Why Does Air Pressure Change with Altitude?

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SUMMARY

The discussion focuses on the relationship between air pressure, density, and temperature at different altitudes, specifically using the ideal gas law. The participants clarify that the mass of gas remains constant between two heights, allowing the use of the equation P1/(ρ1T1) = P2/(ρ2T2) to calculate pressure changes. Key values discussed include sea level pressure at 1,013 x 10^5 Pa, sea level density at 1.293 kg/m³, and altitude temperature at 233 K with a density of 0.51 kg/m³. The confusion arises from understanding how gas density changes with altitude while maintaining constant mass.

PREREQUISITES
  • Understanding of the Ideal Gas Law (PV=nRT)
  • Knowledge of basic thermodynamics concepts
  • Familiarity with pressure, density, and temperature relationships
  • Basic algebra for manipulating equations
NEXT STEPS
  • Study the Ideal Gas Law and its applications in different conditions
  • Research atmospheric pressure variations with altitude
  • Learn about the concept of density and how it changes with temperature and pressure
  • Explore the implications of gas behavior in open systems versus closed containers
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Students in physics or engineering, meteorologists, and anyone interested in atmospheric science or the principles of gas behavior at varying altitudes.

dumbperson
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Hello, I don't need help solving a problem, but need help explaining why the given answer is true because I don't understand.

A guy is jumping from something(outside) and I need to calculate the air pressure at a certain height.

Given is that the temperature on sea level is 1,013*10^5 Pa, and the density of air is 1,293 kg/m³, and the temperature is 273k.

on that specific height(where I need to calculate the air pressure) the temperature is 233k, the density of air is now 0,51 kg/m³, and you need to calculate the pressure at that height .

In the solution it says that the mass of gas are the same on those two heights, so you can solve ((P1*(m/p1))/T1)= ((P2*(m/p2))/T2)

My problem is that I don't understand that the amount of gas is same in both situations? It's outside so yea the total amount of gas is the same, but you're not looking at that are you? because for example the temperature isn't the same in all of that amount of gas so you can't use all of that mass in this equation? I could imagine this being true in for example a closed container, where the temperature changes everywhere in the container by the same amount, but in open air I do not understand. Am I looking at it wrong?

I'm sorry if you do not understand me because of my bad english, I will try explain further and sorry if this is a dumb question.
 
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The molar mass is the same at both height. The ideal gas low says that PV=(m/M) RT, m is the mass of gas contained in V volume and M is the molar mass.

The density is ρ=m/V, so you can rewrite the gas law as P=(ρ/M)RT, or
P1/(ρ1T1)=P2/(ρ2T2).

ehild
 
here u used(or in the solution they used) the eqn. P*V/T = const. as pressure varies according to P = h*rho*g, 'the different pressure at different levels' can easily be understood. for the change of volume for different heights is only due to change in density. as air goes less dense a larger voloume is required for same mass. hence m/rho1 comes into play. i think yr confusion is why u have to take the same mass.
when u construct the eqn. P*V/T = cnst (cnst is actually nR). u have consider only two variables at a time, i.e, when P and T varies V and n (no. of grm-moles present or in other words, mass)remains const. and so on. so when u consider V varies u have to consider mass is const.
 
Okay thanks for the help.

So the amount of gas is the same on all heights, just less dense because it has to be divided over a larger amount of volume(because the higher you get, the bigger the radius of the 'circle' layer the gas is in gets?do you know what I mean? or am I wrong again here )

thanks a lot
 

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