Lesson video

This lesson is called diffusion and surface area to volume ratio, including Fick's law, and is from the unit transport and exchange surfaces in humans.

Hi there.

My name's Mrs. McCready, and I'm here to guide you through today's lesson.

So thank you very much for joining me.

I hope you're looking forward to it.

In our lesson today, we're going to calculate the surface area, volume, and surface area to volume ratio of different sized cubes, and we're going to observe how long it takes for a liquid to diffuse into them.

So I'm gonna come across a good number of keywords in our lesson to today, and they're shown there on the screen for you now.

You may wish to pause the video to make a note of them, but I will introduce them to you as we come across them.

So in our lesson today, we are going to firstly calculate surface area, volume and their ratio, and then we're going to have a look at the rate of diffusion in cubes.

So are you ready? I am.

Let's get started.

Now, surface area is the total area of the surface of the organism, and it can be calculated using the equation width times height times the number of faces, in this case of a cube.

So if we have an organism which we assume to be a perfect cube, such as this bacterium, if it has a width of two micrometres, its surface area is width times height times number of faces, so two times two times six.

So its surface area is 24 micrometres squared.

The volume of an object is the total matter inside the surface, and we can calculate the volume of a cube by timesing width by height by depth.

So if we take our cube shaped bacterium with a width of two micrometres, its volume, width times height times depth, is two by two by two.

So, two times two times two is eight, and so the total volume of our cube shaped bacterium is eight micrometres cubed.

Now, we can compare the surface area and volume as a ratio.

So we have seen that for a bacterium with a width of two micrometres, its surface area is 24 micrometres squared, and its volume is eight micrometres cubed.

So its surface area to volume ratio is 24:8, which we can simplify down to 3:1.

So its surface area to volume ratio in simplest terms is 3:1.

So let's calculate these together then.

Let's calculate firstly the surface area to volume ratio of a bacterium cube with a width of three micrometres.

So the surface area is width times height times the number of faces, which is three by three by six.

Three by three by six is 54 micrometres squared, so the surface area of this bacterium cube is 54 micrometres squared.

Now let's calculate the volume.

So the volume is width times height times depth, so that's three by three by three, and that equals 27.

So the volume is 27 micrometres cubed.

So now let's compare that as a ratio.

A surface area to volume ratio of 54:27, which we can simplify down to 2:1.

So that's my workings.

Now I'd like you to have a go and calculate the surface area to volume ratio for a bacterium cube with a height of six micrometres.

Remember to show all of your workings.

I'll give you five seconds, but you will need longer, so you may need to pause the video.

Okay, let's see how you got on then.

So to calculate surface area, we're doing width times height times the number of faces.

So for a bacterium cube with a height of six micrometres, we're timesing six by six by six, and that equals 216 micrometres squared.

Now, to calculate the volume, we're timesing width by height by depth, so that's also six times six times six, and therefore the volume is also 216, but this time the units are micrometres cubed.

So let's compare the surface area to volume ratio.

Surface area to volume ratio is 216:216, and we can simplify that down to 1:1.

So just check over your workings.

Did you get all of those correct? Well done if you did.

So let's compare these bacterial cubes.

So we're going from a bacterium with a width of three micrometres to a width of six micrometres.

In other words, we're doubling the width.

Now if we double the width, what happens to the surface area? Well, going from 54 to 216 is a multiplication of four, so the surface area has increased by four.

What about the width then? Well, to go from 27 to 216, it's actually an eightfold increase.

We're timesing that by eight.

And what about the ratio then? Well, the ratio of 2:1 reducing to 1:1 means that actually the ratio of surface area to volume is timesing by a half or dividing by two.

So we can see that as we double the width, we quadruple the surface area.

We have an eightfold increase in volume, but the surface area to volume ratio halves.

So what I'd like you to do now is to calculate the surface area to volume ratio of cubes that have widths of seven, 14 and 28 millimetres, and I'd like you to simplify those ratios to x to seven.

So if you divide them down so that the volume has a value of seven, what does that leave the surface area? That's what I'd like you to leave the simplification in.

Once you've done that, I'd like you to say what happens with the surface areas, the volumes and the ratios as the width doubles, and you can compare the seven, the 14 and the 28 millimetre cubes to help you with that description.

Then I'd like you to consider the fact that actually a sphere might be considered a more appropriate fit for some cells rather than a cube.

