Lester_01
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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?
R= 0.08206\frac{L atm}{mol K}
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=\frac{PV}{RT}
= \frac{(5)(1)}{(0.08206)(283)}
= 0.2153 moles N2
ntot = 1 + 0.2153 = 1.2153 moles total
PN2= Ptot /frac{n<sub>N<sub>2</sub></sub>}{n<sub>tot</sub>}
5= P<sub>tot</sub>\frac{0.2153}{1.2153}
\frac{5}{0.1772}= 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?
R= 0.08206\frac{L atm}{mol K}
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=\frac{PV}{RT}
= \frac{(5)(1)}{(0.08206)(283)}
= 0.2153 moles N2
ntot = 1 + 0.2153 = 1.2153 moles total
PN2= Ptot /frac{n<sub>N<sub>2</sub></sub>}{n<sub>tot</sub>}
5= P<sub>tot</sub>\frac{0.2153}{1.2153}
\frac{5}{0.1772}= 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.