Question

The mixtur of 0.5 mole of helium and 0.3 mole of n2 in a container 0.82 litre at 27 c

Answer 1

Given : The mixture of 0.5 mole of Helium and 0.3 mole of N₂ in a container 0.82 litre at 27°C

To find : The partial pressure of each component of mixture and total pressure applied in the container.

solution : here Helium and Nitrogen are two non - reactive gases. so we can find pressure of each separately.

pressure of Helium, P₁ = n₁RT/V

here no of moles of He, n₁ = 0.5

R = 0.082 atm.litre/mol/K

T = 27°C = 300 K

V = 0.82 Litre

so, P₁ = (0.5 × 0.082 × 300)/(0.82) = 15 atm

now pressure of nitrogen, P₂ = n₂RT/V

here no of moles of nitrogen, n₂ = 0.3

so, P₂ = (0.3 × 0.082 × 300)/(0.82) = 9 atm

from Dalton's law of partial pressure,

total pressure, P = P₁ + P₂

= 15 atm + 9 atm

= 24 atm

Therefore the total pressure applied by the gases in the container is 24 atm.