Question

A 7.55 L container holds a mixture of two gases at 25 C. The partial pressures of gas A and gas B, respectively, are 0.348 atm and 0.711 atm. If 0.120 mol of a third gas is added with no change in volume or temperature, what will the total pressure become?

Answer 1

Pressure of third gas:

PV=nRT.

P=nRT/V

P=(0.120)(0.0821)(298)/7.55=0.389

total pressure:

0.348 + 0.711 + 0.389 = 1.448 atm