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Air pressure of guy jumping

  1. Jul 20, 2011 #1
    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 dont 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 cant 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.
     
    Last edited: Jul 20, 2011
  2. jcsd
  3. Jul 21, 2011 #2

    ehild

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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
     
  4. Jul 21, 2011 #3
    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 differnt 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.
     
  5. Jul 21, 2011 #4
    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 alot
     
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