It's possible to use the ideal-gas law to show that the density of the earth's atmosphere decreases exponentially with height. That is, p = p₀ exp (─z/z₀), where z is the height above sea level, p₀ is the density at sea level (you can use the Table 14.1 value), and z₀ is called the scale height of the atmosphere. (b) What is the density of the air in Denver, at an elevation of 1600 m? What percent of sea-level density is this?

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Hey everyone. Let's go through this problem starting off by reading through it. It can be shown using the ideal gas law that the density of a gas compressed by gravity changes exponentially following the equation row equals row knot multiplied by the oiler number. Raise the power of negative Y divided by why not? Where? Why is the displacement from sea level? Where upwards is positive row knot is equal to 1.28 kg per cubic meter and is the density of air at sea level. And why not is the scale height of the atmosphere? Part one of the problem asks use the weight of an air column to determine the value of why not? Part two of the problem tells us that a commercial airline is cruising at 11,500 m above sea level. Express air density at the cruising altitude as a percentage of air density at sea level. Now, the problem provides us with four multiple choice options. Option A for part 18.7 times 10 to the power of three m. And for part 2 86.7% option B says for part 17.92 times 10 to the power of four m. And then for part 2 86.5% option C says for part 17.92 multiplied by 10 to the power of four m. And for part 2 98.6%. And option D says for part 18.7 multiplied by 10 to the power of three m. And then for part 2 24.0% let's start this problem by just trying to solve part one of the problem which asks us to find the value of why not? This problem is a little tricky because since why not is kind of an unusual variable that's defined specifically for this problem, we don't have too many different options on what we can do with it. However, since why not exists in the density equation that is given to us by the problem, the row equals row knot E equation, we can use this density and try and relate it to the pressure and the forces that exist in the column of air that the problem hints that we can create. So the first thing I'm going to do then is I'm going to draw a little diagram here showing an imaginary column of air that I'm kind of arbitrarily picking out for the sake of this problem. This will be a column of air that in theory goes all the way up through the top of the atmosphere and covers some amount of air with some cross sectional surface area. A that can be defined somewhat arbitrarily. This is useful because there are two different ways we can represent the weight of the air in this column. One of those methods will involve density and therefore the density formula. And the other one of those methods will involve other variables and constants that we know. So the way I'm going to solve this problem is to write out both of those methods to solve for the weight of the air in the column in two different ways, set them equal to each other. And then algebraically solve that new equation for why not? And that's how I'm going to solve this problem. So let's begin with the first method and that first method will make use of the fact that pressure recall that pressure is equal to force divided by area. This is the equation I'm going to use for method. One solving this for the force, the weight that we want to find algebraically solving this force is equal to pressure multiplied by the surface area. For us, the force of the weight of the column on the air. I'm just gonna call that W is gonna be equal to the atmospheric pressure from the atmosphere, which is what the pressure is of the, of the whole column multiplied by the cross sectional area of that column. The atmospheric pressure P sub ATM is something that we can just look up and is generally constant all over the earth. The A on the other hand, is something that wasn't given to us in the problem since we arbitrarily defined it. So let's not worry about that right now. And now start looking at method two of finding the weight of the air column. The other method we can use for finding the weight of the air column is to make use of the fact that the pressure of the air is constantly changing. For different individual heights. We can take an integral of these individual height elements in order to integrate this across the entire height of the atmosphere and find the weight. That way, I'm going to define an infinitesimally thin strip of volume of the column. And I'm going to say that has a width of D Y. That means that one of an element of that, that corresponds to an element of volume of the column is going to correspond to D Y, the thickness of that area multiplied by the cross sectional area of that strip. So that means that the volume element D V is going to be equal to the cross sectional area. A multiplied by the thickness of that strip. Element D Y recall from earlier that the weight due to a pressure is equal to that pressure multiplied by the area over which it acts, which is the same a from earlier for our purposes, the pressure is going to be based on the height of the entire column of the atmosphere. So we're going to have to also recall the formula for the pressure