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?

  • $\begingroup$ Have you watched this Youtube mini 8 min lecture, it addresses exactly the same problem on board... youtube.com/watch?v=uFwFxgWDm5U. $\endgroup$
    – AChem
    Commented Sep 29, 2019 at 19:51
  • $\begingroup$ @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. $\endgroup$
    – NoCo
    Commented Sep 29, 2019 at 20:18
  • $\begingroup$ You can determine the moles of Ar from its pressure temperature and volume using PV=nRT. $\endgroup$
    – AChem
    Commented Sep 29, 2019 at 20:28
  • $\begingroup$ @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? $\endgroup$
    – NoCo
    Commented Sep 29, 2019 at 20:43
  • $\begingroup$ 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. $\endgroup$
    – AChem
    Commented Sep 29, 2019 at 20:50

1 Answer 1

  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: $V_B = \pu{6.00 L}$


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