 Hi, I'm Zor. Welcome to Unisor education. I would like to solve two very very simple problems related to molecular movements and the laws which we have learned by now about molecular movements. Now this lecture is basically a continuation of the course Physics for Teens, which is presented on Websiteunisor.com. I do suggest you to go to this website to listen to this and any other lecture because it's a course which has its own logical dependencies etc. So if you found it on YouTube for instance or anywhere else, I would rather suggest you to go to Unisor.com for the same thing because the same lecture is actually accompanied by very detailed explanation in a textual format, like a textbook basically. So you can listen to the lecture, you can read the text and the site is free by the way. There is no strings attached, no advertisement. You don't even have to sign on if you don't want to. So basically that's what I recommend. In addition to this course Physics for Teens, the site contains math for Teens which is actually a prerequisite for the physics because there is no physics without math as you understand. Okay, so these couple of simple problems are about molecular movement and the formulas which we have learned. The only thing is I really hate to present you the formula as a solution basically because both problems actually are solved in like one algebraic operation. So I would like actually to talk about the formula which I'm going to use because any repetition, I think of this particular logical way of derivation of this formula is just useful from pedagogical standpoint. Now obviously everything is explained before in the lecture which is dedicated to kinetics of ideal gas. So I will just very briefly repeat this in relationship to the problem at hand. So the problem is how much kinetic energy are in all the molecules in the room? Well, obviously for this we need certain information about the room, about the pressure, etc. So the room size is 4 by 4 by 3 meters and we're talking about normal atmospheric pressure of 100,000 Pascals, which is newtons by square meters. This is the normal atmospheric pressure which is like 760 millimeters of hydrogen or 30 pounds per square inch something like this. I don't remember actually normal atmospheric pressure, whatever it is that's what it is and I actually round it up obviously. It's not exactly 100,000. So we have the pressure of the air. Well, we assume that the air is ideal gas and we have the size of the room. So question is how much kinetic energy are in this room? Okay. Now before applying a formula, I'll just as I was explaining, I would like to derive it against so to speak, but briefly not as not in as many details as in the corresponding lecture, which is dedicated to this. I think the lecture is called Kinetics of Ideal Gas. But very briefly, excuse me. So very briefly is a relationship between pressure, the volume, and kinetic energy because that's all I need. I need this formula which relates kinetic energy, pressure, and volume. I can write it down right now because it was already written in that particular text. However, I think it's just good if I will discuss it again. So let's talk about reservoir of cubical form with the side equals to L. And let's talk about one molecule of gas which is going back and forth between two opposite walls perpendicularly to these walls and elastically reflected from each wall. So it goes like this all the time. So now let's think about what kind of a pressure this particular one molecule exhorts onto the wall. Well, let's think about the following. The molecule has certain impulse which is m times speed, right? Now if you act during certain time with certain force, onto object, it changes this momentum of the movement, right? So if you subtract from the ending moment, you subtract the beginning moment, that actually is an impulse of the force which is acting on this particular molecule during the time tau. The force is f. Now this obviously is a vector equation. Now what's important in this particular case is the force is not constant during this change. The force actually is equal to zero while the molecule is traveling from one end to another. And then at the moment of hitting the wall during a very small interval of time, the force actually is acting as elastic force which reflects the molecule back to this direction. And we are assuming this is 100% elastic reflection which means that these two speeds are opposite in direction but the same in absolute value. So we are not therefore talking about one particular force which is acting during this time. We are talking about some average force which acts during this time. And what is the time? Well, the time is, let's talk about the time from the when the molecule is at opposite wall, then it goes to this wall, it reflects and goes back because this is basically a periodicity of this movement because then everything starts from the beginning. So during this period of time tau, our force is changing from zero to something and then to zero again but we can talk about the average force which is exorted during this time. And that would be the difference between two momentums, right? Now, since these two are the same of absolute value and opposite in direction, let's just talk about V as the absolute value. So if I will subtract this from this, I will have basically 2m times V, right? Because this is minus this, right? In vector algebra. Now, what is equal to? It's equal to average force. I'll just use it without the vectors. And what is the tau? Tau is the time during which we are measuring the whole thing. So it's a period from the position when molecules is on that wall and goes this way and back. So the length is L and L back, so it's 2L. And the time, therefore, is 2L divided by the same absolute speed V, right? So from this, we do this and we do this equals to F times L, right? V goes here. Now, if you divide it by 2, that would be a kinetic energy of the molecule, right? So the kinetic energy of the molecule equals to this. Now, force and pressure are related very easily to each