Question

at what temperature will be the rate of diffusion of N2 be 1.6 times the rate of diffusion of So 2 at 27 degrees c

Answer 1

Answer:

The molecular mass of N₂, M₁ = 28 g/mol

The molecular mass of SO₂, M₂ = 64 g/mol

Let the rate of diffusion of N2 and SO2 be “r₁” & “r₂” respectively and the temperature of N2 & SO2 be “T₁” & “T₂” respectively.

The temperature of SO₂ gas, T₂ = 27℃ = 27+273 = 300 K

We are given,  

r₁ = 1.6*r₂

⇒ r₁/r₂ = 1.6

Therefore, we can write the eq. as,

 

r₁/r₂ = √[{M₂/M₁}*{T₁/T₂}]

substituting the given values

⇒ 1.6 = √[{64/28} * {T₁/300}]

taking square on both sides

⇒ 2.56 = {64/28} * {T₁/300}

⇒ T₁ = (2.56*28*300) / 64  

⇒ T₁ = 336 K = 336 – 273 = 63℃

Thus, the temperature of N₂ will be 63℃.