Deduce Graham’s law of diffusion from kinetic theory of gases using

1.	Deduce Graham’s law of diffusion from kinetic theory of gases using expression for pressure.
Answer» Let us consider two gases A and B diffusing into one another. Let ρ₁ and ρ₂ be their densities and v₁ and v₂ be their respective r.ms. velocities. Pressure exerted by gas A, \(P_1= \frac{1}{3}ρ_1V_1^2\) and pressure exerted by gas B, \(P_2= \frac{1}{3}ρ_2v_2^2\) When steady state of diffusion is reached P₁= P₂ \( \frac{1}{3}ρ_1v_1^2\) = \( \frac{1}{3}ρ_2v_2^2\) \( \frac{v_1^2}{v_2^2}=\frac{ρ_2}{ρ_1}\) or \( \frac{v_1}{v_2}=\,\sqrt\frac{ρ_2}{ρ_1}\) If r₁and r₂ be the rates of diffusion of gases A and respectively. \(\frac{r_1}{r_2}=\frac{v_1}{v_2}=\sqrt\frac{ρ_2}{ρ_1}\) Thus this law states that the rate of diffusion of a gas is inversely proportional to the square root of its density. r ∝ \(\frac{1}{\sqrtρ}\)

Deduce Graham’s law of diffusion from kinetic theory of gases using expression for pressure.

Answer»

Let us consider two gases A and B diffusing into one another.

Let ρ₁ and ρ₂ be their densities and v₁ and v₂ be their respective r.ms. velocities.

Pressure exerted by gas A,

\(P_1= \frac{1}{3}ρ_1V_1^2\)

and pressure exerted by gas B,

\(P_2= \frac{1}{3}ρ_2v_2^2\)

When steady state of diffusion is reached

P₁= P₂

\( \frac{1}{3}ρ_1v_1^2\) = \( \frac{1}{3}ρ_2v_2^2\)

\( \frac{v_1^2}{v_2^2}=\frac{ρ_2}{ρ_1}\)

or \( \frac{v_1}{v_2}=\,\sqrt\frac{ρ_2}{ρ_1}\)

If r₁and r₂ be the rates of diffusion of gases A and respectively.

\(\frac{r_1}{r_2}=\frac{v_1}{v_2}=\sqrt\frac{ρ_2}{ρ_1}\)

Thus this law states that the rate of diffusion of a gas is inversely proportional to the square root of its density.

r ∝ \(\frac{1}{\sqrtρ}\)

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