Lesson video

Hello there, I'm Mr. Forbes, and welcome to this lesson from the Particle explanations of density and pressure unit.

This lesson's called Explaining Pressure Changes and in it, we'll be looking at the factors that affect the pressure inside a gas.

By the end of this lesson, you're going to be able to explain how the temperature affects the pressure of a gas, and also how a change in volume will affect the pressure of that gas, including some calculations.

Here are the keywords that will help you understand the lesson, and there's only two here, particle model, and that's the scientific model that we use to describe the behaviour of matter based upon the idea of tiny particles moving about, and pressure, and pressures caused by a force acting over an area, and we're going to be calculating some of that using pressure equals force divided by area.

This lessons in three parts, and in the first part, we're going to be looking at the particle model of pressure and trying to describe the behaviour of a gas and how increasing the temperature of that gas, increases the pressure based upon what the particles are doing.

In the second part of the lesson, we're going to be looking at the relationship between pressure and volume, and that will involve some calculations.

And in the third part, we're going to be looking at doing work on a gas, transferring energy to and from a gas using mechanical forces.

So when you're ready, let's begin by looking at the particle model of pressure.

Let's start with a quick recap of the particle model of fluids, and the particles inside a fluid are in constant random motion.

What that means is that the particles are moving very fast in random directions colliding with each other and the container that they're in.

So here's a sort of figure of that and you can imagine those as air molecule or air particles moving around quite fast.

They collide with each other and the surfaces that they come into contact with.

And those collisions produce very small forces that act over an area and produce pressure.

So there's billions of collisions per millimetre in a container every second.

If you imagine a balloon, that's going to contain many billions of particles of gas.

So I've got a balloon here, and I'm just gonna put a few of the particles in there.

So there's a few of the particles and they're moving around in random directions at very high speeds, and they're gonna be colliding with the inner surface of the balloon.

And each collision's gonna produce a small outwards force on that balloon.

And that outward force is acting over the inner surface of the balloon and it's producing a pressure.

So there's a pressure on the inner surface of the balloon because of the gas particles inside it.

Now around the balloon also contains many gas particles and they're moving around randomly as well.

So I've put a few of those on here, and they're going to be colliding with the outer surface of the balloon.

And each time they collide, they produce a tiny force, and that's going to create a pressure because that force is gonna be acting over the area of the balloon, so we've got inwards pressure as well.

So when I've inflated the balloon, what I get is the inwards and the outwards pressures are balancing each other, and the balloon doesn't expand or contract.

If I added more particles to the balloon, then the balloon would expand, and that's because of this, if you add more particles into a sealed container, you're going to have more gas inside.

So I've got a small mass of gas inside the container here, and then I can add more gas to it, so I've added some more gas particles.

What that's going to do is there's going to be more collisions of those moving gas particles with the container walls, more collisions every second, and more collisions every second means there's going to be an overall outward force increase, more of those collisions happening everywhere, so I've got so double the collisions there, there's going to be a greater outward force overall, and that means there's going to be a greater outward pressure acting on it.

So adding more gas to the container is going to increase the pressure that gas is producing on the container.

Going back to the balloon example, I've got an inflated balloon here.

And overall the outward forces and the inward forces balance each other, so the balloon doesn't expand or contract, but adding more air is going to increase the number of particles inside the balloon, and that's going to increase the outwards pressure, there's gonna be more collisions of those particles with the inner surface of the balloon.

So what I get is this situation where there's a greater outwards force overall, a greater pressure pushing outwards, and that's going to cause the balloon to expand, it's gonna stretch the rubber of the balloon, so the balloons are gonna become larger, and as it becomes larger, then it's going to reduce the number of collisions per square millimetre on the inner surface of the balloon that's happening.

So eventually, the balloon stretches to a point where the outward forces and the inward forces are balanced again, the pressures are balanced and the balloon stops increasing its size, so it stops expanding when we've got equalised pressures.

Let's check if you've understood the basics of that.

