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Height related gas pressure function

  1. Sep 23, 2006 #1
    This is not a homework, but it seems that this is may be the most appropriate forum for my question.

    Consider that there is a column of homogen gas, h meter high. Under the assumption that gas is compressible, how can we determine the function of density (or pressure) related to height? Of course gravity is parallel to the column axis. As you may have guessed, I think this is similiar to a question about the atmosphere's pressure.

    I would like to try solving the equation but I don't even know how to start....
     
  2. jcsd
  3. Sep 23, 2006 #2

    Andrew Mason

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    Start with the idea that the pressure at any point is equal to the weight of the air in the column above that point divided by the area of the column (for simplicity, assume a column of 1 m^2). Density follows from the ideal gas law: n/V = P/RT

    AM
     
    Last edited: Sep 23, 2006
  4. Sep 23, 2006 #3
    Do you mean that I can simply get the pressure value at any point by doing this: P=density*g*h? The problem is, the gas' density is not a constant because the gas is compressed by it's own weight (the lower it is, the higher the density).

    Do you think that we must know a "compressibility function", and then somehow "insert" it into the equation?
     
  5. Sep 24, 2006 #4

    Andrew Mason

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    You have to start with the pressure, temperature and density of air at the bottom. Assuming you are talking about the earth's atmosphere, the starting point is the pressure, density and temperature at the earth's surface. Call those [itex]\rho_0, P_0 \text{ and } T_0[/itex] As you rise, the weight of the column above you decreases by the weight of the column below and this provides a new pressure and, therefore, new density ([itex]\rho = n*M/V = PM/RT[/itex] where M is the molar weight of air).

    You have to work out the differential equation (change in pressure / change in height) and find a solution to that differential equation.

    Assume g and T are constant for this to make it simple.

    AM
     
  6. Sep 24, 2006 #5

    Andrew Mason

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    Following up on my previous post, it seems that the pressure after a change in height (ie distance above the earth surface), dh is:

    [tex]P = P_0 - \rho gdh[/tex]

    so:

    [tex]dP = -\rho gdh[/tex]

    So, the differential equation for pressure would be:

    [tex]\frac{dP}{dh} = -\rho g = -\frac{PMg}{RT}[/tex]

    Assume g and T remain constant to simplify the solution to this differential equation.

    AM
     
  7. May 27, 2008 #6
    hello
    i encountered a very similar problem

    i have a cilinder of homogen gas , and i know the pressure at the top, will this equation help me solve the pressure while going down the cilinder?
     
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