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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!

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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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