Hey everyone, welcome to another session of Surroundable Downswell Physics. In today's session guys we're going to be doing a lesson on pressure and temperature. So put down today's title, it's going to be pressure and temperature. And before we get going, make sure you hit the like and subscribe button to keep my channel going and keep my content as free as possible. Right, let's get straight into it.
Okay, so hopefully you have an idea about how gas pressure works. I've drawn a box over here, actually a box of fixed volume, and inside there is a gas. Hopefully you can identify that inside the particles are moving in random directions.
They're colliding with the walls of the container. Every time they collide with the walls of the container, they exert a force. That force will be acting upon an area. Yes. And obviously, you should recognize that pressure is equal to force divided by area.
So as there are forces hitting the wall divided by an area, therefore, the gas exerts a pressure on the container. Let's put that down. Here we are. Particles collide with the wall.
They exert a force as they collide with the walls. Obviously, that's a area. And therefore, hence, the gas exerts a pressure. Right.
So we have an idea that gases exert a pressure. Yes. Due to this principle. But what happens?
if you were to increase the temperature of the gas. Okay, so let's say we were to heat up the gas, but what would happen is the particles would move faster. The particles would gain kinetic energy, they would therefore travel faster, therefore there would be more collisions per second with the walls of the container, and therefore the pressure will increase.
And there we go, guys. Particles gain kinetic energy, there are more collisions per second with the walls of the container, and therefore the pressure increases. So I'm going to put this over here, we can therefore say that pressure increases.
Pressure we can say is proportional to the temperature Okay, so temperature over here Yes So as you increase the temperature the pressure increases and obviously by symmetry if you were to decrease the temperature the pressure would also decrease Okay. Now let's try and plot a graph for pressure versus temperature and let's see what we get. Okay, right in order to do this I've also drawn three balloons over here to help you understand the problem So let's say there are three identical balloons one of them you place in the freezer So this one is placed in the freezer hence why it cools down to zero degrees Celsius. One's at room temperature over here and another one you're going to place let's say you're going to place it on top of a radiator yes so in let's put it in a hot room over here yes 60 degrees Celsius.
Right so first of all if we were to talk about the pressure inside the balloon so don't forget there's particles inside the balloon over here yes which are moving around over here at zero degrees Celsius there is still some pressure so at zero degrees Celsius there is still some pressure over here. At 25 degrees Celsius, obviously there'll be a greater pressure than at zero degrees Celsius, so it'll be over here. Oh, I forgot, let's put zero degrees Celsius at the bottom of this over here. So zero degrees Celsius, okay, we've got this value over here.
And then at 60 degrees Celsius, we know that you've increased the temperature, the pressure will increase. Right, so let's draw a line of best fit through these values over here. There we go, imagine that's a straight line, there we have it, over here.
So you might be thinking, how do I do this? Hang on a minute. Didn't you say pressure is proportional to temperature earlier on? Well, not quite right now because the graph is slightly a bit weird.
We can actually find out which temperature will the pressure drop down to zero. Which temperature will the pressure drop down to zero? So if I was to just extend this line over here, let's say extend this line all the way outwards over here.
Don't forget these are negative numbers now. And then if I was to then extrapolate this line all the way down, there we go over here, we can actually find out... This temperature at which the pressure inside the balloon is zero.
Right, this value is actually going to be minus 273 degrees Celsius. Minus 273 degrees Celsius. This temperature at which the pressure is zero, this is known as absolute zero.
So let's put that over here, let's circle that, so let's put that over here. Minus 273 degrees Celsius is equal to absolute zero. Some people like to call it the true zero, yeah, if you want. And that's when the kinetic energy of the particles is zero.
Kinetic energy of particles is zero. Excellent stuff here. Right, now, from here, we can do something quite cool. We can now try and get a graph which...
goes for the origin. But obviously we're going to have to use a different type of unit. So up to this point you've been talking about temperature in terms of degrees Celsius.
We are now going to introduce a new type of unit for temperature. It's called Kelvin. So Kelvin is an alternative unit of temperature and you can convert from degrees Celsius to Kelvin using the following formula. Right, so the way we do it is the following. Minus 273 degrees Celsius, this is known as 0 Kelvin.
0 Kelvin. 0 and then the K here. Not OK, 0 Kelvin over here.
Right, so now we can now try and plot this graph. We can have a new graph right now. And look, this graph is different. On the x-axis, I'm going to put temperature.
And this one's going to be in Kelvin. And this one's going to be pressure. Yes, still measured in Pascal. Yes, in Pascal over here.
Right, so at zero Kelvin, we know that the pressure is going to be zero. There we go. And now look, we have this kind of graph here. And now we can say, guys, that pressure is directly proportional to temperature in Kelvin.
