Partial Pressure

Question

A 9.60 L9.60 L container holds a mixture of two gases at 21 °C.21 °C. The partial pressures of gas A and gas B, respectively, are 0.181 atm0.181 atm and 0.877 atm.0.877 atm. If 0.130 mol0.130 mol of a third gas is added with no change in volume or temperature, what will the total pressure become?

Answer 1

Total pressure (P_T) = ∑partial pressures

Use the ideal gas law to find the pressure contributed by gas #3

PV = nRT

P = pressure of gas #3 = ?

V = volume = 9.60 L

n = moles = 0.130 moles

R = gas constant = 0.0821 Latm/Kmol

T = temperature in Kelvin = 21ºC + 273 = 294K

Solving for P, we have ...

P = nRT / V = (0.131)(0.0821)(294) / 9.60

P = 0.329 atm

P_T = 0.329 atm + 0.181 atm + 0877 atm = 1.39 atm total pressure