In my book it is given that:

The general form of the Grahams Law of Diffusion can be stated as follows when one or all of the parameters are varied: $$\text{rate} \propto \frac{PA}{\sqrt{TM}},$$ where $P$ - pressure, $A$ - area of the hole, $T$ - temperature, $M$ - Molecular weight.

How can rate be inversely proportional to the square root of temperature? I feel that on increasing temperature it must increase, but how can the rate decrease? Is the formula in my book correct or incorrect?

  • 1
    $\begingroup$ Call it Effusion and you will easily find the answer. Hint: relation kinetic energy and T. $\endgroup$
    – Alchimista
    Aug 29 '19 at 8:53
  • $\begingroup$ Thanks @Alchimista. I am unable to understand you. Could you please explain? Only thing I know is kinetic energy increases with Temp. T. $\endgroup$
    – user81791
    Aug 29 '19 at 10:25

It helps to rewrite $$\text{rate} \propto \frac{PA}{\sqrt{TM}}$$ by assuming the ideal gas law holds, as follows: $$\text{rate} \propto \frac{RA}{V_m}\sqrt{\frac{T}{M}}$$ where $V_m$ is the molar volume or inverse of molar particle density.

Written this way it is clear that if the particles occupy the same volume, increasing their temperature increases the rate at which they effuse through the apperture.


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