If a gaseous mixture is made by combining 3.82g Ar & 2.41g Kr in an evacuated 2.50L container at 25.0∘C

Question

what’s the partial pressures of each gas,𝑃Ar and 𝑃Kr,and what is the total pressure,𝑃total,exerted by the gaseous mixture?

Answer 1

The partial pressure of a gas in a mixture is equal to the total pressure x mole fraction of that gas.

The mole fraction is moles of gas / total moles of gas in the mixture.

moles of Ar = 3.82 g Ar x 1 mol Ar/39.9 g = 0.0957 moles Ar

moles Kr = 2.41 g Kr x 1 mol Kr/83.8 g = 0.0288 moles Kr

Total moles = 0.0957 + 0.0288 = 0.1245 moles of gas

mole fraction Ar = 0.0957 / 0.1245 = 0.769

mole fraction Kr = 0.0288 / 0.1245 = 0.231

To find the total pressure, we use total moles and the ideal gas law equation PV = nRT

P = ?

V = volume = 2.50 L

n = moles = 0.1245

R = gas constant = 0.0821 Latm/Kmol

T = temperature in K = 25.0ºC + 273.15 = 298.15K

Solving for P, we have P = nRT/V

P = (0.1245)(0.0821)(298.15) / 2.50

P = 1.22 atm

Partial pressure Ar = mole fraction Ar x total pressure = 0.769 x 1.22 atm = 0.938 atm

Partial pressure Kr = mole fraction Kr x total pressure = 0.231 x 1.22 atm = 0.282 atm