Now, to calculate the surface area of a sphere, you do four pi r squared, and to calculate the volume of a sphere, you do 3/4 pi r cubed.

So I'd like you to use those equations to show the surface area to volume ratio of a sphere with a radius of one centimetre.

So the radius is the width from the outer edge into the centre, as shown in the diagram on the right, and you can use the value 3.

14 as your value for pi.

So pause the video while you're having a go at these calculations, and come back to me when you are ready.

Okay, let's check our work then.

So firstly, I asked you to calculate the surface area to volume ratio of cubes of 7, 14 and 28 millimetres wide.

So let's start with seven millimetres.

So for seven millimetres, the surface area, which is width times height times the number of faces is seven times seven times six, which is 294 millimetres squared.

To calculate the volume, we're timesing the width by height by depth, so that is seven times seven times seven, which equals 343 millimetres cubed.

And the surface area to volume ratio is therefore 294 to 343, which if simplified down to something to seven makes a ratio of six to seven.

What about then for the 14 millimetre cube? Well, for surface area, it's 14 time 14 times six, which gives us a value of 1,176 millimetres squared.

And for volume, it's 14 times 14 times 14, which gives us a value of 2,744 millimetres cubed.

So the surface area to volume ratio is 1176:2744, and that simplified down to a ratio of seven is 3:7.

Then finally for the 28 millimetre cube, the surface area is 28 times 28 times six, which gives us a value of 4,704 millimetres squared, which gives us a value of 4,704 millimetres squared.

And volume is 28 times 28 times 28, and that gives us a value of 21,952 millimetres cubed.

So the surface area to volume ratio is 4,704:21,952, which simplified down to something to seven gives us a value of 1.

5:7.

So did you get all of those correct? Just check over your work quickly to make sure that you did.

And well done.

So now let's look at the trends in surface area and volumes.

So we've seen the values that we got for surface area volume and surface area to volume ratio for 7, 14 and 28 millimetres cubed.

So let's see the trends.

So when we're going from 7 to 14 and 14 to 28, we're in each case doubling the width.

Now, when we're doubling the width, we can see that we're timesing by four, so we're quadrupling the surface area, and that's true for both cases.

They're both a quadrupling of surface area.

But what about the volume? So the volume is increasing by eight times on both occasions.

Both are an eightfold increase.

And what about surface area to volume ratio? Well, we can see that they're both multiplying by 0.

5.

In other words, the surface area to volume ratio is halving.

So we can describe that in words to say that as the width doubles, the surface area quadruples.

Volume increases eightfold, and surface area to volume ratio halves.

So did you get all of those analysis correct? Well done if you did.

Then finally, I asked you to calculate the surface area to volume ratio of a sphere with a radius of one centimetre.

So let's see these calculations.

So for surface area, we're doing four pi r squared, so that means four times 3.

14 times one squared, and that gives us a value of 12.

56 centimetres squared.

Now to calculate volume, we need to use the equation 4/3 pi r cubed.

So to do that, we do four divided by three times 3.

14 times one cubed, and that gives us a value of 4.

19 centimetres cubed.

So that means that our surface area to volume ratio is 12.

56:4.

19, which is more or less simplified to 3:1.

So just check again your workings and make sure you got your answers correct, and well done again for doing that.

So there's some very obvious trends going on when we're increasing width, especially with cubes, and that's what we're going to explore next.

So, and now we're going to have a look at the rate of diffusion in cubes.

Now, the surface area and volume of the cube affects the diffusion rate, so the surface area impacts the diffusion rate.

In other words, how fast diffusion is happening.

And the volume of the cube affects the distance over which diffusion has to occur.

So both surface area and volume will impact diffusion.

Now, it's possible to see the effects of both the surface area and the volume on diffusion by observing the rate of diffusion of an acid through cubes of agar jelly.

And if those cubes are of different sizes, and we've got three here in the picture of one centimetre, one and a half centimetres and two centimetres, then we will see a difference in the rate of diffusion into the centre of these cubes.

So what I'd like you to do is just to watch this experiment video to observe diffusion of an acid in a single cube of agar jelly, so you've got a rough idea of what you are looking for.

Okay, so we've seen that in that example, the one centimetre agar cube completely discoloured in about 15 minutes.

