Can someone walk me through this gas mixture question?
Solve for the amount of gas in A: $n_A=p_AV_A/RT=\pu{0.727 mol}$. You'll need to perform necessary unit conversions or use R in units that match the given T, p and V units. 
Write the volume of chamber B in terms of knowns: $V_B=n_BRT/p_B=n_B \times\pu{4.13\times 10^3 m^3/mol}$.  
Write the total volume when mixed: $V_{tot}=n_{tot}RT/p_{tot}=n_{tot} \times\pu{6.89\times 10^3 m^3/mol}$. 
Solve this last expression for $n_B$ after inserting the equalities $V_{tot}=V_A+V_B$, $n_{tot}=n_A+n_B$  and the result of step 2: 
$$\begin{align} V_A+V_B &amp;=(n_A+n_B)\times \pu{6.89\times 10^{-3} m^3/mol}\\ &amp;=V_A + n_B \times\pu{4.13\times 10^{-3} m^3/mol}\\\rightarrow n_B &amp;= \frac{V_A - n_A\times \pu{6.89\times 10^{-3} m^3/mol}}{\pu{2.75\times 10^{-3} m^3/mol}} = \pu{1.45 mol}\end{align} $$
Compute the volume in chamber B: $V_B=\frac{n_{B}RT}{p_B}=\pu{6.00\times 10^{-3} m^3}$


It's usually a good idea to perform a check to see if we did things right. Compute back the final pressure: $p_{tot}=\frac{n_{tot}RT}{V_{tot}}=\pu{3.60 bar}$.

Answer: $V_B = \pu{6.00 L}$