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Calculations using Saturated Vapour Pressure

  1. Dec 24, 2008 #1
    Hi all, new here, so dont know if this is the right place. All the other places in the forum seemed like advanced stuff!

    Anyway, i dont understand why when calculating the partial pressure of a gas in SVP conditions, you subtract 6.28kPa from the normal atomospheric pressure of 101.1kPa? Could someone explain the rationale of why thats done please?

    If it helps, you can email me the answer to my email: blue.rascal [at] hotmail.com

    Thank you all and Merry Xmas!

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  3. Dec 26, 2008 #2

    Andrew Mason

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    What temperature are you talking about? What substance has a vapour pressure of 6.28 kPa at normal atmospheric pressure and at that temperature?

  4. Dec 26, 2008 #3

    Hi Andrew, thanks for the reply.

    Sorry for not making it clear. I meant water vapour has a SVP of 6.28kPa at 37 deg C.

    When calculating the partial pressure of a particular gas, say Oxygen, in the atmosphere which is saturated with water vapour, you minus 6.28kPa from the normal atmospheric total pressure of 101.1kPa and then find 21% (oxygen content in air) of that.

    So I want to know why it is that you subtract the svp (6.28kPa) from the total pressure (101.1kPa) in such a calculation?
  5. Dec 26, 2008 #4

    Andrew Mason

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    Ok. I see what you are trying to do.

    The concentration of Oxygen, O2, in the dry air is 21% by volume. The concentration of Nitrogen, N2 is atmosphere is 78% by volume of dry air. This relative proportion of O2 to N2 does not change. As water vapour concentration increases, the air becomes heavier as water molecules are added to the same population of O2 and N2 molecules. But the total pressure does not change. There are just more molecules which take up more volume at the same pressure. So the partial pressures contributed by O2 and N2 must decrease.

    Oh, and welcome to PF, by the way!

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