So it took about 15 minutes for the liquid to diffuse half a centimetre, which is half the width of the cube.

15 minutes, that's quite a long time, isn't it? Now, we could predict how long it would take for a two centimetre wide cube to fully discolour.

So what do you think? How long is that going to take? It took 15 minutes for a one centimetre cube to fully discolour.

So what do you reckon it would take, how long do you think it would take for a two centimetre wide cube to fully discolour? Now we can observe those discolorations in the one centimetre, the one and a half centimetre and the two centimetre wide agar cubes.

How long do you think each one will take to discolour? So make your predictions before playing the video.

So in that experiment, we saw that the one centimetre cube fully discoloured in about 15 minutes, and the two centimetre cube fully discoloured in about 30 minutes.

Did you get it right? Did you make your prediction accurately? Well done if you did.

So we can see that by doubling the width, we've approximately doubled the time it takes to diffuse to the centre.

So here's another trend that we're observing in our data.

So why is that the case though? As the volume increases, we know that the surface area is increasing as well, so why does the time taken increase as well? So let's look at those numbers again.

We've got a width of times two, a surface area increase of times four, a volume increase of times eight, but a surface area to volume ratio of times half.

So now that means that there is four times the surface area, but eight times the volume in the larger cube.

In other words, twice as much volume through which the substance has to diffuse.

And you can see that perhaps in the little squares drawn on the cubes on the diagrams, there's just much further, much greater volume through which diffusion has to pass.

And therefore, because it is twice the volume, it's going to take twice as long, and that is why the time doubles.

Now if we go back to our model bacterium, we can compare our model bacterium to the agar jelly cubes that we've just been investigating.

So if the bacterium was two centimetres wide and it took about 30 minutes to diffuse nutrients to the centre, do you think it would survive? Hmm, it probably would not.

No, because 30 minutes is a very long time to wait for nutrients to get from the outer part of your body, well, from the outside actually, right into the centre of your cell.

So what can we do about that then? Well, we could increase surface area without losing volume, and we can do that by cutting the large cube up into smaller cubes, which would expose more surface area, and therefore increase the rate of diffusion, reduce the length of time it takes for substances to diffuse into the centre of each of the smaller cubes.

So what I'd like you to do is just to summarise our conclusions so far using these bullet points and the words increases or decreases and only those two.

So as width increases, what happens to surface area, volume, surface area to volume ratio and the time taken to diffuse into the centre? So I'm going to give you five seconds to think about it.

Okay? Let's see then.

So as width increases, surface area increases, volume increases, surface area to volume ratio decreases, and the time taken to diffuse into the centre of the cube increases.

Did you get all of those correct? Well done if you did.

So we've seen how rate of diffusion can be affected by surface area and volume, but the rate of diffusion is affected by three key factors, the concentration difference, that is how many particles are on one side of the membrane compared to the other, the surface area, and also the thickness of the membrane through which those particles are diffusing.

Now, if we have a look at the relationship between the rate of diffusion and those three factors, we can see that if we look at concentration difference, that as the concentration difference increases, so that is that there's a steeper gradient between those particles on one side of the membrane compared to the other, as the concentration difference increases, so does the rate of diffusion.

And we can describe this as a directly proportional relationship.

As one factor increases, so does the other.

So as the concentration difference increases, so does the rate of diffusion.

And we see the same relationship between rate of diffusion and surface area.

As the surface area increases, so does the rate of diffusion.

But if we have a look at the thickness of the membrane, we can see that the opposite is true.

As the membrane thickness increases, so the rate of diffusion decreases because the particles have a much greater distance over which they need to diffuse.

So as the thickness of the membrane increases, the rate of diffusion decreases, and we can call this an inversely proportional relationship because as one factor increases, the other one decreases.

So there are these three specific relationships between the rate of diffusion and these three factors.

Now we can summarise this relationship between all of these factors in an equation called Fick's law.

Fick's law states that the rate of diffusion is proportional to the surface area times concentration difference, all divided by membrane thickness.

So you can see the little symbol between the rate of diffusion on the left and the right hand side of the equation.

That means proportional to.

And the rate of diffusion is proportional to this arrangement of factors, surface area times concentration divided by membrane thickness.

Now, we can see this happening in actuality if we look in depth at certain examples in biology where diffusion is taking place.

So if we look at the alveoli in the lungs, for instance, we can see firstly that there is a vast surface area.

