We can use ideal gas law here assuming both gases act as ideal gasses.

First we should find the moles of each gas before mixing.

`PV = nRT`

`n = (PV)/(RT)`

Assume the temperature is T in Kelvins.

For gas A;

`n_A = (1xx2.5)/(0.08206T) = 30.466/T`

For gas B;

`n_B = (0.75xx0.5)/(0.08206T) = 4.569/T`

`A+3B rarr AB_3`

To complete the reaction we need 3times B than A. But as you can see above it doen't have that much of B in the flask to react with all the A.

So what will happen is all the B will be reacted.

Amount of B reacted `= 4.569/T`

Amount of A reacted `= 1/3xx(4.569/T) = 1.523/T`

So the rest A in flask `= 30.466/T-1.523/T = 28.477/T`

Amount of `AB_3` formed = reacted amount of A

Amount of `AB_3` formed `= 1.523/T`

So in the final mixture we have excess A and produced AB3.

Total moles in the mixture `= (1.523/T+28.477/T)`

Volume of flask `= (2.5+0.5) = 3L`

Using PV = nRT

`P = (nRT)/V`

`P = (1.523/T+28.477/T)xx0.08206xxT/3 = 0.8206atm`

So the total pressure of the system after joining the flasks is 0.8206atm.