atmospheric-pressure-leftrightarrow-atmospheric-pressure-p-in-kilopascals-mathrmkpa-at-altitude-h-in

Question

Atmospheric Pressure $\Leftrightarrow$ Atmospheric pressure $P$ (in kilopascals, $\mathrm{kPa}$ ) at altitude $h$ (in kilometers, $\mathrm{km}$ ) is governed by the formula
$$
\ln \left(\frac{P}{P_{0}}\right)=-\frac{h}{k}
$$
where $k=7$ and $P_{0}=100 \mathrm{kPa}$ are constants.
a. Solve the equation for $P$.
b. Use part (a) to find the pressure $P$ at an altitude of $4 \mathrm{km}$.

Ankit Gupta · Accepted Answer

at most Eric pressure is given by equation Allen Be overpay Nord equals minus Etch Viking. There, at 10-K I can&#x27;t be known are constants. So we were KNP. Nor are constants and the value off case given to us at seven. On value off penal is 100 kilo Paschal No, in part a. We need to find our pressure. Atmospheric pressure. So equation four be that is atmospheric pressure at altitude edge, we will write Add a question here which is given to us Ln p l and b overpay nor equals minus vay cay Where at his altitude taking exponential both side bigger, bi, bi, bi nor equals it is to about minus at white gay. Now for the multiplying this equation throughout Bipin or Bigger beak will spin or in tow years to part minus X bike even substituted the value off B bean, R and K here, so bigger p equals 100 into eaters to pound minus at by seven. So this is that equation four at most. Eric pressure at altitude. It now for part B, we need to find the pressure at most. Eric pressure at on altitude off four kilometer. That is that is full. Now we will put the value at its port in equation number one and get at most Eric pressure. So P equals 100 in tow. Eaters to par minus pulled by seven. Since it is food now, calculating this, we get 100 in tow. We will use a calculator to solve this. So we get 0.56472 They&#x27;re from a deep Langford that we get 56.472 Therefore, we have God that our atmospheric pressure like an altitude of four kilometers 56.472 kilo Pastor.

Sam Sohn · Answer

So we&#x27;re given that the natural log of P divided by P zero is equal to negative a trove. Okay, so part is asking us to solve the equation for P as in leave the P on one side and move everything else the other side. We&#x27;re going to do that first, but taking the exponents of both sides, as in we&#x27;re going to do e to the natural log of p. The vote a P zero is equal to e to the negative h over K. Now we got e to the natural log and remember e to the national log of X is just equal to X, which means we can simplify this left hand side. As p, divided by p zero, is equal to e to the negative h over K. And now this offer P only to do is multiply both sides by p zero and get P is equal to p zero times e to negative h overcame. So that&#x27;s our answer. Depart A. And now part B is asking us to plug in our values where p zero is 100 k p a Caisse seven and h is for So we do that P is equal to 100 times e to the negative four over seven, and that&#x27;s going to turn out to be approximately 56 0.47 k p a. KP is We&#x27;re pill. Kill a Pascal, by the way. So that&#x27;s it. We&#x27;re done.

Ryan Mcalister · Answer

this question. Ask you about air pressure depending on your elevation. It gives you the equation that air pressure P is defined in terms of peanuts, a constant que a constant and your height h the equation is the natural log of P overpay not equals negative H over K. The first part of this question wants you to rearrange this equation to solve for P. So the first thing we&#x27;ll do is we&#x27;re gonna try to get P and Peanut out of this equation out of this longer them. To do that, we&#x27;re gonna need to use the exponential form of this law grammar. So as we know we&#x27;re going from a longer than to his exponential form, we take the base of the log, which in this case is E raise it to the power of the answer, which is negative h over K and that will be equal to the inside, which is P over peanut from here to get pee on his own. It looks like all we had to do is multiply both sides by peanut. This will give us the answer that p equals peanut times e to the negative h over K next it asks us to find the pressure when the height of seven kilometers. Now, for this, we&#x27;re gonna need to know what these constants are. We knew that peanut is 100 killer pass scales, so I&#x27;m just gonna write in 100. We also know that from the problem that h is Excuse me. Four kilometers. So agents four and it K, is it a constant seven? So now that&#x27;s left is to play this into our calculators and we will get unanswered of 56 0.47 Kila pass scales and those are your final answers.