If we took our lungs and spread them out so that the alveoli were lying one layer thick, it would cover about half the size of a tennis court.

It truly is a very vast surface area.

In addition to that, we can see that there is a steep concentration gradient between the gases within the alveoli and the gases within the bloodstream flowing through the capillaries.

So there is a high concentration of oxygen in the alveoli compared to a much lower concentration of oxygen within the bloodstream, and a greater concentration of carbon dioxide in the bloodstream compared with the carbon dioxide within the alveoli.

So there are great concentration differences, very steep concentration gradients, which also means that a fast rate of diffusion will be maintained.

Then finally, if we look at the cells through which the diffusion has to take place, so the cells lining the alveoli and those lining the capillaries, and if we at them under the microscope, we'll see that they look as if they've been rolled flat, and that's because they are incredibly thin.

They're as thin as they can possibly be, and that means the distance between the alveoli and the blood in the capillaries is as short as it can be because those cells are as thin as they can be.

And when we put all of those three things together, the large surface area, the steep concentration difference, and the very thin cells through which the diffusion must occur, we can see why there is such a high rate of diffusion of gases between the lungs and the bloodstream.

So let's summarise this then.

So which of these factors will increase the rate of diffusion according to Fick's law? Is it A, reduction in membrane thickness, B, reduction in concentration difference, or C, increase in surface area? I'll give you five seconds to think about it.

Okay, so did you spot that a reduction in membrane thickness will increase the rate of diffusion? I hope you also spotted that an increase in surface area will also increase the rate of diffusion.

Well done if you got both of those.

So what I'd like you to do now is to consider these two questions.

Firstly, a two centimetre wide cube takes about 30 minutes to fully discolour.

If we cut a two centimetre wide cube into eight times one centimetre wide cubes, I would like you to suggest how long it would take the cubes to fully discolour, and then I'd like you to explain why that is the case.

Then I'd like you to suggest why an organism which does not have any mechanisms for reducing the diffusion distance, in other words, they're not chopped up into small pieces, will not be able to grow very large.

And I'd like you to use the ideas of diffusion distance and time taken to diffuse to help you write your response.

So pause the video now and come back to me when you're ready.

Okay, let's see what you've written.

So the two centimetre wide cube takes about 30 minutes.

How long would it take for each cube if we chopped those into eight times one centimetre wide cubes instead? So how long would it? Well, they would all discolour in about half the time, and half the time is 15 minutes.

Why? So you might have said for this that this is because there is a larger surface area to volume ratio, and after cutting, the equivalent area of surface has half the volume to serve in each small cube, compared to the original large cube, and therefore diffusion to the centre will take half the time.

Now, you may have phrased that slightly differently, but have you got the right gist of what we're trying to say here? Make sure your explanation is clear though, however you've phrased it.

And well done for having a go, 'cause this is actually a really difficult thing to try and explain.

Then I asked you to suggest why an organism which does not have any mechanism for reducing the diffusion distance will not be able to grow very large.

So your answer might have said that the diffusion distance will increase as the size of the organism increases.

And the time taken for diffusion into the centre of the organism will double as the width doubles, and therefore at some point, the time taken for diffusion will be just too long and the organism will not survive at any greater size.

So again, check your work over.

Have you got those salient points? Again, you might well have phrased it differently, but make sure your meaning is clear and accurate.

Well done again.

That's really quite tricky thing to try and explain, so well done if you've even got some of that right.

That is hard.

Okay, we've reached the end of our lesson today, and I hope you've enjoyed it.

So in our lesson today, we've seen that surface area and volume of cubes can be calculated and compared as a ratio.

And we saw that as the cube width doubles, the surface area quadruples and the volume increases eightfold.

We've also seen how we can observe the rate of diffusion using agar jelly cubes.

And we saw when we did that that as the width of the cube doubles, the time taken for the substance to diffuse into the centre also doubles.

Now, this has significant ramifications for living organisms because if living organisms increase their volume without significantly increasing their surface area, this will absolutely limit their viability because of the time taken for substances to diffuse over the large distances involved.

And the relationship between the rate of diffusion, surface area concentration difference, and membrane thickness can be summarised in Fick's law.

So I hope you've enjoyed our lesson today.

Thank you very much for joining me and I hope to see you again soon.

Bye.