Question

16 mL of He gas effuses through a pin hole
in 4 sec from a container having PHe equal
to 1 atm. If same container is filled with CH4
having pressure 2 atm, how much volume
(in mL) of CH4 will be leaked through same
pin hole in 2 sec ?​

Answer 1

Given info : 16mL of He gas effuses through pin hole in 4 seconds from a container having pHe is 1 atm. The same container is filled with methane having pCH4 is 2atm.

We have to find the volume of methane leaked through the same pin hole.

solution : using formula,

V₁/V₂ = P₁/P₂√(M₂/M₁)

Where, V represents volume , P pressure and M molar mass.

Here, V₁ = 16mL, V₂ = ?, P₁ = 1atm , P₂ = 2atm, M₁ = 4g/mol and M₂ = 16g/mol

So, 16/V₂ = (1/2)√(16/4)

⇒16/V₂ = (1/2) × 2 = 1

⇒V₂ = 16 mL

Volume of methane gas leaked through pin hole is 16mL