A #6 L# container holds #5 # mol and #6 # 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 #480^oK# to #270^oK#. How much does the pressure change?

1 Answer
Jul 1, 2017

the pressure change from #P= 7312800 N/m^2# to #P= 1121850 N/m^2#

Explanation:

To calculate the initial pressure you can use the general law of perfect gasses: PV = nRT .

#P= (nRT)/V = (11 mol 8,31 J/(K mol) 480 K)/(0,006 m^3) = 7312800 N/m^2#
to know the final pressure, i suppose that there are still 3 moles of A, 3 moles of B and it was formed 1 mol of gas from the union of the moles reacted toward a total of 7 gass moles. The new pressure will be:

#P= (nRT)/V = (7 mol 8,31 J/(K mol) 270 K)/(0,006 m^3) = 2617650 N/m^2#
if instead all the molecules (6moles) of B react with the maximum possible number of the moles of A (4 moles, as the ratio is 3:2) at the end you will have still a mole of A and it was formed 2 moles of #A_2B_3# toward a total of 3 gass moles, the final pressure will be:
#P= (nRT)/V = (3 mol 8,31 J/(K mol) 270 K)/(0,006 m^3) = 1121850 N/m^2#
I think that this second hypothesis is what is required