A 8.75-L container holds a mixture of two gases at 23 °C. The partial pressures of gas A and gas B, respectively, are 0.317 atm and 0.755 atm. If 0.100 mol of a third gas is added with no change in volume or temperature, what will the total pressure become?

Answer 1

p of gas C = nRT/V. Substitute and solve for pressure of gas C.
Then Ptotal = pA + pB + pC = about 1.35 atm.