A 9.80-L container holds a mixture of two gases at 39 °C. The partial pressures of gas A and gas B, respectively, are 0.426 atm and 0.832 atm. If 0.250 mol a third gas is added with no change in volume or temperature, what will the total pressure become?

Answer 1

The easy way to do this is
p 3rd gas = nRT/V, then
pA + pB + p3rd = Ptotal.

You can do it the long way this way.
n gas A = PV/RT.
n gas B = PV/RT
n third gas = 0.250
ntotal = nA + nB + n3rd.
Then Ptotal = nRT/V
You should get the same answer either way. I believe the answer is close to 1.91 atm but check that. I may have punched in a wrong number.