The total pressure #P# inside the #V = 5 L# container is due to the pressures of both gasses #A# and #B#:
#P = P_A + P_B#.
According to the ideal gas law:
#PV = nRT#.
If we consider gasses #A# and #B# to be ideal, then this equation must apply to each one of them. For the total pressure, this means that:
#P = (n_ART_A)/V_A + (n_BRT_B)/V_B#.
If we also assume that both gasses can occupy all the container's volume and that they will always be at thermal equilibrium with each other, then:
#P = (RT)/V(n_A + n_B)#.
This means that we can obtain the pressure's value if we know the above quantities. Since #R = 8,2 J/(molK)# is constant and we know the values for #T# and #V#, we must figure out what happens to the number of mols of each gas after the reaction takes place.
Since it takes 5 molecules of gas #B# to bind to 2 molecules of gas #A#, we can extend this proportion to their corresponding number of mol. Since initially we have #n_(1A) = 8# mol and #n_(1B) = 10# mol, then, after the reaction takes place, we will end up with #n_(2A) = 4 mol#, #n_(2B) = 0# mol and #n_C = 2# mol of the new gas, the product of the #A# and #B# reaction.
Then, our final pressure will be described by:
#P_2 = (RT)/V(n_A + n_B + n_C)#.
(Remember that #P_1 = (RT)/V(n_A + n_B)#.
Taking the values to perform the calculations:
#P_1 = (0.082 (atm cancel(L))/(cancel(mol)cancel(K)) * 360 cancel(K))/(5 cancel(L)) * (8 + 10) cancel(mol)#;
#P_1 = 106.3 atm#.
After the reaction takes place:
#P_2 = (0.082 (atm cancel(L))/(cancel(mol)cancel(K)) * 270 cancel(K))/(5 cancel(L)) * (4 + 0 + 2) cancel(mol)#;
#P_2 = 26.6 atm#.
Therefore, the change in pressure, #DeltaP#, is:
#DeltaP = P_2 - P_1#;
#DeltaP = -79,7 atm#.
