A 9.80 L container holds a mixture of two gases at 55 ° C. The partial pressures of gas A and gas B, respectively, are 0.298 atm and 0.589 atm. If 0.130 mol of a third gas is added with no change in volume or temperature, what will the total pressure become?

1.24 atm is the new pressure

Explanation:

We may solve this question with the Ideal Gases Law that must be used, twice. → P . V = n . R . T

Total pressure = Partial pressure of gas A + Partial pressure of gas B

Total pressure = 0.298 atm + 0.589 atm → 0.887 atm

We convert the T° to Absolute value  → 55°C + 273 = 328K

0.887 atm . 9.80L = n . 0.082 . 328K

(0.887 atm . 9.80L) /(0.082 . 328K) = 0.323 moles

These are the moles from the initial mixture, but we add 0.130 moles

Total new moles are 0.323 + 0.130 = 0.453 moles

P = (0.453 mol . 0.082 . 328K) / 9.80L

P = 1.24 atm

Notice, that the pressure was increased. As we add a third gas, the pressure is correctly increased because the molecules from all of the gases   collide more with the walls of the vessel.

the correct answer is:

a. formation of precipitation