Stephen Hobbs · Answer

given this differential equation, then we could derive that the original function is going to be an exponential. And so this will be a function of height and this will give us the pressure. And so it&#x27;s going to be equal to the initial pressure multiplied by an exponential times, the decay factor multiplied by not not time but height. And then we&#x27;re given some initial conditions here. When the height of zero or at sea level, we have a pressure of 10 13 millibars and we could go ahead and substitute these in. And so we&#x27;ll end up with 10. 13 on the lot. Site is equal to peanuts, multiplied by the exponential to the power zero since h zero and then that knocks out the exponential. And therefore, peanut is 10 30 and next we&#x27;re gonna solve four K. So we&#x27;re gonna use that pressure function, and we&#x27;re going to use the information that the pressure is 90 millibars. And of course, we have the initial pressure here, but the pressure is 90 millibars at 20 kilometers. So we&#x27;ll go ahead, but that as r h. And then multiply that by RK and we want again sulfur RK value here and we could do this and two steps here, we&#x27;re gonna first divide the 10 13 over. So we&#x27;ll have 90 over Tom 13. Then we have e on the right side. So this at this point, we&#x27;re gonna take the natural log of both sides, and that will knock out the exponential on the right side. And so then that just leaves us with 20. Okay. And then, of course, we want a sulfur case. So we&#x27;re gonna divide by 20 and just for reference, this is equal to about negative 0.1 to 1. But I&#x27;m just gonna go ahead and leave that as this here. So this is our K value. And so we now have all the information for the, um, original function. And so, for part B, we could go ahead and solve a specific situation such as Part B is asking, What is the pressure at a height of 50 So p of h, what is this pressure? We know that we have our initial on 13 millibars by by the exponential plans K, what we just found over here, all of that is okay. Thank you. And then multiplied by our height, which is 50 kilometers. And so you could just use some graphing utility here to solve for this, but p of 50 will equal approximately 2.38 Bella bars. So we&#x27;ll just put em be Carmela bars. And then lastly, part C we want to solve. For what altitude does the pressure of 900 exists? So we&#x27;ll set the pressure on the left side equal to 900 and then keeping everything that we had originally. Number right side. This process is going to be very similar to how we solved for K. Except we&#x27;re Soling for H this time. Uh, sorry. He to the K times each. Okay, dividing both sides by 10. 13 we&#x27;ll have 900 over 10 13 and then we&#x27;re going to take the natural law. Good. Both sides. Ellen, that gets rid of the E on the right side. So then we have k times h. But of course we know. OK, so that is unknown. So we could just divide that OK over to both sides and so finally or H is going to be equal to whatever that&#x27;s quantity here gets us and that ends up being approximately equal to zero 0.9 77 kilometers

Brittany Knowlton · Answer

So for this problem, we&#x27;re looking atmospheric pressure and so were given that the Earth&#x27;s atmosphere of pressure P is often model by assuming that the rate dp th which p changes without h of sea level is proportional to pee, and we&#x27;re given some initial values. So for the party, we&#x27;re supposed to solve the initial value problem. So we&#x27;re given that DPD ish is equal to K P. So we already know from book. That&#x27;s so this is an exponential decay problem. So write another The solution. Tio these tax problems, it&#x27;s just going to be p is equal. Tio he not he to the K H. McGinnis is seen in your section in the book. So now, looking at some initial values they&#x27;ve given us, I first see that says suppose of the pressure Psi level is 1,013 millibars. So pressure at sea level, that&#x27;s going to be my initial value. So, given we ready know that our initial value uh, C level is 10 13. So it&#x27;s an answer to one of my problems that we&#x27;re supposed to find next to says that the pressure, an altitude of 20 kilometers is 90 millibars, so that is telling us that we solve p for on altitude 20 kilometers. We get that the pressure is 90 millibars. So we&#x27;re in uses, given information to vines everything else that we need. So the first thing I wanna do is express it in terms of h. So we&#x27;re almost there and we just now plug in some things that we know. So I know what my peanut is. So I have p equal to 10. 13 eat to the K H. And so the last thing I need to find is what this value of K is. So to find that I&#x27;m gonna use this other given information that we were given in the problem. Someone plug in 20 for my H and 90 for my p about 90 is equal to 10. 13 e to 20. Okay, so as we learned in the last section saw this I need a first take the natural log of both sides have natural log of 90 sequel to Natural Like Of this, which product rule is just 10 13 plus the natural log of he&#x27;s a 20 k and now let&#x27;s attract over that natural log of 10 13 to the other science. I get National 90 minus the natural log of 10 15 single too. This one we learned last time that this is actually just equal to 20 k. So want to go ahead and buy that 20 as well? So I came by itself. And if you put this in a calculator, is she get that? Kay is approximately negative. 0.121 So for final answer, Yes, But my answer is this Kay people that you know And here it is in terms of Sam H ands. So sorry. Now going on, too. Part B. That&#x27;s asking us what that Miss trick pressure is that H is equal to 50 co monitors. So we&#x27;re given h is equal to 50. We want to find the pressure p. So we&#x27;re gonna do is just plug in that 50 into our equations sort of peace equal to 10. 13 e to the 50 times Where are we now? Found Arcade B, which was negative 0.121 It&#x27;s not a first. Just multiply out my exponents and I get negative 6.5 and putting us in a car killer issue. This is approximately 2.38 nine Metal bars you in an Alfa parts. See, we&#x27;re kind of going the opposite way were given the altitude is 900 millibars. So that&#x27;s, um, now we&#x27;re going to The pressure is 900. Oliver Soapy is equal to 900 we want to find the altitude that gets that. So now we have 900 is equal to 10. 13 e two negatives. Air 20.121 each and a little bit good about same way. Take natural log on both sides and I get it. This again right here. I&#x27;m using my product rule when I subtract this natural log of 10. 13 to the other side looking here, This is really the same thing as negative 0.121 h someone. I go ahead and divide this onto the other side as well. So I just have aged by itself. And when you put that into your calculator, you should get that This is approximately 0.9777 Christ

