Question

If a gaseous mixture is made by combining 1.55 g Ar and 1.80 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

Partial Pressure Ar = 0.379 atm

Partial Pressure Kr = 0.210 atm

P total = 0.589 atm

Step-by-step explanation:

First let's convert the masses of both gases to moles, using their respective atomic weight:

• 1.55 g Ar ÷ 39.948 g/mol = 0.0388 mol Ar

• 1.80 g Kr ÷ 83.798 g/mol = 0.0215 mol Kr

Now we can calculate the partial pressure of each gas, using PV = nRT:

• 25 °C ⇒ 25+273.16 = 298.16 K

• Ar ⇒ P * 2.50 L = 0.0388 * 0.082 atm·L·mol⁻¹·K⁻¹ * 298.16 K

P = 0.379 atm

• Kr ⇒ P * 2.50 L = 0.0215 * 0.082 atm·L·mol⁻¹·K⁻¹ * 298.16 K

P = 0.210 atm

Finally, the total pressure is the sum of the partial pressure of each gas:

• Total Pressure = 0.379 atm + 0.210 atm = 0.589 atm