Answer:
P(mixture) = 1.92 atm
Explanation:
Given data:
Mass of H₂ = 0.200 g
Mass of N₂ = 1.00 g
Mass of Ar = 0.820 g
Volume = 2 L
Temperature = 20°C
Pressure of mixture = ?
Solution:
Pressure of hydrogen:
Number of moles of hydrogen = mass / molar mass
Number of moles of hydrogen = 0.200 g / 2 g/mol
Number of moles of hydrogen = 0.1 mol
P = nRT / V
P = 0.1 mol× 0.0821 atm. L.mol⁻¹ .k⁻¹ × 293 K / 2L
p = 2.41 atm. L /2 L
P = 1.2 atm
Pressure of nitrogen:
Number of moles of nitrogen = mass / molar mass
Number of moles of nitrogen = 1 g / 28 g/mol
Number of moles of nitrogen = 0.04 mol
P = nRT / V
P = 0.04 mol× 0.0821 atm. L.mol⁻¹ .k⁻¹ × 293 K / 2L
p = 0.96 atm. L /2 L
P = 0.48 atm
Pressure of argon:
Number of moles of argon = mass / molar mass
Number of moles of argon = 0.820 g / 40 g/mol
Number of moles of argon = 0.02 mol
P = nRT / V
P = 0.02 mol× 0.0821 atm. L.mol⁻¹ .k⁻¹ × 293 K / 2L
p = 0.48 atm. L /2 L
P = 0.24 atm
Total pressure of mixture:
P(mixture) = pressure of hydrogen + pressure of nitrogen + pressure of argon
P(mixture) = 1.2 atm + 0.48 atm + 0.24 atm
P(mixture) = 1.92 atm
