A 3 L container holds 12  mol and 9  mol of gasses A and B, respectively. Groups of three of molecules of gas B bind to two molecules of gas A and the reaction changes the temperature from 160^oK to 240^oK. How much does the pressure change?

Jan 26, 2017

The difficult aspect of this problem is finding the value of ${n}_{2}$, the moles of gas remaining in the container after the reaction is complete. Once that is done, we find the pressure will decrease to 64.3% of its initial value.

Explanation:

This is an application of the ideal gas law in which $n$ and $T$ change, but not $V$.

So, the ideal gas law (in ratio form) becomes

$\frac{{P}_{1}}{{n}_{1} {T}_{1}} = \frac{{P}_{2}}{{n}_{2} {T}_{2}}$

Since we are not given the initial pressure, the best we can do is find the fractional change in pressure, that is the ratio ${P}_{2} / {P}_{1}$

Filling in what we are given

$\frac{{P}_{1}}{\left(21\right) 160} = \frac{{P}_{2}}{\left({n}_{2}\right) \left(240\right)}$

To find ${n}_{2}$, we note that 9 mol of gas B will bind with only 6 mol of gas A, to form 3 mol of the new product, and 6 mol of gas A will remain. So, after the reaction, we will have a total of 9 mol of gas in the container. This is ${n}_{2}$.

$\frac{{P}_{1}}{\left(21\right) 160} = \frac{{P}_{2}}{9 \left(240\right)}$

Rearranging:

$\frac{9 \left(240\right)}{\left(21\right) 160} = \frac{{P}_{2}}{{P}_{1}}$

$\frac{{P}_{2}}{{P}_{1}} = 0.643$

The pressure will decrease to 64.3% of its initial value.