General Chemistry 1

Question

A gaseous mixture is made by combining 2.06g Ar and 3.26g Kr in an evacuated 2.50 L container at 25.0 °C. Calculate the partial pressures of each gas and the total pressure exerted by the gaseous mixture.

Answer:

𝑃Ar= ______bar

𝑃Kr= ______bar

𝑃total= _____bar

Answer 1

PV = nRT

P = pressure (atm); V = volume (L): n = moles; R = gas constant (Latm/Kmol); T = temperature(K)

For Ar:

2.06 g x 1 mol/40 g = 0.515 moles

P = nRT/V = (0.515)(0.0821)(298) / 2.50 = 5.04 atm = P_Ar (unless you use 1.01325 bar = 1 atm) = 5.11 bar

For Kr:

3.26 g x 1 mol/84 g = 0.0388 moles

P = nRT/V = (0.0388)(0.0821)(298) / 2.50 = 0.380 atm = P_Kr x 1.01325 = 0.385 bar

Total pressure = 5.11 bar + 0.385 bar = 5.50 bar