I've got three containers here, all of them are the same volume and they're all at the same temperature.

So in which one does the air exert the highest pressure on a container surface, and what the difference is if they've got different masses of air inside? So I'd like you to pause the video, make your decision, and then restart, please.

Welcome back, hopefully you selected C, that one contains 1 kilogramme of air or 1,000 grammes of air, and that means there's gonna be more air particles inside there, many more than the other two containers.

Those particles are gonna be colliding with the surface of the container on the inside, and that's gonna create the greater pressure because there's gonna be more collisions every second.

So well done if you selected that.

When you heat a gas, the average speed of the particles in that gas are going to increase.

So heating any gas, will make those particles move faster on average.

So that's going to have any effect on the pressure because those particles are going to be moving faster and hit the container walls with a greater force, and that's gonna increase the pressure.

So if I've got a gas at low temperature inside the container, those particles are moving quite slowly and they hit the container walls, then they are gonna produce forces, but those forces are going to be smaller.

Then if I get another gas and increase the temperature, those particles are gonna be moving faster.

So the faster particles are gonna hit the wall harder, and that's going to create bigger forces, and the effect of that is that you're going to get more collisions, and more force, and that's going to increase the pressure.

So the hotter the gas is inside the container, the more collisions there are per second, the harder the collisions are, and therefore, the greater pressure is produced by the hotter gas.

Let's check your understanding of that.

I've got three containers again here.

All of the containers have the same volume and they contain the same mass of gas.

So we've got an identical number of particles in each one.

In which one of those does the gas exert the highest pressure on a container's surface? Pause the video, make your selection, and restart.

Welcome back, well hopefully you've selected option C there, the temperature of 30 degrees.

And that's the correct answer 'cause that's the highest temperature the particles are going to be moving fastest in that, more collisions per second with the container walls, and therefore, a greater pressure.

It might have been fooled by option A, but remember, that's a minus 60 degrees Celsius and that's a cold temperature, so those particles are moving slowest.

Well done if you picked C.

Another factor that will affect the pressure in a gas is the size of the container that the gas is contained in it.

If you decrease the volume of the container, you're going to increase the pressure inside.

So imagine I've got some gas in a container here, and they're in a large volume, and they're going to be colliding with the sides and the surfaces, but they're going to do that at less regularly because they've got a quite a large distance to travel once they've collided with one surface, bounce off it and hit a second surface.

So because the distances are quite large, the number of collisions per second is quite low.

In a smaller container, I've got much smaller distances.

So the particles have a shorter distance to move before colliding, so they're going to be colliding with the container walls more frequently, we're gonna get more collisions per second for the gas when it's in a small container than when it was in a large container, and that means there's more collisions per centimetre squared of the container every second, and that gives a higher pressure.

So the smaller the container is, the higher the pressure if you've got an equal amount of gas in it.

And another check to see if you've understood that.

I've got some containers here, and they all contain the same mass of gas at the same temperature, but they've got different volumes.

In which one is the gas exerting the highest pressure on the container? So I'd like you to pause the video, make your selection and restart , please.

Welcome back, hopefully you selected option B, that's got the smallest volume, it's going to have the most collisions per square centimetre per second, and so it's gonna have the highest pressure.

Well done if you selected that.

And now it's time for the first task of the lesson.

And I've got two questions for you here.

I'd like you to use the particle model to explain why each of the following cause a change of pressure inside the balloon.

So why does increasing the air temperature change the pressure? Why does pumping more air into the balloon increase the pressure? And why does squashing the balloon as shown in the figure there increase the pressure inside that balloon? And then the second one, I'd like you to explain why a balloon expands when you add more air to it, okay? So pause the video, write out your answers and then restart, please.

Welcome back, well, here's the three factors and the effects.

If you increase the air temperature, the particles are moving faster, they'll collide with the balloon's inner surface more often, and with a greater force, and that will increase the pressure.

If you put more air into the balloon, there's more air particles, so there's going to be more collisions with the balloon's surface every second.

So more particles mean more collisions.

