Question

A gas mixture contains each of the following gases at the indicated partial pressures: N2, 231 torr ; O2, 101 torr; and He, 155 torr .
What is the total pressure of the mixture?

What mass of each gas is present in a 1.00 -L sample of this mixture at 25.0 degrees Celsius?

Enter your answers numerically separated by commas.

Answer 1

Answer:

P = 487 Torr

mN₂ = 0.347 g

mO₂ = 0.174 g

mHe = 0.0333 g

Explanation:

Step 1: Calculate the total pressure of the mixture

The total pressure of the mixture (P) is equal to the sum of the partial pressures of the gases.

P = pN₂ + pO₂ + pHe

P = 231 Torr + 101 Torr + 155 Torr = 487 Torr

Step 2: Calculate the moles of each gas

We have 3 gases in a 1.00 L container at 25.0 °C (298.2 K). We can calculate the number of moles using the ideal gas equation: P × V = n × R × T.

nN₂ = pN₂ × V / R × T

nN₂ = 231 Torr × 1.00 L / (62.4 Torr.L/mol.K) × 298.2 K = 0.0124 mol

nO₂ = pO₂ × V / R × T

nO₂ = 101 Torr × 1.00 L / (62.4 Torr.L/mol.K) × 298.2 K = 0.00543 mol

nHe = pHe × V / R × T

nHe = 155 Torr × 1.00 L / (62.4 Torr.L/mol.K) × 298.2 K = 0.00833 mol

Step 3: Calculate the mass of each gas

The molar mass of N₂ is 28.01 g/mol.

mN₂ = 0.0124 mol × 28.01 g/mol = 0.347 g

The molar mass of O₂ is 32.00 g/mol.

mO₂ = 0.00543 mol × 32.00 g/mol = 0.174 g

The molar mass of He is 4.00 g/mol.

mHe = 0.00833 mol × 4.00 g/mol = 0.0333 g