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?
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?
via Physics Forums RSS Feed http://www.physicsforums.com/showthread.php?t=720096&goto=newpost
0 commentaires:
Enregistrer un commentaire