Using Henry's law for dissolved methane in a well

[zaboda42]

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 10^o C (at the bottom of the well) and 4.13 x 10⁴ atm at 25^o C. The density of water is 1000 kg/m³.

So we have:

x_methane = y_methaneP/H_methane(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 H_methane(10_o C), but how do I obtain a value for y_methane so that I can plug it into the above equation to get the desired value for x_methane?

 

[Chestermiller]

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 10^o C (at the bottom of the well) and 4.13 x 10⁴ atm at 25^o C. The density of water is 1000 kg/m³.

So we have:

x_methane = y_methaneP/H_methane(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 H_methane(10_o C), but how do I obtain a value for y_methane so that I can plug it into the above equation to get the desired value for x_methane?

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

 

[zaboda42]

So then my answer simple becomes x_methane = P/H_methane(T)?

...

P = 8453 mmHg = 11.12 atm

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

Does this process appear correct?

 

[Chestermiller]

So then my answer simple becomes x_methane = P/H_methane(T)?

...

P = 8453 mmHg = 11.12 atm

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

Does this process appear correct?

Yes. Now, what do you do next?

 

[zaboda42]

Convert the 2 L given into moles of methane. I suppose a pretty straightforward process from here. Thank you for your help!

 

[Chestermiller]

Convert the 2 L given into moles of methane. I suppose a pretty straightforward process from here. Thank you for your help!

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).