A 8.95 L container holds a mixture of two gases at 21 °C. The partial pressures of gas A and gas B, respectively, are 0.373 atm and 0.650 atm. If 0.220 mol of a third gas is added with no change in volume or temperature, what will the total pressure become?

The total pressure is 1.616 atm

Explanation:

First of all we say:

In a closed system, sum of partial pressures is the value for the total pressure.

In our system, we have gas A and B

Total pressure without the third gas is = 0.373 atm + 0.650 atm = 1.023 atm

As we add a third gas, with no change in volume or T°, let's find out by the Ideal Gases Law, its pressure:

P . V = n . R . T

P = ( n . R . T) / V

P = (0.220 mol . 0.082 . (273 + 21°C)) / 8.95L = 0.593 atm

273 + 21°C → Absolute value of T°

Let's sum the partial pressures, then:

0.373 atm + 0.650 atm + 0.593 atm = 1.616 atm.

It's ok to say that the total pressure was increased, because we have more gas now.