In the troposphere, the part of the atmosphere that extends...

In the $troposphere$, the part of the atmosphere that extends from earth's surface to an altitude of about 11 km, the temperature is not uniform but decreases with increasing elevation. (a) Show that if the temperature variation is approximated by the linear relationship $$T = T_0 - \alpha y$$ where $T_0$ is the temperature at the earth's surface and $T$ is the temperature at height $y$, the pressure $p$ at height $y$ is $$ln ( {p\over p_0} ) = {Mg \over R\alpha } ln ( {T_0 - \alpha y\over T_0} )$$ where $p0$ is the pressure at the earth's surface and $M$ is the molar mass for air. The coefficient $\alpha$ is called the lapse rate of temperature. It varies with atmospheric conditions, but an average value is about 0.6 C$^\circ$/100 m. (b) Show that the above result reduces to the result of Example 18.4 (Section 18.1) in the limit that $\alpha \rightarrow 0$. (c) With $\alpha$ = 0.6 C$^\circ$/100 m, calculate $p$ for $y$ = 8863 m and compare your answer to the result of Example 18.4. Take $T_0$ = 288 $K$ and $p0$ = 1.00 atm

In the $troposphere$, the part of the atmosphere that extends from earth's surface to an altitude of about 11 km, the temperature is not uniform but decreases with increasing elevation. (a) Show that if the temperature variation is approximated by the linear relationship $$T = T_0 - \alpha y$$ where $T_0$ is the temperature at the earth's surface and $T$ is the temperature at height $y$, the pressure $p$ at height $y$ is $$ln ( {p\over p_0} ) = {Mg \over R\alpha } ln ( {T_0 - \alpha y\over T_0} )$$
where $p0$ is the pressure at the earth's surface and $M$ is the molar mass for air. The coefficient $\alpha$ is called the lapse rate of temperature. It varies with atmospheric conditions, but an average value is about 0.6 C$^\circ$/100 m. (b) Show that the above result reduces to the result of Example 18.4 (Section 18.1) in the limit that $\alpha \rightarrow 0$. (c) With $\alpha$ = 0.6 C$^\circ$/100 m, calculate $p$ for $y$ = 8863 m and compare your answer to the result of Example 18.4. Take $T_0$ = 288 $K$ and $p0$ = 1.00 atm

Transcript

00:01 So essentially, the question is asking us to write the proof for the change in pressure as you add a certain elevation above earth's surface.

00:13 So it would be the pressure as a function of the height.

00:20 So it would be d .p.

00:24 D .y.

00:25 And this equals, this is a formula that equals the density of the air times the gravity.

00:30 And then they're saying to approximate the temperature change as this t equals to minus alpha y.

00:43 So t o would be temperature at y equals zero meters.

00:52 And then t would be temperature at y.

00:56 And then y would be height.

00:57 And then alpha they're saying is a lapse rate of temperature.

01:07 So it's simply just a constant, simply a constant that will keep it.

01:20 Simply constant to model how temperature changes.

01:23 So we're saying that temperature changes linearly as you increase, as you go higher into earth's atmosphere.

01:31 So let's define a y1 and a y2, and then we can go ahead and try to approximate this.

01:42 So we can say that y1 will be t minus to over alpha, and then we can say that, whoops, these are just defining the variables in order to make the integral work out.

02:08 So at this point we can say, okay, we need to find density.

02:10 So density, we can say pv equals nrt and then n equals the mass divided by the molar mass.

02:19 So now we can say pv equals mrt over m and then m over v, which is the definition for density.

02:28 This is going to be equal to pm over rt.

02:33 So at that point we can say, okay, we can plug this into, we can plug this into our equation right here.

02:43 Okay, so let's get a new workbook and say, okay, d .p over d, y equals negative p.

02:55 Whoa, apologize.

02:58 Not b, p.

03:03 So that would be negative.

03:04 P, m, g over r t.

03:12 And now we're going to have to put our p's on one side and move.

03:18 This d -y to the other side so we'll have d -p over p equals negative m -g over r -t -d -y and at that point we can say plug we can stop we can plug in for t and so that this is going to be equal to mg over r times and then it'll be dy over t sub o minus alpha y.

03:52 At this point, now we can actually integrate.

03:56 So we're going to integrate from p o to p, and this would be of dp over p.

04:03 And then we're going to integrate from y sub 1 to y sub 2.

04:08 And this would be we can just say d -y over t -o minus alpha -y is unconventional but save time and then we're going to actually factor this fact we can factor that out we can say negative mg over r from y1 to y2 of d -y over t -o minus alpha -y at this point we can use u substitution.

04:52 So we can say u equals t -o minus alpha -y d -u -d -y equals negative alpha and that means that d -y equals negative d -u over alpha.

05:09 And then if we were to substitute this, we can say this equals negative m -g over r and then this would be this is why it works because this now becomes t -o to t of 1 over u negative d -u over alpha so this negative cancels out with this negative and then we can move this alpha here and then erase it from the original equation so this these two negatives go out this so that's what that becomes so that would be what's that that integral becomes and then at that point this equals we know that this equals ln so we can just say equals negative m g over our alpha and then it would be ln of u ln of absolute value of you and then from t o to t at this point we can say okay let's get a new workbook here p o is a p of d p over p this is a p of d p over p this is a going to equal the l n of p from p o to p which is going to equal ln of p over p o and then this is going to be equal to negative m g over r alpha times l n of of t over t over t o we know that t is going to be equal to and this equation looks exactly like what they gave us.

07:26 So that would be the answer for part a.

07:28 That would be rather the proof.

07:30 So we found a density.

07:31 We found a density from the ideal gas law.

07:34 We figured out some bounds for why.

07:37 We had to use substitution in order to figure this, in order to figure this out, factor out the 1 over alpha, cancel out the negative with this negative right here, this d -y negative, and then find the integral of 1 over u.

07:55 Of course, it's a natural log.

07:58 D -p over d -p is also a natural log, and then we can simply plug in for our t, and t -o, of course, stays the same.

08:07 So natural log of the pressure at y meters over the pressure at the earth's, surface equals negative m g over r alpha times the natural log of the temperature at earth surface minus the lapse rate of temperature times the height divided by the temperature at earth's surface so at that point it's saying what would happen if all of this was um if this was uh alpha approach zero so let's actually get a new be so limit as alpha approaches 0 of 1 over t0 minus alpha y is going to equal 1 over t o.

08:56 And this is just taken from this from the first slide where we have first sorry from the second slide where we have this right here.

09:06 So instead of it being negative mg over r times dy over t o minus alpha y, it's simply going to be 1 over to for that t temperature.

09:16 So it would be d .p over d .y equals negative p m .g over rto, like that.

09:26 And so at that point, we can say, okay, this is going to be d .p over p...

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