And finally, squashing the balloon decreases the volume of the balloon.

So the particles travel less distance between the collisions, so there's more collisions with the balloon's inner surface every second, well done if you've got those.

And here's your explanation of why adding more air to the balloon causes it to expand.

So we've got several points to mention there.

You're putting more air in it, and so you're getting more particle collisions on the inner surface every second.

The outwards pressure increases and that causes the rubber of the balloon to stretch, expand, as the balloon's volume increases, that then increases the surface area, and that decreases the number of particles hitting that surface per square centimetre per second.

And so the pressure's going to drop, and eventually, the balloon's going to stop expanded when the pressures are equal again, well done if you've got an explanation, something like that.

And now it's time for the second part of the lesson, and this is about pressure and volume calculations.

And we'll see there's a relationship between the pressure of a gas and the volume of a gas for a fixed amount of gas in a container.

So let's have a look at that.

If you have a container of gas and you can increase the volume of that container, what that's going to do is decrease the pressure of the gas inside it.

So I've got a container here with some gas particles in, it's got a smallish volume, there's going to be quite a lot of collisions per second between those particles and the container walls.

If I stretch to that container, make it larger somehow, then the particles have got further to go, and that means there's gonna be less collisions with them and the container walls per second, and overall, the pressure's going to drop.

They've got a greater distance to travel and so there's less collisions per second.

So every centimetre squared of that container surface is gonna have a lower pressure on it because of that increase in volume.

If I've got a certain massive gas inside the container, a fixed mass of it, then the pressure's going to be inversely proportional to the volume of the container.

What that means is, I start off with a container of volume V, it's gonna have a pressure p in it.

If I double the volume of that container, that's going to half the pressure.

So the pressure's fallen to p/2, it's half of what it was initially because there's less collisions with the surface per second.

On the other hand, if I squash that container and reduce the volume to a half of its original volume, that's going to double the pressure, there's gonna be twice as many collisions per second and that's gonna double the pressure inside the container, so I've got two times the pressure now.

So if I've got a fixed mass of gas inside a container and it's kept at a constant temperature, I can write out the relationship like this.

The pressure times the volume is a constant.

And if I use symbols, I can use p for pressure, V for volume, and I'll use a lowercase k for that constant.

So I've got pressure, p, and that's measured in Newton's per square metre or pascal's.

The volume, V is measured in cubic metres, and the constant is is measured in Newton metres or pascal metres cubed.

That equation allows the pressure of the gas to be calculated when it's been compressed or it's allowed to expand, as long again, as is a fixed mass of gas and its temperature doesn't change.

So let's have a look at an example of that.

I've got a cylinder of gas at a pressure of 40 kilonewtons per metre squared and I've got a volume of 1.

5 cubic metres.

What's the pressure when the gas is compressed to 0.

5 metres cubed? So I work out the solution to this in different stages.

First of all, I wanna find the value for that constant k.

Remember, that's a fixed value of that k, so I'm gonna try and find it.

So I write out the original equation, k = p x V, the constant is the precious times the volume, and I substitute in the numbers from their question, I've got the pressure, and I've got the volume, and then I can find the value for k and k is 60 kilonewtons per metre.

What I do then is I can use that value, remember it's a constant, it doesn't change when I compress the gas, so I'm going use that k to find the new pressure when the volume's decreased, I write out the expression again, but this time, I substitute that k, and I substitute in the new volume, which is not 0.

5 metres cubed, and I've got the new pressure in there, which is that p, and I find that value for p, p is equal to 60 kilonewtons per metre divided by 0.

5 metre cubed, and that gives me a new pressure of 120 kilonewtons per metre squared.

Okay, I'd like you to repeat that process following the same methods.

So at a volume of 0.

2 metres cubed, the pressure of a gas is 60 kilonewtons per metre squared, what's the pressure of the gas when its volume is 1.

2 metres cubed? So the volume has increased substantially there.

So pause video, work out your answer, and restart, please.

