A 7.30-L container holds a mixture of two gases at 45 °C. The partial pressures of gas A and gas B, respectively, are 0.410 atm and 0.620 atm. If 0.150 mol of a third gas is added with no change in volume or temperature, what will the total pressure become?

Answer 1

Total Pressure= PressureA + PressureB.....
when you add .150 mol you have to find pressure using PV=nrT sooooo.... P(7.30)=(.150)(.08206)(318) for P=.536 where n is moles, r is a constant and T is in kelvin... now add up all the Pressures
Ptotal= .410+.620+.536= 1.566