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How to find water vapor volume at different pressures than atm?

  1. Aug 23, 2013 #1
    1. The problem statement, all variables and given/known data
    If a known amount, lets say 300 liters of water gas (water vapor) is flowing from an atmospheric (14.7 psi) source/chamber, into a chamber of reduced pressure of 10 psi, does the volume amount and/or moles change?


    2. Relevant equations
    So I believe I would multiply the 300 liters by the pressure ratio, but not sure which pressure value is the denominator, which is the numerator


    3. The attempt at a solution
    I believe its either: 300 L * (10psi / 14.7 psi) which is 204.1 liters,
    or it is: 300 L * (14.7 psi / 10 psi) which is 441 liters

    Thanks for any help, I am leaning more towards 441 liters as my answer
     
  2. jcsd
  3. Aug 23, 2013 #2
    Hi rhinohugger. Welcome to Physics Forums.

    If the new chamber were initially empty (vacuum) and you could get all the moles into the new chamber at 10 psi, and, if the temperature were held fixed, then using the ideal gas law the way you did would give you a pretty accurate answer. If you needed a more accurate answer, you would have to use a more accurate equation of state, or you could use the steam tables (which have the specific volumes at various temperatures and pressures tabulated for you).

    Chet
     
  4. Aug 23, 2013 #3
    Thanks for the quick reply Chet,
    1.So to clarify, the ideal gas method of 300 L * (14.7 psi / 10 psi) [NOT 10 psi /14 psi], would provide an accurate answer?
    2. Also, I am looking at a steam table currently on engineering toolbox.
    To use one to find a more accurate answer, I would look at the pressure on the steam chart (10 psi, in my case) and look find the specific volume which is in m^3/kg, so I would have to first calculate how many kg are in 300 liters of water vapor at 14.7 psi, and multiply it by the specific volume?
     
  5. Aug 24, 2013 #4
    Yes, as long as it is at the same temperature.
    No. At 14.7, you take the specific volume at the particular temperature (≥100C) and divide it into 300 liters (0.3 m^3) to get the kg.
     
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