Heather Zimmers · Answer

in this problem. We&#x27;re told that the rate of change of the atmospheric pressure with respect to altitude, which we would describe as D p. D. H is proportional to pee and is proportional. The P means it equals some constant K Times peak. And whenever we have a relationship like this, what we&#x27;ve seen in this lesson is that it translates to the exponential growth and decay equation, which would be P equals p, not times e to the k times h so we can use this equation to solve the problem. In addition to the initial conditions that were given. We&#x27;re told that at a height of zero, the pressure is 101.3 Killer Pascal&#x27;s and we&#x27;re told that at a height of 1000 meters, the pressure is 87.14 Killer Pascal&#x27;s. So let&#x27;s use that information and find the value of K. All right, so we&#x27;re going to substitute 87.14 in for the final pressure and 101.3 and for the initial pressure and will substitute 1000 in for the height and we&#x27;ll solve this for K, so we&#x27;ll divide both sides by one a 1.3, and as a fraction that ratio is going to simplify to be 4357 over 5065. So that equals E to the 1000 K. Then we take the natural log on both sides, and then we divide both sides by 1000 and will have the value of K. All right, now, this is not a very user friendly, user friendly number. It&#x27;s around negative 1.5 times 10 to the negative fourth. So what I decided to do is store this in my calculator so that I don&#x27;t use any so that I don&#x27;t lose any accuracy when I go to use it again. All right, so now we want to use that value in our model, and our model is P equals one. A 1.3. The initial pressure at zero degrees of altitude, times E to the K. I&#x27;m just gonna call it K. For now, we know the number, but I don&#x27;t want to write that number every time. So K times h. So let&#x27;s use that model to do parts A and B. So for party were finding the pressure at an altitude of 3000 meters. So we substitute 3000. And for H, we&#x27;re going to use the K value that was stored. Put that in the calculator and we get approximately 64.48 killing Pascal&#x27;s. Now. The book answer is very close. They have 64.59 could be that a rounded value was used to get the book answer rather than the exact value that I used. So I suspect that mine might be slightly more accurate because I didn&#x27;t round anything. Now we&#x27;re gonna go on to Part B and we&#x27;re doing the same thing. But we&#x27;re finding the altitude. We&#x27;re finding the pressure at an altitude of 6187 meters, and that would be the top of Mount McKinley. So we substitute 6187 in for the height. Use your stored value of K, put that in the calculator and we get approximately 39.91 Killer Pascal&#x27;s of pressure and the book has 40.5 again, probably just a rounding thing

Problem

Suppose you're driving your car on a cold winter …

Problem 99 Hard Difficulty

Answer

(a) $P=P_{0} e^{-h / k}$
(b) $56.47 \mathrm{kPa}$

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(a) $P=P_{0} e^{-h / k}$(b) $56.47 \mathrm{kPa}$

College Algebra

Grace H.

Alayna H.

Caleb E.

Kristen K.

Absolute Value - Example 1

Absolute Value - Example 2

James Stewart, Lothar Redlin, Saleem WatsonCollege Algebra

Grace H.

Alayna H.

Caleb E.

Kristen K.

(a) $P=P_{0} e^{-h / k}$
(b) $56.47 \mathrm{kPa}$

James Stewart, Lothar Redlin, Saleem Watson

College Algebra