Gas Speed/KE constant temp Please Clarify

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Monkey12

Monkey12
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I was reviewing gas speed, and I know fromt that Boltzman curve that KE and Avg Speed is constant for a given temperature...But is this for every gas? Lilke every gas has the same speed/KE at a given temp? But then I start reading about grahams law stating that r1/r2=(MM2/MM1)^.5 so it would seem that the rate is changing as function of weight? So then does that disagree with the Boltzman logic stating that all gases have the same avg speed/KE at a given temp? Or is okay since its not specifically stating for every single gas...KE=.5mv^2 that mass term would seem to contradict it?
Clarification please!
 
At a particular temperature, any gas will have the same kinetic energy.

But since their masses differ, their velocity will differ. The larger the molecule, the slower it will travel..
 
dude what bonez said

btw bones do u like bangle..ur icon says rally all the way ...but im just curious how u feel
 
DieselPetrolGrl said:
dude what bonez said

btw bones do u like bangle..ur icon says rally all the way ...but im just curious how u feel

mmm. i think that as a designer, Bangle has some pretty progressive ideas. but i am not sure how that meshes with BMW's desire to increase market share... ie: BMW should listen to what the market wants to hear.. controversial designs won't help sell cars. the z4 is actually selling really poorly and word is that BMW has already started doing rebates (something they RARELY do).. so for them to provide rebates this early in a model's lifecycle isn't boding well for bangle.. i am curious to see how the next generation of the 3 series will look, i think the current 3 rocks..

i think BMW should go back to their motto of designing 'the ultimate driving experience' and build great cars to drive (a la e30 M3).. but their desire to increase market share is winning out, so they are making their cars alot less 'drivable'.. but then again, the ultimate drivers car won't exactly win BMW any market share wars..

are you a bimmer fan?
 
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