# Pressure variations with altitude

## Main Question or Discussion Point

Sorry if this is in the wrong category but I had an exam question today that I couldn't figure out.

The problem states that at an altitude of of 9600m the air pressure is measured to be 25 kPa. The question is what is the pressure at sea level?

I feel like to answer this question you need to know something about temperature. Is it possible to get the pressure at sea level with the given information?

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I can't recall the correct formula but if memory serves me correct its a inverse direct ratio.

One thing to keep in mind on exam questions if temperature or gravity is a facter then they would include that information. Though temperature and gravity does affect pressure for this question its assumed to be identical as no other information is provided with that detail.

25/9600=x/1 this is direct to make it inverse we rotate one term
9600/25=x/1 now its an inverse direct ratio

then we cross multiply and devide

x=9600/25
x=384 kpa

this only works for direct raitio based questions

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That seems a little off though since 1 atm is about 101 kPa, so should the answer be in that ball park?

http://en.wikipedia.org/wiki/Atmospheric_pressure

the question itself is off as it does not provide the details to utilize the formula in the link I provided.
yes sea level standard pressure is 101325 pascals
however this has factors such as average sea level temperature, molar mass of dry air, universal gas constant and temperature lapse rate none of these details are provided so you must work with the details that are.

A similar question I had on an exam was the volume of an air bubble is 25 cm at 1000 ft below sea level what is the volume of the air bubble at sea level.

Good luck finding a formula for that one lol.
NOt all exam questions are real world examples or consider all the real world factors you must work with the terms provided. Others may correct my solution to this problem and I welcome that.

Ken G
Gold Member
I agree the problem seems to suggest you should imagine the temperature stays constant, but that last solution has a lot of problems. The equation of hydrostatic equilibrium (this one is worth committing to memory) says dP/dz = -rho*g and at constant T this says dP/dr = -P/H*(R/r)2), where H is a constant and gets called the "scale height". You can see what it has to be-- H=kT/mg, where m is the mean mass per particle and g is at the surface of the Earth, where r=R. We don't really care what H is here though, because we can redefine a new hight variable, say z=r/R, and get dP/dz = -P*R/H*(1/z)2). That's still not simple enough, so define y = R/H*(1 - 1/z), such that y goes from 0 to H/R. Now we have dP/dy =-P. That has a simple solution-- P/Po = e-y, where y separates P and Po. (Most likely on the scale of 9.6 km, they want you to approximate this with y ~ h/H, where h is the height above the surface, but I give you the full solution as if the constant T could really be continued arbitrarily out into space.)

Since you know P=25, and you want to infer Po at sea level, you need to know the value of y at 9600 km. That requires knowing R and H, so that you can find y=R/H*[1-R/(R+9600)], which will be quite close to 9600/H (that's the usual way people talk about the "number of scale heights above the surface"). So you don't really need to know R, but you'll need to calculate H. Apparently, the problem does need you to know what the T is (and what mg is also, remember m is for a mix of 80% N2 and 20% O2). Note also the curious aspect of this solution-- since y only goes up to R/H, not infinity, there is a finite pressure at infinite r, equal to Poe-R/H[/sub]. When considered for the Sun, this is related to why there is a solar wind. For Earth, e-R/H[/sub] is vastly small, and so escape of Earth's atmosphere depends more on things like ozone heating and interactions with the solar wind.

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As you can see from Ken's post that the real factor is providing the answer that the nstructor is looking for dependant on what has been taught to you. Exam questions are often poorly put together regardless of field of study. The unforunate part is the student often has to look at what the instructor wishes rather than what one would think if they happen to have a better understanding of the problem presented. This is due to the fact that all questions asked are set to an average student standard. As best as possible in decent education systems.

By the way the direct/indirect ratio mathematics will serve you well in real life applications and is definetely one that cannot be stressed enough. Not all situations can be described accurately mathematically as factors may be unknown but if you know the relation of one factor to another you can get a close approximation. Ratio mathematics is often used in applications such as motion controls, pressure temperature etc where not all the factors is fully understood, but rather inferred through baseline testing results and graphing of conditions.

Ken G
Gold Member
I agree the problem seems to suggest you should imagine the temperature stays constant, but that last solution has a lot of problems. The equation of hydrostatic equilibrium (this one is worth committing to memory) says dP/dz = -rho*g and at constant T this says dP/dr = -P/H*(R/r)2), where H is a constant and gets called the "scale height". You can see what it has to be-- H=kT/mg, where m is the mean mass per particle and g is at the surface of the Earth, where r=R. We can redefine a new hight variable, say z=r/R, and get dP/dz = -P*R/H*(1/z)2). That's still not simple enough, so define y = R/H*(1 - 1/z), such that y goes from 0 to H/R. Now we have dP/dy =-P. That has a simple solution-- P/Po = e-y, where y separates P and Po. (Most likely on the scale of 9.6 km, they want you to approximate this with y ~ h/H, where h is the height above the surface, but I give you the full solution as if the constant T could really be continued arbitrarily out into space.)

Since you know P=25, and you want to infer Po at sea level, you need to know the value of y at 9600 km. That requires knowing R and H, so that you can find y=R/H*[1-R/(R+9600)], which will be quite close to 9600/H (that's the usual way people talk about the "number of scale heights above the surface"). So you don't really need to know R, but you'll need to calculate H. Apparently, the problem does need you to know what the T is (and what mg is also, remember m is for a mix of 80% N2 and 20% O2). Note also the curious aspect of this solution-- since y only goes up to R/H, not infinity, there is a finite pressure at infinite r, equal to Poe-R/H. When considered for the Sun, this is related to why there is a solar wind. For Earth, e-R/H is vastly small, and so escape of Earth's atmosphere depends more on things like ozone heating and interactions with the solar wind.
(ETA: fixed typo, clarify you need H)

Sorry if this is in the wrong category but I had an exam question today that I couldn't figure out.

The problem states that at an altitude of of 9600m the air pressure is measured to be 25 kPa. The question is what is the pressure at sea level?

I feel like to answer this question you need to know something about temperature. Is it possible to get the pressure at sea level with the given information?
Given only the information in your post, it is not possible to answer the question without making certain assumptions. I assume that what the exam wants you to do is to apply the barometric formula to an idealized atmosphere.

As Ken G pointed out, this formula requires the mean molecular mass of the air to be constant from 9600 m to sea level (never happens in the real world), the temperature to be constant over that range (never happens in the real world), and the gravitational constant to be constant over that range (never happens in the real world, but doesn't make much difference given the accuracy of most real world atmospheric pressure readings). If you want to make all of those assumptions, then you can follow Ken's instructions and get an idealized answer.

As Mordred pointed out, this is a badly-worded question.