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Topic: Graham's Law  (Read 3758 times)

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Offline jehu3

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Graham's Law
« on: December 10, 2006, 12:39:46 AM »
I am studying for a test and came across a question using Graham's Law.

I won't bother copying out the whole question.  We are given the time for an unknown element to effuse through a barrier and the time for Cl to effuse at same temp. and pressure.

My understanding is Graham's Law is:

Rate A/Rate B = square root [(Molecular Weight B/ Molecular Weight A)]

So in this case:

Rate Cl/Rate unknown = square root [(Molecular Weight unknown/ Molecular Weight Cl)]
4.3 min/ 5.2 min = square root [(Molecular Weight unknown/ 35.47)]
and so on .

But, the Prof. said that rate is the inverse of time so, the equation should be :

Rate Cl/Rate unknown = square root [(Molecular Weight Cl/ Molecular Weight unknown)]
4.3 min/ 5.2 min = square root [(35.47/Molecular Weight unknown)]


My question is - aren't rate and time the same thing?


Offline Yggdrasil

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Re: Graham's Law
« Reply #1 on: December 10, 2006, 01:56:12 AM »
If something occurs at a greater rate, it will take less time to finish.  For example, speed is a rate.  If you are traveling at a faster speed, you will reach your destination in less time than someone traveling at a slower rate.

Offline Donaldson Tan

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Re: Graham's Law
« Reply #2 on: December 13, 2006, 03:56:29 PM »
But, the Prof. said that rate is the inverse of time so

Your Prof is right!
"Say you're in a [chemical] plant and there's a snake on the floor. What are you going to do? Call a consultant? Get a meeting together to talk about which color is the snake? Employees should do one thing: walk over there and you step on the friggin� snake." - Jean-Pierre Garnier, CEO of Glaxosmithkline, June 2006

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