# Why is the rate inversely proportional to the square root of temperature in Grahams Law of Diffusion?

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?

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