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?

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?

Charlotte Aaron · Accepted Answer

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.

This is an application of the ideal gas law in which #n# and #T# change, but not #V#. Thus, the ratio form of the ideal gas law 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# Completing what is provided #(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))# Organizing: #(9(240))/((21)160)= (P_2)/(P_1)# #(P_2)/(P_1)=0.643# By then, the pressure will drop to 64.3% of its starting point.

Leo Abeyta · Answer

undefined The pressure changes by a factor of $ \frac{3}{2} $.