due to a difference in height. So substituting in, for P, the density of the air row multiplied by the gravitational acceleration G multiplied by the height, the entire height of the column of atmosphere H and then we're just gonna multiply by A again as to continue and finish off the weight formula. But one way we can consolidate this down is instead of writing uh H multiplied by A, instead we can write V to represent the full volume of that column of air as we established earlier. However, in order to find the way using this method, we're going to have to integrate a small element along the height of the column. So the actual formula we're gonna want to use instead of W is D W, the element of weight that's based on whichever variable is changing. We have a row which is given to us in the problem as a function and multiplied by G which is constant, at least we're assuming it to be. And the volume strip is the thing we're integrating along. So instead of V, you write D V substituting in the D V equals AD Y formula. From earlier, we can write this more efficiently as row multiplied by G multiplied by AD Y to make it even more clear that it's the Y element that we're integrating along the height of the column. One final thing we need to account for is the fact that row, the density of the air as given to us by the problem is not constant. It is a function of Y. And since Y is the variable of integration, we need to include that the equation. So instead of writing just row, I'm gonna write row knot multiplied by E to the power of negative Y over Y divided by, why not then multiplied by G AD Y. Now we're ready to set up the integral. So the weight W is gonna be equal to the integral of row knots E to the power of negative Y divided by Y not multiplied by G multiplied by a multiplied by D Y. And the final thing we need to consider when setting up the integral are the bounds of integration. We're going to be integrating this along the height of the column of air. So the lower bound is going to be zero, the upper bound is a little trickier. However, because we're not necessarily given the height of the column of, we don't actually know how high up the air itself goes. We know that the pressure is going to be decreasing like a gradient as you move higher and higher up in the atmosphere. We don't know where exactly the pressure reaches zero or even if there's a point where the atmosphere ends at all necessarily. So for the upper bound, I'm going to use infinity, let's try doing that first and see if it works. So the first thing I'm going to do in actually solving this integral is pulling up the constants that I know aren't going to change and only get in the way. So the row knot is not going to change that's constant, the G is also constant and the cross sectional area A is constant. So the only thing in the integral itself is E to the power of negative Y divided by, why not? And we're integrating with respect to D Y. So I'm gonna continue writing this out and this integral in terms of actually solving it. This is a fairly simple chain rule type of integral where all we have to do is apply the fact that the derivative of E to the power of something doesn't change. And then we divide the whole thing by the derivative of the inner function of the thing that's being raised to the power of. So that would be E to the power of negative Y is divided by why not multiplied by one over? And then the inner derivative here and the derivative of negative Y over why not is equal to negative one divided by? Why not? So much more simply this is just written as uh negative row knot multiplied by G A. Why not E to the power of negative Y divided by Y? Not with the bounds of integration just being zero to infinity. So in order to get a numerical answer for this, we'll now have to apply the fundamental theorem of calculus. So we're going to plug infinity in for the equation and then, and then subtract what we would get from plugging in zero into it before I plug anything in though, first thing I want to do is make this look a little bit neater. I'm gonna write this as negative row knot G A, why not? And then instead of multiplying this by E to the power of a negative Y divided by Y not instead, I'm going to make use of the fact that something raised to the power of a negative exponent is the same as the whole thing of being divided by that thing er race, the power of the positive exponent. So instead of multiplying the whole thing by E to the power of a negative Y divided by? Why not? Instead I'm gonna take the expression so far and just divide it by the power by E to the power of positive Y divided by, why not? I'm doing it this way, partly because it looks neater and partly because it's going to make what I do next a little bit easier to see what's going on. So first we're gonna plug infinity into this. So that's negative row knot multiplied by G A Y knot. And then we're plugging the infinity in for the Y. So in the denominator, we get E raised to the power of infinity divided by why not? Now what's going to happen here is that the infinity is being put into the exponent of E. So this E term, this entire E term in the denominator is going to be made infinitely massive. And when you have something in a denominator being made infinitely massive, then the entire term essentially converges to zero. So this entire term is