Question

A quantity of N2 gas originally held at 5.26atm pressure in a 1.20−L container at 26 ∘C is transferred to a 13.5−L container at 20 ∘C. A quantity of O2 gas originally at 5.27atm and 26 ∘C in a 5.20−L container is transferred to this same container.
What is the total pressure in the new container?

Answer 1

The question might seem tricky, but it's actually not too bad! I'll try to explain slowly, and if you have a question let me know:

The pressure in the new container will be the sum of the pressures of the N2 and O2 IN THAT CONTAINER (that is, using that container's T and V). First remember the deal gas law: PV=nRT. If you rearrange to solve for P, you get P=nRT/V. This is important because we need to express the pressures of N2 and O2 as described in the question in this form:

\[P_{new}=P_{N_2}+P_{O_2} \]\[P_{new}={n_{N_2}RT_{N_2} \over V_{N_2}}+{n_{O_2}RT_{O_2} \over V_{O_2}}\]
In the new container, N2 and O2 are occupying the same volume, so V(N2)=V(O2). Similarly, in the new container, both N2 and O2 are at the same temperature, so T(N2)=T(O2). That means we can make the above equation a bit simpler by factoring out all the shared factors:

\[P_{new}={RT_{new} \over V_{new}}(n_{N_2}+n_{O_2})\]
Now we just need ways to find the moles of N2 and O2 to plug into this equation. Luckily the question gives us information to do so - just apply PV=nRT to the conditions of the ORIGINAL containers for N2 and O2. You can find each of these values separately then plug them into the n's in the equation above, but I like coming up with one big expression and solving it all in one step.

If you rearrange PV=nRT to solve for n, you get n=PV/(RT). By plugging in this expression instead of n to the equation above, I get:

\[P_{new}={RT_{new} \over V_{new}} \left( {P_{N_{2_o}}V_{N_{2_o}} \over RT_{N_{2_o}}} + {P_{O_{2_o}}V_{O_{2_o}} \over RT_{O_{2_o}}} \right)\]
The little "o" just means the original container. And once again, just to make things a bit easier, you can factor out the R's in the denominators, which actually allows it to reduce out the R outside the brackets, leaving you with:
\[P_{new}={T_{new} \over V_{new}} \left( {P_{N_{2_o}}V_{N_{2_o}} \over T_{N_{2_o}}} + {P_{O_{2_o}}V_{O_{2_o}} \over T_{O_{2_o}}} \right)\]
If you plug all the values given in the question into this equation and solve, you will have your answer!

The most important, and trickiest, thing to do is to come up with the right equation. I know my last equation there looks kind of scary, but you just need to work through it step by step. Read my answer carefully and hopefully this will make sense to you! And again, please let me know if you have any questions!