# Can someone walk me through this gas mixture question?

Argon $$(\ce{Ar})$$ and helium $$(\ce{He})$$ are initially in separate compartments of a container at $$\pu{25 °C}.$$ The $$\ce{Ar}$$ in compartment A which has a volume $$V_\ce{A}$$ of $$\pu{9.00 L}$$ and a pressure of $$\pu{2.00 bar}.$$ The $$\ce{He}$$ in compartment B of unknown volume $$V_\ce{B}$$ has a pressure of $$\pu{6.00 bar}.$$ When the two compartments are connected and the gases allowed to mix, the total pressure of gas is $$\pu{3.60 bar}.$$ Assume both gases behave ideally.

$$\begin{array}{|l|l|} \hline \text{Chamber A} & \text{Chamber B} \\ V_\ce{A} = \pu{9.00 L} & V_\ce{B} = \pu{??? L} \\ p_\ce{A} = \pu{2.00 bar} & p_\ce{B} = \pu{6.00 bar} \\ \hline \end{array}$$

[4 marks] Determine the volume of compartment B.

I've been stuck on this problem for a while. Could someone walk me through the beginning steps? I should be good to go with some guidance. Would I use Boyle's law in attempt to find the final volume once the partition is removed?

• Have you watched this Youtube mini 8 min lecture, it addresses exactly the same problem on board... youtube.com/watch?v=uFwFxgWDm5U. – M. Farooq Sep 29 '19 at 19:51
• @M.Farooq I watched the lecture, but didn't know how to approach without any values for n. I can determine the number of moles in chamber A, but I'm not sure how I would do that for chamber B or the final mixture. – NoCo Sep 29 '19 at 20:18
• You can determine the moles of Ar from its pressure temperature and volume using PV=nRT. – M. Farooq Sep 29 '19 at 20:28
• @M.Farooq Using the equation Pfinal = RTfinal[(n1+n2)/(V1+V2)], I still have 2 unknowns, n2 and V2. Is there a way to find n2 with my current information? – NoCo Sep 29 '19 at 20:43
• You need one more equation to set-up. Two unknowns require two equations. What if you set up PV=nRT for helium? P is given, V is unknown, n is unknown and RT are known. See if that solves the problem. – M. Farooq Sep 29 '19 at 20:50

1. 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.
2. Write the volume of chamber B in terms of knowns: $$V_B=n_BRT/p_B=n_B \times\pu{4.13\times 10^3 m^3/mol}$$.
3. Write the total volume when mixed: $$V_{tot}=n_{tot}RT/p_{tot}=n_{tot} \times\pu{6.89\times 10^3 m^3/mol}$$.
4. 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 &=(n_A+n_B)\times \pu{6.89\times 10^{-3} m^3/mol}\\ &=V_A + n_B \times\pu{4.13\times 10^{-3} m^3/mol}\\\rightarrow n_B &= \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}
5. 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: $$V_B = \pu{6.00 L}$$