1 mol He and 3 ~mol ~N ₂ exert a pressure of 16 atm . Due to a hole in the vessel in which mixture in placed, mixture leaks out. What is the composition of mixture effusing out initially?

Step-by-step solution

r₁ r₂ = p₁ p₂ M₂ M₁ aligned & P_ He =x_ He P_ total &= 1 4 16=4 atom aligned aligned & P_ N₂ =x_ N₂ p_ total &= 3 4 16=12 atom = (p_ total ⁻ p_ He ) aligned aligned r_ He r_ N ₂ & = p_ He p_ N ₂ M ( ~N ₂ ) M( He ) & = 4 12 28 4 =0.88 aligned Hence, moles of He and N ₂ effusing out initially are in the ratio 0.88: 1 .