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Hydrostatic pressure distribution for an ideal gas

  1. Nov 26, 2014 #1
    Considering a unit mass of an ideal gas in a cylindrical container of volume 'V' at temperature 'T' the pressure exerted by the gas at the walls of the container is given by the ideal gas equation as,

    pV=RT where 'R' is the characteristic gas constant for the particular gas.

    Under equilibrium conditions this pressure should have equal values as measured at every point within the container.

    Again considering the gas in the container as a fluid at rest there should be a vertical pressure gradient satisfying the relation as given by the hydrostatic law,


    where, 'z' is the elevation measured in a vertically downward direction,
    'γ' is the weight density and 'ρ' is the mass density of the gas,
    and 'g' is the acceleration due to gravity.

    Now for the gas at rest within the cylinder if we assume uniform density everywhere within the container i.e, the variation of density with spatial location to be absent we may integrate previous equation as,


    Thus, pressure should vary linearly with depth for the gas within the cylinder whereas we know that the pressure at all points within the cylinder should be the same as given by the ideal gas model!

    I fail to understand where am I possibly wrong.Any views on this?
    Last edited: Nov 26, 2014
  2. jcsd
  3. Nov 26, 2014 #2

    Jano L.

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    Gold Member

    The equation of state


    is valid for gas volume only if it has the same pressure and temperature everywhere. In gravitational field this is valid only for horizontally placed slices that are thin enough to make the variation of pressure negligible.
  4. Nov 26, 2014 #3
    Does that mean one could only apply the ideal gas equation for containers that are not significantly large in vertical dimensions?
  5. Nov 26, 2014 #4
    You can apply it only for volumes of gas which have uniform pressure and temperature within the volume.
    Think about it, if the pressure or temperature is not uniform, what values would you use in formula? What p, what T?

    If the volume of gas does not satisfy this condition, you can divide it into small volumes so that within each one the p and T are approximately constant.
  6. Nov 26, 2014 #5
    The ideal gas law is correct locally at each depth for the situation you are looking at. The local density of the ideal gas is given by:


    where ρ is the local mass per unit volume at depth z and M is the molecular weight. So,


    where z is measured downward. So,


  7. Nov 27, 2014 #6

    Philip Wood

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    Gold Member

    A couple of remarks.
    (a) These days it's generally considered better style not to work with unit quantities. If you work with a mass m of gas, the equations manifestly retain their homogeneity of units. You can always put m = 1 kg when you put the figures in near the end.

    (b) When dealing with gases the symbol R is reserved for the molar gas constant, which applies to all (ideal) gases alike, and is defined by the equation pV = nRT in which n is the amount of substance (number of moles [x mol]). That's the way Chestermiller is using R in his post.

    The constant you've called 'R' is equal to [itex]nR = \frac{mass\ of\ gas}{molar\ mass}\times R[/itex].

    The molar mass (mass of a mole of the gas!) is what Chestermiller is calling M.
    Last edited: Nov 27, 2014
  8. Dec 2, 2014 #7
    Excellent clarification!!!!!

    The ideal gas equation is only valid for a certain quantity of a gas where the thermodynamic parameters are invariant with location.

    While for a fluid at rest both pressure and temperature are expected to vary with location(i.e, elevation) and consequently density varies with location according to the ideal gas law applied locally at any point of the containing vessel for the gas.

    Thank You once again
    Sir Chet:)
  9. Dec 2, 2014 #8
    Those are certainly some very important remarks one should be careful about!

    The first point you made is indeed quite necessary to avoid a serious malpractice.

    Thank You :)
  10. Dec 2, 2014 #9
    I understand now what you meant :)
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