Am I doing this right? (pressure, constant density, mount everest)

Click For Summary

Homework Help Overview

The discussion revolves around calculating the barometric pressure at the summit of Mt. Everest, utilizing the assumption of constant air density. Participants are exploring the implications of this assumption in the context of atmospheric physics.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants are attempting to compute pressure using the formula p = density * g * h and questioning the validity of assuming constant density at different altitudes. There are discussions about the height of the air column and its effect on pressure calculations.

Discussion Status

There is an ongoing exploration of the implications of negative pressure results and the assumptions regarding air density. Some participants are questioning the initial assumptions and discussing the variability of air density with altitude, indicating a productive direction in the conversation.

Contextual Notes

Participants are grappling with the concept that air density is not constant throughout the atmosphere, which is a key factor in their calculations. There is also mention of the specific values used for density and gravitational acceleration, as well as the height of Mt. Everest.

endeavor
Messages
174
Reaction score
0
"A student decides to compute the standard barometric reading on top of Mt. Everest (29,028ft) by assuming the density of air has the same constant density as the sea level. Try this yourself. What does the result tell you?"

Am I doing this right?
h = 8847.7344 m
density of air = 1.29kg/m3
pressure = density * g * h = 111.853 kPa

Since the by using constant density, we see that the pressure is higher on Mt. Everest. Of course, we know this to be false. Therefore, the density of air must be less on Mt. Everest than at sea level.

I'm not sure this is right, but please check.
 
Physics news on Phys.org
What is h here. Did you think about that. The column of air above us is important. Not the one below us. Pressure is exerted due to the exertion of force over an area. The Force is applied due to the weight of the air above us and thus the atmosphere experience a pressure due to the weight of the air above it.
So let the height of atmosphere from the surface of Earth be R. Thus the chane in height of air column is R - h, where h is the height of Mt. Everest.
 
is my equation correct? p = d * g * h?
if h is R - h, then
p = pat sea level - density * g * h

but density * g * h = 111.853 kPa, and then p would be negative!

I'm confused.
 
I didn't plug in the numbers, but if your pressure is negative, what does that tell you about either the assumption that density of air at that altitude is the same as sea level, or g, or both?
 
Why should you think it is negative. In fact now I am confused abt what you are confused?
R - h > 0 since R>h as R is the height of air column above Earth surface -sea level - while h is the height of Mt. Everest. dgR is 101.325KPas, while dgh is less than that.
P = dg(R - h) just as P(at sea leavel) = dgR = 101.325KP
 
sorry, this is my first time learning physics...

if P(at sea level) = 101325 Pa and P = dgR, and d = density of air = 1.29 kg/m3[/sub], then
R = 8015m?

ptotal = density * g * (R - h)
ptotal = (density * g * R) - (density * g * h)
density = 1.29kg/3, g = 9.8m/s2[/sub], h = 8847.7344 m
then (density * g * h) = 111853.0583 Pa, which is greater than 101325 Pa, and therefore ptotal < 0.
 
I got disconnected yesterday before your reply.

Density of air is not constant along the whole column of air at atmosphere. That is why the result is interesting. I have not looked it. Thus the density of air input should be proper enough. 1.29 is the density of air nearby Earth's surface.
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 7 ·
Replies
7
Views
9K
  • · Replies 2 ·
Replies
2
Views
5K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 5 ·
Replies
5
Views
6K
  • · Replies 4 ·
Replies
4
Views
3K
Replies
2
Views
1K
  • · Replies 5 ·
Replies
5
Views
2K