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#.