Determining pressure at various altitudes using integration

AI Thread Summary
The discussion focuses on calculating air pressure at high altitudes, specifically at the summits of Mt. McKinley and Mt. Everest, and determining the elevation where air pressure is one-fourth of sea level pressure. Participants explore the relationship between air density and pressure, noting that density is variable and proportional to pressure at different heights. Integration is suggested as a method to solve the problem, but there are challenges in understanding its application, particularly regarding the correct formulation of the equations. Errors identified include the need to consider the direction of pressure change with altitude and the omission of the constant of integration. The conversation emphasizes the importance of correctly applying the principles of physics and mathematics to derive accurate results.
joseph_kijewski
Messages
12
Reaction score
0
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.
 
Physics news on Phys.org
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.
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Back
Top