Question

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?

Answer 1

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

`(P_1)/(n_1T_1)= (P_2)/(n_2T_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

`(P_1)/((21)160)= (P_2)/((n_2)(240))`

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`.

`(P_1)/((21)160)= (P_2)/(9(240))`

Rearranging:

`(9(240))/((21)160)= (P_2)/(P_1)`

`(P_2)/(P_1)=0.643`

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