- #1
Lester_01
- 2
- 0
1. The problem/question is as follows:
1 mole of O2 mixed with N2 gas (PN2= 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]
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.
Ptot= ∑PO2 + PN2
nN2=[tex]\frac{PV}{RT}[/tex]
= [tex]\frac{(5)(1)}{(0.08206)(283)}[/tex]
= 0.2153 moles N2
ntot = 1 + 0.2153 = 1.2153 moles total
PN2= Ptot [tex]/frac{nN2}{ntot}[/tex]
[tex]5= Ptot\frac{0.2153}{1.2153}[/tex]
[tex]\frac{5}{0.1772}[/tex]= Ptot= 28.22 atm
PO2= 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 ntot, it will result in a negative total pressure of -18 atm, and also interestingly result in PO2= -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.
1 mole of O2 mixed with N2 gas (PN2= 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
Ptot= ∑PO2 + PN2
nN2=[tex]\frac{PV}{RT}[/tex]
= [tex]\frac{(5)(1)}{(0.08206)(283)}[/tex]
= 0.2153 moles N2
ntot = 1 + 0.2153 = 1.2153 moles total
PN2= Ptot [tex]/frac{nN2}{ntot}[/tex]
[tex]5= Ptot\frac{0.2153}{1.2153}[/tex]
[tex]\frac{5}{0.1772}[/tex]= Ptot= 28.22 atm
PO2= 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 ntot, it will result in a negative total pressure of -18 atm, and also interestingly result in PO2= -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.