Recent content by sgstudent

Prof chestermiller, could I continue on with this on the next weekend as my midterms are next week and this chapter isn't tested. Just wanted to let you know that I haven't ghosted you for this problem.

I don't see where there will be a point like that though. I thought of this nA/L*1.5+(1-nA/L)*0.031=1.3 but I can't seem to think of the other equation to solve the simultaneous equations though. Could I get a nudge in the right direction?

Let's say we have 10 moles of water vapour. So by Raoult's law, P(A)=38.76/58.76*1.5=0.989atm and P(water vapour)=0.011atm. So by Avogadro's principle we would have 89.9 moles of A. Would this be correct?

If we had 20 moles of water, (0.031)(20/20+b)+(1.5)(b/20+b)=1atm so solving for B gives us 38.76 moles. Would be this correct?

Hmm, so everything would just liquefy?

Haven't even gone there yet, we just got introduced to the topic of mixing but I've been thinking a lot to make sure I understood what's going on in the lectures. Sometimes I feel like I overthink too much but I'm just unsure if what I'm overthinking will be looked at further in the module or...

Let's say we have 20 moles of H2O in the liquid phase and 5 moles of A in the liquid phase initially. And from the graph, the partial pressure of water would be 0.0248atm and A would be 0.9752atm. Would this be alright?

Oh, wait we can tell it from the graph right? So we would know that the exact pressures of water and A at a total pressure of 1 atm? So it will look something like this right? So we can get the mole fractions of either component? But continuing on, when we compress the gas with 2atm of...

P(H2O(g))=P*(H2O(g))X(H2O) and P(A)=P*(A)X(A), but we only know the values of P*(H2O) and P*(A) so I don't get how we can predict their mole fractions though.

I was taught that an ideal dilute mixture would have the solvent follow Raoult's law, and the solute follow Henry's law. So I presumed that A being the added gas should follow Henry's law. But even if I were to use assume a Raoult's law for both, when I first introduced the gas water's mole...

Sorry for the wrong terminology used, I meant that the mixture is compressed with 2atm of pressure. But since gas A undergoes a phase transition at 1.5atm, and water undergoes a phase transition at 0.031atm in their pure states how can their gaseous pressures ever be equal to 2atm? Edit: it...

Yes I would assume so for this scenario. So I guess I'd use Raoult's law for water and Henry's law for A to calculate their partial pressures at each point? So initially, when I first introduced A to the pure water system, how would I know how much it dissolves? We would just know that the total...

Yes so in a phase diagram for A it would transition at 1.5atm. Does that matter if it's a mixture? I get that the final internal pressure has to be 2atm. But I don't see how that can be achieved with the parameters I've set because A would liquefy once it's pressure is at 1.5atm, and the water...

For example we had a closed system with water at 298K, so it's pressure is 0.031atm. To it we add a gas, A such that we get 1atm of total pressure. That gas undergoes a phase transition at 1.5atm and 298K. So initially, the partial pressure of water is a little smaller than 0.031atm due to the...

I think I understand what's going on now. The pressure just stays at 0.031 atm because that's its vapour pressure so it can't go any higher than that at 298K. However, I have another question regarding this example now. For example, if the air added undergoes a phase transition at 1.5atm. And...