Pressure-Volume Formula Question

I have a chemistry homework problem that states the following:



> 
> $6.2$ L of $N\_2$ at $0.74$ bar is mixed with a $15.2$ L sample of $O\_2$ at $0.35$ bar. The gaseous mixture is placed in a container of volume $12.0$ L at the same temperature. What is the pressure of the system, in bar?
> 
> 
> 


I answered the question and I **believe** I have the right answer, but I'm unsure as to whether it's a fluke or not. Here's my approach:


Use the formula $P\_1 \* V\_1 = P\_2 \* V\_2$.


$6.2 \* 74 = 21.4 \* P\_2$, $P\_2 = 21.4$ kPa.


Repeating the process for $O\_2$, I get $P\_2 = 46.26$ kPa.


I add the volumes together since the gases are now mixed, and I also add the pressures. Now that the mixture is placed into a $12.0 L$ container, we use the formula again:


$46.26 \* 21.4 = P\_2 \* 12$. $P\_2 = 83$ kPa. This is equal to 0.83 bar.


I'm still not 100% sure of my reasoning for this question, wondering if someone could verify my logic and provide perhaps a better way of doing this question that I might've overlooked.


Thank you!




I arrived at the same conclusion, but I am not sure your approach is optimal, for efficiency and conceptual reasons. The exercise states nothing about where the two samples are mixed, and I think it is reasonable to simply mix them in the final container:
$$
p\_i \cdot V\_i / V\_{\text{final}} = p\_{i,\text{final}},
$$
where $i$ runs over $\ce{O2}, \ce{N2}$. Add the $p\_{i,\text{final}}$ to obtain total final pressure $p\_\text{final}$.


The actual calculation (I dropped the units) is thus: 
$$
( 0.74 \cdot 6.2 + 0.35 \cdot 15.2) / 12 \approx 0.83,
$$
confirming your result.


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