And welcome back, and the answer here is 10 kilonewtons per metre squared.

First of all, I find a value for k, which I've done here, and then I use that value in the equation again, substituting it in to find a new pressure, and I find a new pressure is 10 kilonewtons per metre squared.

Well done if you've got that.

It's time now for the second of the tasks.

So this is based around what we've just seen about compressing a gas and changing its volume.

So a sample of gases in a container is compressed into a smaller volume while it's temperature is kept constant.

Explain why the pressure of the gas is increased as it's compressed.

And the second question is a calculation.

I've got a sample of gas at a pressure of 150 kilonewtons per metre squared, it's got a volume of 0.

10 metres cubed.

Calculate a new pressure when the gas is allowed to expand to a volume of 3.

0 metres cubed.

Pause the video, work out your answers to hose two questions, and restart, please.

And welcome back again.

And here's the explanation of why the pressure of the gas increases as it's compressed and it's based around those collisions.

So the pressure is caused by collisions of the particles with the wall.

As the volume decreases, the distance the particles travel between collisions decreases so we're getting more collisions with the container wall more often.

There's a smaller surface area and the same number of particles, so there's more collisions per centimetre squared every second, and both of those factors result in more collisions per centimetre squared every second, and therefore, the pressure increases.

Well done if you've got an explanation, something like that.

And here's the answer to the calculation and basically again, you find the value of that constant k, and then you use that in the expression again, substitute it in to find the new pressure, which was 5.

0 kilonewtons per metre squared, well done if you got that.

And now it's time to move on to the final part of the lesson, and in it, we're going to look at the work done by a gas, and that basically means how a gas can produce a force that can do some mechanical work on some object, so let's do that, If you have a gas inside a cylinder, it's going to cause an overall outward force on the surfaces of that container that it's in.

So I've got a container here, and I'll put some gas in it.

All that's going to do is produce outward forces like this acting in all directions, I've just drawn a few of those directions.

If that gas is inside a container that can expand in some way, so in here, I've got it inside a piston, then the gas will be producing a force on the fixed walls, but also force on that movable wall, that part which I've labelled a piston.

And what can happen there is, that piston can be pushed outwards allowing the gas to expand.

So the volume of the gas is increasing, but it's doing that by producing a force moving that piston.

So the gas is producing that force, and so doing mechanical work, mechanical work is any force acting on an object and causing it to move.

So the gas can produce mechanical work when it expands.

We can calculate the mechanical work that a gas does by using the equation for work done.

Work done is normally written, as work done is a force times distance moved.

In this case, I'm gonna use average force because the force acting on the piston actually changes as that piston's moving.

So I've got my diagram here, the work done is the force times the distance, or the average force times the distance moved by that piston.

So as the gas is doing work, it's going to be transferring energy.

So some of the internal energy of the gas is transferred, and so I'm gonna have a reduction in the energy of the particles in the gas.

What that reduction means is the average speed of the particles is decreasing, the kinetic energy is decreasing, they're gonna be moving slower on average after the gas has expanded.

So the average speed of the particle decreases and that means that the temperature of the gas is going to decrease as well, the gas is going to cool as it expands.

Let's have an example of using that equation.

And expanding gas produces an average force of 40 newtons on a piston, as it's pushed through a distance of 20 centimetres.

Calculate the work done by the gas.

So all I need to do here is write up the equation.

Work done is average force times distance, substitute those two values in, the force.

40 newtons, the distance was 0.

2 metres, that's 20 centimetres, so I get work done of 8.

0 Joules.

Now you can have a go.

So an expanding gas does 60 Joules of work when it pushes a piston through a distance of 1.

2 metres.

Calculate the average force produced by the piston during the expansion.

Pause the video, work out your answer, and restart, please.

Welcome back, well, you start with the original equation work done is force times distance, you'll need to rearrange that to find a force because that's what I've asked for, so the force is the work done divided by the distance, I substitute the values in, that's 60 Joules divided by 1.

2 metres, and that gives a force of 50 newtons.

