before and say it's a lot easier, I can make these q's go away if I ask for a relative probability. So the probability of being at height h relative to the probability of being on the ground or the probability of being at height zero, probability of being at sea level, whatever height zero means in this particular problem, probability of being at h relative to the probability of being at zero is just using this equation twice, 1 over q e to the minus m g h over k t divided by 1 over q e to the minus m g times zero over k t. Then the reason we used two different probabilities stacked on top of one each other is purely because this 1 over q term is going to cancel out. Another useful thing happens because my denominator has h equals 0, e to the 0 is equal to 1, and I don't need to worry about that exponential in the denominator. And the result here is just e to the minus m g h over k t tells us the probability of being at height h relative to being at height zero. So let's see how that works with some actual numbers. Let's say the thing we're interested in being at some particular height is the air pressure, the pressure of a certain gas as we go up to altitude. So this question is motivated. I was hiking recently out west if we calculate an example of how thin the air gets when you go to the top of a mountain. Let's say oxygen as the gas I'm most interested in breathing. If I say let's say I'm at an altitude of 10,000 feet, some mountain out in Colorado for example, and we're going to want to do this in metric units of course. So 10,000 feet works out to be 3,050 meters. The other information I have to give you is the temperature. So if we're, it's a nice pleasant day with a temperature of 20 degrees Celsius or 293 Kelvin, that's now enough information to ask how thin is the air that I'm breathing when I'm at an altitude of 10,000 feet. So the probability of finding one of these oxygen molecules at an altitude of 10,000 feet relative to the probability of finding it at sea level, an altitude of zero, is going to be this exponential, e to the minus mgh over kt. So let's start plugging in those numbers. M is the mass of the oxygen molecule that I'm interested in. O2, one oxygen atom weighs 16 grams per mole, two of them weigh 32. So 32 grams per mole for the mass. G, the gravitational constant. Again, as you may remember from a physics course, 9.8 meters per second squared. H is 30, 50 meters. I'm going to divide that mgh by kt. Boltzmann's constant, 1.38 times 10 to the minus 23 joules per Kelvin. And the temperature, 293 Kelvin. So all that with the negative sign is up here in the exponential. Be careful, don't plug that into a calculator just yet. If we stop and look at the units, some of the units work out, Kelvin cancels Kelvin, but none of the rest of these look like they're canceling correctly. So we need to be a little more careful and make sure our units cancel. One number one is this moles up here in the mass of these molecules. I need to get rid of moles by recalling that one mole is Avogadro's number of molecules. So that will cancel moles top and bottom. And then the last thing I need to remember is that grams, if I convert those into kilograms, then a kilogram meter squared per second squared will cancel this joules. So I have to convert grams into kilograms in order to do that. So still up here in the numerator, let me multiply by 1 kilogram over 1,000 grams. So then grams here will cancel grams here. And then as we said, a kilogram meter squared per second squared, that's the same thing as the joules. So kilogram meter squared per second squared cancels joules and the units and the exponent all go away as they should. Now we can go ahead and plug that into a calculator. If I do the arithmetic up in the exponent, that works out to be 0.39. So this probability, e to the minus mgh over dt works out to be e to the minus 0.39. If I do one more step of the math, e to the minus 0.39 works out to be 0.68. So we're not quite done. I'll interpret this answer a little more. But what that means, what that tells us right now is the probability of finding oxygen molecules at the top of this mountain at 10,000 feet relative to the probability that we find an oxygen molecule at sea level is 0.68. So there are only 68% as many oxygen molecules in each lung full of air that I breathe at 10,000 feet as there are at sea level. So that's why the air feels thin on top of a mountain that's harder to breathe because you're only getting 68% as much oxygen. And that's probably the most intuitive number to understand. If I actually want to know what is the partial pressure of O2 at sea level, air is 21% oxygen. So out of the one atmosphere, air pressure at sea level, 21% of that or 0.21 atmospheres is oxygen. If I actually do want to know what is the partial pressure of oxygen, let's say at 3,050 meters, that's going to be 68% of the value at sea level based on this result. So 68% of 0.21, that works out to be just 0.14 atmospheres. So when you're breathing at sea level, partial pressure of oxygen is 0.21 atmospheres at the top of a mountain at around 3,000 feet. That works out to be 0.14 atmospheres. So it's, again, been reduced by 68%. So the two things to point out here. Number one, if we're specifically interested in pressure of gases at altitude or air pressure varying with altitude, notice that this calculation depended on mass. If I had done this calculation for a different molecule other than O2, it would have dropped by some different amounts. So air thins out as we go up in altitude. But the amount of oxygen in the air thins out at one rate. The amount of nitrogen in the air thins out at a slightly different rate. The amount of CO2 or water vapor or argon or other gases in the air will all thin out at various different rates. So if we're really interested in the composition of air at altitude, we have to do more than just this one calculation. But in a broader sense, this example points out, if you compare this with the example we did previously of the butane conformational probabilities, once we have the Boltzmann distribution, any type of energy a molecule has or a system has, whether it's gravitational energy, conformational energy, some other type of energy, a lot of what we do going forward in this course will be using the Boltzmann distribution with different types of energy that molecules can have to explain different types of behavior of different chemical systems.