Question

A gaseous mixture in a 3 litre vessel contains 0.1 mol H2, 0.3 mol He at 27 °C and 2 bar.
Find the partial pressures of H, and He.​

Answer 1

152

Answer:

Moles of N2= 0.3

Moles of O2= 0.1

Moles of He = 0.1

Mole fraction of O2=Λ

$\frac{no2}{nn2 + no2 + nhe} = \frac{0.1}{0.3 + 0.1 + 0.1} = 0.20$

Total pressure = 1 atm

Partial Pressure of O2=Λ02×Total pressure = 0.02 × 760 mm of Hg = 152 mm of Hg

Answer 2

Given :

Volume (V) = 3 L

Moles of H₂ = 0.1

Moles of He = 0.3

To Find :

Partial pressures of H and He.​

Solution :

Mole fraction of H₂ ($X_{H_2}$)  = $\frac{0.1}{0.1 + 0.3}$

                                          = $\frac{0.1}{0.4}$

                                          = 0.25

Mole fraction of He ($X_{He}$) = $\frac{0.3}{0.4}$

                                          = 0.75

Partial pressure of H₂ = Total pressure × $X_{H_2}$

                                     =   2 × 0.25

                                     = 0.5 Bar

Partial pressure of He = Total pressure × $X_{He}$

                                      = 2 × 0.75

                                      = 1.5 Bar

Hence, the partial pressure of H₂ and He is 0.5 Bar and 1.5 Bar respectively.