Calculating the total and partial pressure of gas mixture

[Lester_01]

1. The problem/question is as follows:

1 mole of O₂ mixed with N₂ gas (P_N2= 5 atm at 10 degrees celcius in a 1 L flask. What is the total pressure after 2 moles of gas is allowed to escape? How about the partial pressure of O₂?

[tex]R= 0.08206\frac{L atm}{mol K}[/tex]

Homework Equations

Using the ideal gas law, I have calculated the quantity of moles of nitrogen gas in the mixture, I add one mol of oxygen and it results in 1.2153 moles total. The problem states that 2 moles of gas are allowed to escape, yet there is not even 2 moles of gas in the system. We can't have negative moles so the situation seems impossible, or a poorly designed question at best.

The Attempt at a Solution

P_tot= ∑P_O2 + P_N2

n_N2=[tex]\frac{PV}{RT}[/tex]

= [tex]\frac{(5)(1)}{(0.08206)(283)}[/tex]

= 0.2153 moles N₂

n_tot = 1 + 0.2153 = 1.2153 moles total

P_N2= P_tot [tex]/frac{n_N2}{n_tot}[/tex]

[tex]5= P_tot\frac{0.2153}{1.2153}[/tex]

[tex]\frac{5}{0.1772}[/tex]= P_tot= 28.22 atm

P_O2= 23.22 atmThese values seem to make sense, but I'm not sure how to reconcile for how 2 moles are escaping.

Interestingly, if I subtract 2 moles from the 1.2153 and in corporate that into the procedure as n_tot, it will result in a negative total pressure of -18 atm, and also interestingly result in P_O2= -23.22 atm (the opposite of the result above)


Any information regarding how 2 moles can escape out when there is only 1.2153 moles contained?

PS Sorry about the coding, it for some reason isn't working very well.

 

[Lester_01]

Or I'm thinking the answer could simply be zero for both cases because all particles of gas have escaped, all the moles are gone and there's no pressure in the flask in that case.