## Section21.5Ideal Gas Bootcamp

### Subsection21.5.5Miscellaneous

A metallic container of fixed volume of $35\text{ L}$ immersed in a large tank of temperature $27^{\circ}\text{C}$ contains two compartments separated by a freely movable wall. There are $1.2\text{ moles}$ of nitrogen on one side and $1.5\text{ moles}$ of oxygen on the other side of the wall. Find the temperature, pressure and volume of the two sides when equilibrium has reached. Assume ideal gas behavior.

Hint

At equilibrium, the pressure on the two sides will be same. Also, the volumes of the two add up to the unchanging total volume.

$1.9\text{ atm} \text{.}$

Solution

Since the separating wall is freely movable, the two sides will come to the same pressure at the equilibrium in addition to being at the same temperature. Therefore the ratio of volume to number of moles on the two sides will be equal. Denoting the Nitrogen side by a subscript 1 and the Oxygen side by the subscript 2 we have

\begin{equation*} \frac{V_1}{n_1} = \frac{V_2}{n_2} = \frac{RT}{p}=\text{ Same}. \end{equation*}

In addition the sum of the volumes is equal to the total volume, $V_0\text{.}$

\begin{equation*} V_1 + V_2 = V_0. \end{equation*}

Therefore,

\begin{align*} \amp V_1 = \left( \frac{n_1}{n_1 + n_2} \right) V_0, \\ \amp V_2 = \left( \frac{n_2}{n_1 + n_2} \right) V_0. \end{align*}

Putting in the numbers we get

\begin{align*} \amp V_1 = \left( \frac{1.2}{1.2 + 1.5} \right)\times 35 \ \text{L} = 16\ \text{L}, \\ \amp V_2 = \left( \frac{1.5}{1.2 + 1.5} \right) \times 35 \ \text{L} = 19 \ \text{L}. \end{align*}

Now, since we know the volume, and the temperature, which is 300.15 K, we can find the common value of the pressure by using the properties of either side. For instance, using the properties of the nitrogen side, we have

\begin{align*} p \amp = \frac{n_1 RT}{V_1} = \frac{1.2\ \text{mol}\times 0.082\ \text{L.atm/mol.K}\times 300.15\ \text{K}}{16\ \text{L}}\\ \amp = 1.9\ \text{atm}. \end{align*}