Solve Air Pressure Problem: 500m Skyscraper, 1.3 kg/m3

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Homework Help Overview

The problem involves calculating the air pressure at the top of a 500-meter skyscraper, given a constant air density of 1.3 kg/m³ and an initial pressure of 1 bar at ground level. Participants are exploring the relationship between pressure, density, and height in the context of atmospheric pressure changes with elevation.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the formula relating pressure to density and height, questioning the interpretation of force versus pressure. There is an exploration of unit conversions, particularly from bar to pascal, and the implications of these conversions on the calculations.

Discussion Status

The discussion is active, with participants providing guidance on unit conversions and clarifying the distinction between pressure and force. Some participants have shared calculations and results, while others are questioning the assumptions made in the initial setup.

Contextual Notes

There is a noted discrepancy in understanding the units of pressure versus force, and participants are addressing this as part of their reasoning process. The original poster expresses confusion regarding the application of the formulas and the resulting calculations.

edanzig
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The question asks; "Assuming that the density of air is a constant 1.3 kg/m3 and that the air pressure is 1 bar, what is the pressure at the top of a 500 meter high skyscraper?"

I'm having trouble setting something up but here is what I do have;
Regarding liquids we know that F = (Rho)(height)(g) so I want to say that whatever "causes" the "air to push down" on the ground at 1 bar of pressure should have 500 meters worth of "height" removed from it. So force pushing down = 1 bar and (force pushing down) - 500 meters = answer.
So (Difference in force) = (density)(g)(height difference) where given density is in Kg/m^3, height is in m and g is m/s^2. Summing up these values we get Kg/m*s^2 Which is equal to a Newton. The atmospheric pressure is Newtons/m^2.
If I ignore this discrepancy (because the pressure is measuring F/A, and I'm solving just for the force) and solve to problem I end up with a huge number 1.3*8.9*500 which is obviously incorrect. Can someone please steer me in the right direction? thanks
 
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rho*g*h gives you a pressure not a force.

The question is only asking for the pressure so forget everything about forces. Can you convert 1 bar into kPa?
 
To elaborate on what paisiello said, your 1.3*8.9*500 is N/m2, or Pa. This is the difference between the pressure at ground level and the pressure at 500 m. What is 1 bar in Pa?

Chet
 
1 bar in Pa is approx is 100,000. The difference is 6370. Therefore the answer is .93 bar. (thanks) Just not sure when you say that rho*g*h gives pressure and not force, the units sum to "N" and not "N/m^2."
 
If you do the calculation with all of the units then you will see that it gives you a pressure F/A. And you don't sum the units, you multiply them.
 
edanzig said:
1 bar in Pa is approx is 100,000. The difference is 6370. Therefore the answer is .93 bar. (thanks) Just not sure when you say that rho*g*h gives pressure and not force, the units sum to "N" and not "N/m^2."
\frac{kg}{m^3}\frac{m}{s^2}m=\frac{kgm}{m^2s^2}=\frac{N}{m^2}
 

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