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.