Question

If a gaseous mixture is made by combining 3.78 g Ar and 3.27 g Kr in an evacuated 2.50 L container at 25.0 ∘ C, what are the partial pressures of each gas, P Ar and P Kr, and what is the total pressure, P total, exerted by the gaseous mixture?

Answer 1

The partial pressure of Ar = 0.925 atm

The partial pressure of Kr = 0.381 atm

The total pressure = 1.306 atm

Step-by-step explanation:

Step 1: Data given

Mass of Ar = 3.78 grams

Mass of Kr = 3.27 grams

Volume = 2.50 L

Temperature = 25.0 °C

Molar mass of Ar = 39.95 g/mol

Molar mass of Kr = 83.80 g/mol

Step 2: Calculate moles of Ar

Moles Ar = mass Ar / molar mass Ar

Moles Ar = 3.78 grams / 39.95 g/mol

Moles Ar = 0.0946 moles

Step 3: Calculate moles of Kr

Moles Kr = 3.27 grams / 83.80 g/mol

Moles Kr = 0.039 moles

Step 4: Calculate partial press of each gas

p*V = n*R*T

⇒ p =the partial pressure of the gas

⇒ V = the volume in the container = 2.50 L

⇒ n = the number of moles of the gas

⇒ R = the gasconstant = 0.08206 L*atm/K*mol

⇒ T = the temperature = 25°C = 298 K

The partial pressure of Ar = (n*R*T)/V

p(Ar) = (0.0946*0.08206*298)/ 2.5

p(Ar)= 0.925 atm

p(Kr) = (0.039*0.08206*298)/ 2.5

p(Kr) = 0.381 atm

Step 5: Calculate the total pressure

The total pressure = p(Ar) +p(Kr)

Total pressure = 0.925 atm + 0.381 atm = 1.306 atm