other because this is basically the whole side of the reservoir. And this force actually is acting on the whole reservoir in this particular case, although it's one molecule but it's still acting on the whole wall. So we can always say that considering the pressure is actually the force divided by area. So instead of the force, we can put pressure times L squared, which is area, times L divided by 2, which is P times volume. L squared and L is L cubed. L cubed is a volume divided by 2. So this is a very important equation for one particular molecule. Now, if I have lots of molecules which are flying parallel to each other, then this actually would be the kinetic energy of the entire set of molecules which are actually acting in the same direction. And if I consider that the molecules are not flying in the same direction but in all three dimensions, which actually means that I have to divide it by 3. So in this particular case, it's 2 kinetic energy is equal to P times V. Now, if I will consider all the molecules which are flying in all the three dimensions, then I will have to basically divide it by 3. So this is how I briefly derive the formula which I'm going to use in this particular case. So if this is true, then my problem is actually very easy because I have everything I need. All I need to do is to find out kinetic energy and it's equal to 3 second P times V. Now, I know the P. That's the pressure. I know the volume. So all I have to do is basically substitute all these values into this formula. Now, again, I could have just presented this formula which is borrowed from the previous lecture. I just wanted to talk about where exactly this formula comes from because it's very, very important to understand that it's not the formula which you have to remember. It's the logic behind it which I would like you to be comfortable with. So this is how we calculate this particular value. And it happens to me, I calculated it's 7,200,000 J. So this is my total energy, kinetic energy of all the molecules in this average size room. Under normal pressure. Now, is it a lot? Well, let's consider the second half of this problem. Second half of this problem is, consider you have an average car which weighs, let's say, 2,000 kg. Now, what should be the speed of this car for this car to have this kinetic energy? So you have mass of the car, square of the speed divided by 2, that's kinetic energy. And it's equal to this 7,200,000 J. And instead of m, I will put 2,000 kg. What is the V? What is the speed? Well, my calculations show that the V is supposed to be 306 km per hour. Which is about 190 miles per hour. That's fast. 190 miles an hour, 300 something km per hour. That's really fast. So, if the car has this speed, then its kinetic energy is equal to kinetic energy of all the molecules in this room. Well, not in this particular room. In the room of this size, under normal atmospheric pressure. So, it's pretty much a lot. I mean, it's really a very, very high number. Alright, so that's it for my first problem. Now, my second problem is related to a fundamental equation of the kinetic theory of the ideal gas. And I will put this equation here. P times V divided by G equals constant of Boltzmann number, number of molecules. Which is constant for a specific amount of gas. Specific number of molecules. This is Boltzmann's constant. And this is pressure times volume divided by temperature in Kelvin, by the way. In absolute degrees of Kelvin scale. So, we have this formula. Now, it's very important actually to remember it. I mean, this is something which even I remember. I hate to remember any kind of formulas. Now, based on this formula, my problem is how many molecules of gas are in the room. Again, under normal conditions which have been described before. I have the pressure, I have the volume. And the temperature, let's assume, is again normal temperature like 20 degrees Celsius. Which is like 293 degrees Kelvin. And pressure is, so this is the temperature. Pressure is 100,000 Newton square meter. And the volume is equal to 4 times 4 times 3, I think I put. Which is what, 48 cubic meters. Well, if you substitute it, and you put the Boltzmann's constant, whatever the constant is, and in my lecture actually put this value. I will have that number of molecules, N, which is equal to P times V divided by T times constant, Boltzmann constant. It's equal to 0.12, 10 times, in a 28th degree. It's a huge number. I mean, 10 to the 28th power, it's really a lot. So, we have a lot of molecules. I mean, that's basically all I wanted to present here. And again, it's just a direct derivation from the main formula of the kinetic theory of gases. And again, let me just tell you that this is something which is probably good to remember this type of a formula. Now, from my school years, I remember just this one. P V divided by T is constant. Then a little later, I have added that this constant is actually dependent on the number of molecules. Yes, it is constant. But obviously, if you change the number of molecules, it will be a different way. But for a specific number of molecules, pressure, volume and temperature are very much related. And in the exams for this particular part of the course, molecular movements, I actually am trying to present different problems based on different conditions. For instance, temperature is the same. We are changing volume, how the pressure will change. Or pressure is the same. We are changing volume, how the temperature is changed, etc. So all these problems are in the exams for this particular topic. And I do recommend you to take this exam just to check yourself how you master this. Okay, so that's it for today. I do recommend you to go and read the text for these particular problems, because the text actually has a solution. And I also recommend you to take exam. Exam is really a good thing just to check yourself. It's usually I have like six different problems with multiple choices, six different answers for each problem. So it's a good exercise, which I definitely recommend you to take. That's it. Thanks very much and good luck.