gonna be basically equal to zero. So that's the upper bound. And then we sub, then according to a fundamental theorem of calculus, we then subtract what we'd get for plugging in the lower bound. So lower bound is going to have zero, plugged in for Y. So that's negative row knot G A, why not? And then we're plugging a zero and for Y so it's E raised to the power of zero divided by why not? So first off, I want to point out that these negative signs are gonna cancel out. And the second thing I wanna point out is that now our E is being raised to the power of zero. And another important rule to know about exponents is that when you have something raised the power of zero, it always just becomes one. So this entire E term is just equal to one. So this entire final term here just becomes equal to whatever's in the numerator. So what we find is that this method gives us that the weight is equal to row knot multiplied by G multiplied by A Y knot. And that is the expression we get for the weight using method two. So now we have two different methods and two different expressions for the weight of the column due to its pressure. But now we, what we can do is set both of these terms equal to one another. So the weight from our method, one is equal to the weight that we found from our method to. So in other words, the row knot G A Y knot is going to be equal, the atmospheric pressure multiplied by a, the first thing we can do to simplify this is notice that since there's an A in both terms, they both cancel out. So we just have that P ATM the atmospheric pressure is equal to just row knot multiplied by G Y knot. And part one of the problems asking us to solve for. Why not? So what I'm going to do is algebraically solve this for why not by dividing both sides of the equation by row and not G. And what we find is that why not is equal to the atmospheric pressure divided by row, not multiplied by G. Now we just gotta plug the stuff we're given into a calculator. So for P atmosphere that's gonna be equal to again, we can just look this up. Recall that atmospheric pressure is equal to about 1.13 multiplied by 10 to the power of five has scales. And then we divide by row knot, which is given to us in the problem as 1. kilograms per cubic meter. And then we multiply this by the acceleration of gravity G which is equal to 9.81 m per second squared. And if we just put this into a calculator, then we find a value of about 8. multiplied by 10 to the power of three m. And if we check our multiple choice options up above, we can see that options A and D both include that as the answer for part A. So we can cross off options, B and C as possible answers because neither of them, they both have different answers for part one of the problem. Now let's look at the second part of the problem which A asks us to express the air density at the cruising altitude as a percentage of the air density at sea level. So recall that a percentage is basically a ratio multiplied by 100. In this case, the percentage, the ratio that we want to find is the air density at the cruising altitude divided by the air density at sea level. So the way the problem is defined the variables, the air density at the cruising altitude is just row and the air density at the sea level as again, as defined by the problem is row knot. So we want to find a valid ratio for these two variables and then multiply by 100. And what's incredibly convenient to us is that we're actually given a formula that we can use to find this very easily. The problem tells us that row is equal to row knot multiplied by E raised to the power of negative Y divided by Y not notice there is a row and a row knot in this equation. And if we divide both sides of the equation by row knot, then we find that ratio. And we're told that that ratio row divided by row knot is equal to E raise the power of Y divided by why not? The problem statement directly tells us to use 11,500 m as our value for the cruising altitude. So for our purposes, why is equal to 11,500 m? And as we discovered in part one of the problem, why not is equal to about 8. multiplied by 10 to the power of three m. So now let's just plug those two values in for why and why not? And get ratio. So row divided by row knot multiplied by 100 is equal to E raises the power of a negative 11, meters divided by 8.7 multiplied by 10 to the power of three m. And to turn it into a percentage. Again, we're multiplying the whole thing by 100. And if we put this into a calculator, then you'll get a value of approximately 24.0%. So this is the percentage that we find as the answer to this problem. And if we look at our multiple choice options, once again, we can very nicely see that one of our options does indeed include 24% for the second part of the problem. And this is contained in option D. So that means that the answer to this problem is option D. I hope this video helped you out. I hope things made sense. If you're still confused on some things I recommend you check out some of our other videos and that's it for now. If you check out some of our other videos, then hopefully they will give you a better idea of how to tackle problems like this in the future, but that's all for now and I hope you all have a lovely day. Bye bye.

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Key Concepts

Ideal Gas Law

Exponential Decay

Scale Height