[nertil1]

I am getting confused. I have a solubility chart of methane in n-heptane. It states that it is for when the vapor phase fugacity of methane is .0101 bar.  It is a plot of Temperature vs. mole fraction methane. My question is would the mole fraction be moles methane/1 L of heptane. And if so, would the moles of methane be vapor and the Liter of heptane be liquid.

Please Help
Thanks

[macman104]

mole fraction of methane would be (moles of methane)/(total moles in mixture)

[nertil1]

would that be the moles of methane in the vapor phase/moles of methane and heptane in the liquid phase?

[macman104]

It "sounds" like it is showing you the mole fraction of methane in the liquid phase as a function of temperature.  So the mole fraction would be

mole methane(liquid)/(mole methane(liquid) + mole heptane(liquid))

Is there a way you can post the table?

[nertil1]

The plot look like this. Right click and save the image to you desktop first. It will look better.

[Gerard]

The plot look like this. Right click and save the image to you desktop first. It will look better.

friend this is the mole fraction of methane in your liquid (your liquid consists of mole methane + moles of n-heptane), your fugacity is set to constant by setting the pressure constant at 0.101 bar. you got confuse by the fugacity...this is not vapor of methane in your moles of methane and n-heptane mixture.

its like the moles of methane per total moles in the mixture (that is moles methane plus moles n-heptane)
the graph tells us that as you increase the temperature the molar amount of methane in the mixed organic liquid decreases, this is because as temeprature increases most of your methane content vaporizes.

[Gerard]

i recommend you read Chemical,Biochemical and Engineering thermodynamics by STanley Sandler...
he has one whoel chapter dedicated to this kind of subject

[nertil1]

Thank you for your reply, now it makes sense. The reason I am asking about this is because I need to figure out how much methane is in a stream of methane and heptane flowing through a pipe(in vapor and liquid form). The contents then go through a valve, which drops the pressure and temperature. So according to my calculation the methane content after the valve increases in the liquid and decreases in the vapor, which means that after the valve the heptane content in the liquid decreases and in the vapor it increases. Can you verify this?

Thanks