A 3 L3L container holds 12 12 mol and 9 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^oK160oK to 240^oK240oK. How much does the pressure change?

1 Answer
Jan 26, 2017

The difficult aspect of this problem is finding the value of n_2n2, 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 nn and TT change, but not VV.

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

(P_1)/(n_1T_1)= (P_2)/(n_2T_2)P1n1T1=P2n2T2

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_1P2P1

Filling in what we are given

(P_1)/((21)160)= (P_2)/((n_2)(240))P1(21)160=P2(n2)(240)

To find n_2n2, 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_2n2.

(P_1)/((21)160)= (P_2)/(9(240))P1(21)160=P29(240)

Rearranging:

(9(240))/((21)160)= (P_2)/(P_1)9(240)(21)160=P2P1

(P_2)/(P_1)=0.643P2P1=0.643

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