00:01
Here we're going to be looking at how pressure varies with the height internal to a fluid.
00:08
So here's kind of the little picture.
00:11
There's a bottom of the fluid.
00:12
We'll call that z -equal zero.
00:16
And the reason why pressure varies is because as you go further and further down to the bottom of the fluid, there's more and more weight above you.
00:31
One of the things to recall is that pressure is force per unit area.
00:38
So it's f over area.
00:43
And so the difference in pressure, so there's a slab that's showing a cross -sectional area.
00:50
And the reason why there's a difference in pressure between the bottom of this slab and the top, so we'll call that p bottom and the top, is because there is the weight of the fluid inside of that slab that's sitting on the bottom, but not the top.
01:16
So the difference in pressure, p top minus p bottom, is equal to, it's less on the top, equal to minus the weight.
01:33
That slab has mass m, m times g is the weight per unit area.
01:39
So the force is the weight of the slab.
01:42
And of course the cross -sectional area.
01:45
And if we want to write the mass is equal to the density of the fluid times the volume of that slab.
02:01
So i'll try to make my density symbol look quite different from pressure.
02:06
That's one of the problems that comes into being.
02:09
So the slab has not only a cross -sectional area, but a volume, which comes about through the cross -sectional area times the thickness of that slab.
02:24
We'll call that delta z.
02:26
So the difference in pressure is equal to minus row times volume g per unit area.
02:37
And now we'll go ahead and substitute for the volume.
02:45
Delta pressure is equal to minus row area, delta z.
02:51
G over area.
02:55
And so notice that the cross -sectional area will cancel out.
03:00
And we have delta p pressure is minus the density row g times delta z.
03:12
And it's this equation that we can turn into a different differential equation for the pressure as a function of height within the fluid.
03:24
So we're delta z approaches zero, and we'll get a very infinitesimal small change in the pressure with an infinitesimal change in z is minus row times g.
03:41
Okay.
03:42
Now, often when you're looking at things like water over a short amount of distance, you can treat the density as constant.
03:52
And then you have a very simple differential equation to solve.
03:59
Typically use separate and integrate with this.
04:08
But before we do that, we are going to look at density as a function of z inside of a gas column.
04:22
And we are going to treat that gas as an ideal gas.
04:31
So the nitrogen that makes up our atmosphere is by and large an ideal gas.
04:42
So we will go ahead and look at the ideal gas law, which says that pressure times volume is equal to nrt.
05:03
And remember that n is the number of moles of the gas.
05:08
And we don't want to.
05:10
To count that through the atmosphere.
05:12
So we're going to divide both sides by the volume of a sample, and then we don't care how big the sample is or how small the sample is.
05:23
But we're going to represent the number of moles as the mass of the sample, little m, so it's kind of like the mass of that slab, divided by the mass per mole.
05:36
So big m is the mass per mole.
05:46
And that is another way to represent the number of moles.
05:52
And then we can see that the pressure is directly related to the density.
06:07
So r is the gas constant.
06:09
M is a constant that depends on the species of gas.
06:16
If we're talking on atmosphere, it's mostly nitrogen.
06:19
And t is temperature.
06:22
And so we can see that the density is directly related to pressure.
06:33
And this is the model that we're going to put into our differential equation.
06:46
So rather than thinking of the density as constant, we're going to think of it as a function of pressure.
06:54
And of course, the more pressure there is, the bigger the density.
06:58
The gas will get squeezed into a more compressed state.
07:02
So we're going to sub this into the differential equation.
07:15
Okay, where that belongs.
07:20
And we get d .p.
07:28
By dz is equal to minus mg pressure times r.
07:48
And this is the differential equation that we want to separate and integrate.
08:02
Okay, so separate.
08:04
We just simply have to cross -multiply the dz and the p.
08:10
And we'll have the pressure on one side.
08:15
And we will want to take a look at the right -hand side as well.
08:21
And most of the stuff in there are constants, but the one thing that is potentially not a constant is the temperature.
08:30
So a reminder that as you go up in height, say, climbing a mountain, usually your temperature decreases due to the greenhouse effect, essentially.
08:44
But we can go ahead, whoops, and i forgot my dz as yaking too much there.
08:49
So let's get the dz in there.
08:51
And that's what we want to integrate.
08:55
And i'm going to go ahead rather than put an arbitrary constant of integration in, i'm going to put in two limits, a p bottom, we'll call it pb, and a p top, and a z bottom, which we're going to set equal to zero and z top.
09:21
Okay.
09:22
So that's one way to avoid the constant is to put actual limits in.
09:30
And on the left we have the logarithm of p top over p bottom.
09:38
I won't write out those words.
09:40
That's kind of awkward.
09:42
Is equal to minus, let's take all the constants out, m g over r times the integral of temperature of z, from z bottom.
10:00
We'll call that zero to z top...
