Determining pressure at various altitudes using integration

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Homework Help Overview

The problem involves determining air pressure at various altitudes, specifically at the summits of Mt. McKinley and Mt. Everest, as well as finding the elevation where air pressure equals one fourth of sea level pressure. The context includes the assumption that air density is proportional to pressure at different heights.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between pressure and density, with some attempting to derive equations involving integration. Questions arise regarding the application of integration in this context and the handling of variable density.

Discussion Status

Some participants are exploring the implications of constant temperature on pressure and density relationships. There are indications of progress, but also confusion regarding the integration process and the correct application of signs in the equations. Errors have been identified, but no consensus on a resolution has been reached.

Contextual Notes

Participants are working under the assumption that air density varies with height and are attempting to apply the ideal gas law. There is mention of needing to consider constants of integration and the direction of pressure change with altitude.

joseph_kijewski
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1. The problem statement

Solve the following problems assuming air density is proportional to respective pressure at each height: What is the normal pressure at the atmosphere at the summit of a. Mt. McKinley, 6168m above sea level and b. Mt. Everest, 8850m above sea level c. At what elevation is the air pressure equal to one fourth of the pressure at sea level (assume the same air temperature).

Homework Equations



The Attempt at a Solution



I started by considering a block of air with the bottom at sea level and the top at the top of each of the mountains, such that (P+dp)A+dy*A*density*g-PA=0, which simplifies to dp=-dy*density*g. I can't figure out how to go on from here though, as the density is variable.
 
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joseph_kijewski said:
the density is variable.
Yes, and it tells you how to handle that:
joseph_kijewski said:
assuming air density is proportional to respective pressure at each height:
 
I just don't really understand how to do that. I've been trying to formulate an equation but I assume it involves integration which I can do in some cases but don't conceptually understand how to apply to this problem.
 
joseph_kijewski said:
I just don't really understand how to do that. I've been trying to formulate an equation but I assume it involves integration which I can do in some cases but don't conceptually understand how to apply to this problem.
$$\rho=kp$$ where k can be determined from the ideal gas law using the pressure and temperature at the Earth's surface.
 
I might (?) be making progress here. Assuming temperature is constant, nRT should be constant as well, thus P is inversely proportional to V, and hence directly proportional to density. Thus I can substitute in 1.29*P/P(sea level) for density, yielding the equation dp=g*(1.29P/101,325)*dy. From there I get dp/P=1.247668*10^-4. I then integrate from 0 to 6168 meters and get lnP=1.247668*10^-4*6168. Unfortunately, the answer I get is way off. Where am I going wrong here?
 
joseph_kijewski said:
Where am I going wrong here?
Two errors.
You need to pay attention to signs. Does the pressure go up or down as y increases?
You forgot the constant of integration.
 

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