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Using Henry's law for dissolved methane in a well

  1. Oct 31, 2013 #1
    Suppose water at the bottom of a well contains dissolved methane. When the water is pumped up, the dissolved gas is released. How many moles of methane gas will be released if the well is 115 m deep and holds 2 L.

    Assume a Henry's law constant of 2.97 x 104 atm at 10o C (at the bottom of the well) and 4.13 x 104 atm at 25o C. The density of water is 1000 kg/m3.

    So we have:

    xmethane = ymethaneP/Hmethane(T) where y is the mol% and x is the mol fraction.

    We can find P for a column of water by using the relation P = ρgh where ρ = 1000 kg/m3, g = 9.8 m/s2, and h = 115 m. Thus P = 1.127 x 106 Pa = 8453 mmHg.

    I have Hmethane(10o C), but how do I obtain a value for ymethane so that I can plug it into the above equation to get the desired value for xmethane?
     
  2. jcsd
  3. Oct 31, 2013 #2
    If there were methane gas trapped (say under a cavity) at depth in equilibrium with the water, it would be virtually pure methane (y =1), since the vapor pressure of water at 10 C is very low compared to the total pressure. Assume that the water is saturated with methane, and use y = 1.

    Chet
     
  4. Oct 31, 2013 #3
    So then my answer simple becomes xmethane = P/Hmethane(T)?

    ...

    P = 8453 mmHg = 11.12 atm

    xmethane = 11.12/(2.97 x 104 atm) = 3.744 x 10-4

    Does this process appear correct?
     
  5. Oct 31, 2013 #4
    Yes. Now, what do you do next?
     
  6. Oct 31, 2013 #5
    Convert the 2 L given into moles of methane. I suppose a pretty straightforward process from here. Thank you for your help!
     
  7. Oct 31, 2013 #6
    One more thing. Is the Henry's law constant that you are working with equal to the pressure divided by mole fraction in the liquid? Many times, H is given as the pressure divided by the concentration in the liquid (moles/liter).
     
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