Graham's Law of Diffusion

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10+ Year Member
Oct 5, 2011
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Just need clarification on this please:

So I remember Graham's Law as this:


What is the speed of a gas particle at 125° C, if it has a speed of 100 m/s at 25° C?

A. 500 m/s
B. 223 m/s
C. 133 m/s
D. 114 m/s


D) 114 m/s

The speed is greater at higher temperatures. This doesn't help, because all of the answer choices are greater than 100 m/s. To determine the exact value, the temperature must be converted to kelvins. The temperature increase is from 298 to 398, which means the temperature is 1.33 times greater. Because it is a square root function, the speeds increases by a factor of √(1.33), which when multiplied by 100 m/s yields a value less than 133 m/s.
V2/V1 = √(T2/T1)


Why in this example does TBR say that the rate law is: V2/V1 = √(T2/T1)??

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You are dealing with one molecule only, so there is no diffusion happening. You use the root mean square velocity which is a little thing that is very easy to forget.
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