Well done if you've got that.

As I mentioned earlier, if a gas is allowed to expand quickly, it's going to cool down and we can actually use that effect.

The cooling effect is used in aerosols, so if I get an aerosol and I press the top button, that allows the gas to escape quite quickly, and as it does so, it's going to cool quite rapidly and we can use that effect on sports injuries.

You may have seen when somebody's had a collision playing football, they're sprayed with some sort of spray, and what that spray is doing is actually cooling the part that's been bruised.

That cooling effect is because the gas coming out of the aerosol is actually quite cold 'cause it's been allowed to expand very rapidly, and that cooling effect will help reduce the pain of the injury and reduce the bruising.

I said earlier that when you allow a gas to expand, it does work on its surroundings, it can do mechanical work on something, but the opposite's also true, when you compress a gas, when you squash a gas, you have to do mechanical work on it.

You have to put a force on it to compress it to make it occupy a smaller volume.

So you're doing work on the gas, you're providing it with extra energy, so you're going to increase the internal energy of the gas that you're compressing, and what that does is it causes the particles to speed up, their kinetic energy increases.

So if you compress a gas, the kinetic energy of the particles will increase, and that's going to cause the temperature of the gas to increase as you're compressing it.

And you can see this effect when you're pumping up a tyre.

If you've got a flat tyre on your bicycle, you can pump it up, as you pump it up, you are doing mechanical work on the gas, you are compressing it so that it goes into the tyre, and if you feel the pump after you've tried to pump it, then it's actually warmer, you've warmed up things because you're doing mechanical work.

So the pump becomes hot as you compress the gas in it.

Okay, let's see if you've understood the ideas about allowing gas to expand.

A fire extinguisher is filled with carbon dioxide gas under high pressure.

Why does the gas cool down when it's released and used to put out a fire? So pause the video, make a selection from those three, and restart, please.

Welcome back, hopefully, you selected this answer, the carbon dioxide gas does work as it expands, so the energy of its particles decreases, causing its temperature to fall.

So if you set off a carbon dioxide fire extinguisher, you actually get cold carbon dioxide gas coming out.

Well done if you got that.

And now it's time for the final task of the lesson, and what I'd like you to do is this, I'd like you to answer these two questions, A technician using a can of compressed air to remove dust from inside a computer, why does that can start to feel cold when it's been used? And the second question is a calculation.

I'd like you to calculate the work done by a gas as it expands based on that data, and suggest why the force on the piston that's moving, decreases as it's moved by that gas expanding.

Pause the video, work out your answers to those, and restart, please.

Welcome back.

well, here's the explanation about the can feeling colder.

So as the air is released from the can, it expands, and does work on the surrounding atmosphere, and that rapid expansion reduces the temperature of the released air and it cools the surroundings, so you'll feel the surroundings cool.

The pressure of the gas inside the can is also reduced, and that causes that to cool as well, and that will cool the can down as well.

So well done if you've got that.

And the second question, calculate the work done, or we can use work done is force times distance, that gives 100 Joules, and suggest why the force decreases.

What's happening is the pressure of the gas is gradually decreasing so as the gas moves, the piston expands, the volume of the gas increases and the pressure decreases.

The force on the person is given by the force equals pressure times the area,.

And so as the pressure decreases, that force is also going to decrease.

Well done if you've got that.

Now for a summary of the lesson and everything we've covered.

The pressure of a gas in a container can be increased by: increasing the mass of the gas in the container, increasing the temperature of the gas, or reducing the volume of the gas.

And we can use the particle model to explain why those things change thee pressure of that gas.

For a fixed mass of gas kept at a constant temperature, the pressure times the volume is equal to a constant.

An expanding gas can do work.

Work done is average force times the distance.

So as we allow our gas to expand, it does work.

The temperature of the gas decreases as it expands.

Also, if we compress a gas, it increases the temperature of that gas because work is done on the gas, increasing its internal energy.

Well done to reaching the end of the lesson, I'll see you in the next one!.