So there we go. We now have the following relationship, that pressure is directly proportional to the temperature. Yes, and I'm going to put in brackets if it's measured in Kelvin. over here if it's measuring kelvin if you use degree celsius it won't be proportional but now guys because we're using kelvin over here we can say that at zero kelvin the pressure will be zero right so you might be wondering hang on a minute how do i convert from degree celsius to kelvin it is a very simple thing we're going to take um so minus 273 minus 273 degrees celsius that is known as being equal to zero kelvin over here Okay, so this is going to be our starting point. Minus 273 degrees Celsius is known as 0 Kelvin over here.
Right, if you want to convert any value from degrees Celsius to Kelvin, you simply add 273. Simply add 273. And if you want to convert Kelvin to degrees Celsius, all you simply got to do is subtract 273 from the value. Okay, so some of you might struggle with that, so let's do a couple of practice calculations. Alright, so here we have a table of degrees Celsius to Kelvin.
Let's convert from degrees Celsius to Kelvin. So first of all I simply add 273, so 25 plus 273, my value is going to be 298, yes, over here. If I have 30 degrees Kelvin, how much is that going to be in terms of degrees Celsius?
So I simply subtract 273 from this value, so it's going to be minus 243 degrees Celsius. If I have minus 5 degrees Celsius as a converter, so minus 5 is simply plus 273, I'm going to get the value of... 2, 6, 8. And the last one guys, I'll subtract it, 280 minus 273. I'm going to get the value of 7. 7 degrees Celsius is the same as 280 Kelvin. Fantastic stuff. So hopefully that now puts it all into context in which we can have a graph of pressure versus temperature, where temperature within degrees Celsius, the graphics like this.
If I extrapolate the line backwards, the true zero, which the particles have zero kinetic energy, this is called absolute zero. And this is the starting point for the Kelvin scale over here, 0 Kelvin over here. Fantastic stuff. Okay, now from here, the last thing we're going to do is we're going to look at pressure versus temperature in Kelvin.
We're going to look at this graph one more time. Okay, so we've got the following statement. Pressure is directly proportional to temperature over here.
Pressure is directly proportional to temperature. And look, we end up with the following graph. Notice that my x-axis is in Kelvin, hence why it's going through the origin directly proportional. Right, so hopefully you can identify that. These variables are proportional, so therefore you could almost say that there's a constant here.
So we can say that P is proportional to T. Let's get rid of the proportional sign. We can therefore say that P is equal to a constant K.
It could be any variable you can put there, times by T. Okay, right. So this means that the constant is the same everywhere along that line. The constant is the same everywhere along that line.
So let's just say that if we had, this is the original temperature, T1. Yes, and this is the pressure originally is p1. Yes, so T1 stands for initial temperature, p1 stands for initial pressure. And let's say you were to increase the temperature over here.
This one's going to be T2 and the pressure will be p2 over here. Right, because we have this relationship, pressure is equal to a constant times by the temperature. Yeah, therefore, we notice that the constant will remain the same at this point and this point.
And obviously that's the reason why it's called a constant. So if you were to put P1 is equal to K times by T1 over here, this point over here, and this point over here is going to be P2 is equal to K times by T2 over here. Right, don't forget the constant is the same for both of them.
So rearranging this one over here, P1 divided by T1 is therefore equal to, and rearranging K, and therefore this one as well is going to be p2 divided by t2 is now equal to k as well. That constant is the same for both of them. The constant here is the same as the constant there. And therefore, finally, because you know the constant is the same for both of them, you can then equate both of them together. So therefore, we can say that P1 over T1 is equal to P2 divided by T2.
So we have this relationship now as well. Okay, let's test our understanding with the following question. The pressure inside a car tyre is 6.10 times by 10 to the power of 10 Pascal when the temperature is 280 K.
Okay. After the car has made a journey, the temperature of the air in the tyre is 300k. Calculate the new pressure in the tyre, assuming the volume remains constant.
Okay, so the first thing I would say is first of all identify what you've been given. P1, P2, T1 and T2. So P1 is going to be the initial pressure, which is 6.10 times by 10 to the power of 10. The final pressure... We don't know what that is, so we're trying to work that out. The initial temperature is 280K.
Yes, the final temperature is going to become 300K. Yes, then we're going to take this and rearrange it to make P to the subject. P1 over T1 is equal to P2 over T2.
And then from here, all I'm going to do is I'm going to move the T2 upwards over here. So P2 is equal to P1 over T1 and the top line times by T2. over here.
Right, let's all plug it in now. So P2, the final pressure is initial one, 6.10 times by 10 to the power of 10 times by the final temperature, which is 300k over here, divided by T1, the initial one of 280. I'm going to get the following value, guys. It's going to be equal to 6.54 times by 10 to the power of 10 Pascal. Easy. stuff here.
There we go, that's how you do it. That's one example of using that formula here over here. Make sure you brush up on your algebra in order to rearrange that to make different quantities of the subject of the formula. And that's it guys for another session of Surrounded or Downs with Physics. Make sure you hit the like and subscribe button to keep my channel going and good luck in your studies.
Ciao, ciao and goodbye.