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anonymous
 4 years ago
I am trying to prove E(kin)=(3/2)RT (T in kelvin)
in order to prove pV=nRT (so i cant use that equation)
I have proved E(kin)=(1/2)mv^2
anonymous
 4 years ago
I am trying to prove E(kin)=(3/2)RT (T in kelvin) in order to prove pV=nRT (so i cant use that equation) I have proved E(kin)=(1/2)mv^2

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y2o2
 4 years ago
Best ResponseYou've already chosen the best response.0So, do you know that pV = NKT ?!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0no that is what i am trying to prove

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0or is N mole and K=R here or is it something else? Where R is molar gas constant = 8.314472 m2 kg s2 K1 mol1

y2o2
 4 years ago
Best ResponseYou've already chosen the best response.0Actually \[E_{K.E} = {3 \over 2} KT \ \ Not \ \ {3\over2}RT\] where K is Boltzmann constant , not the Universal gas constant. and K = R/A where A is Avogadro's number.

Mani_Jha
 4 years ago
Best ResponseYou've already chosen the best response.0Do u know that PV=\[mnc ^{2}/3\]? From here u get K.E=3(PV/n)/2 per molecule. Combining boyle's and charle's law and avogadro's law, u will get \[P1V1/nT1 = CONSTANT\] for any gas, which is called R. Thus PV=nRT. Put this value and u will get K.E=3RT/2 per molecule

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i wanted to prove it without thoose laws like avogadros law since to me it seems like they are based on empirical data and I wanted to prove it theorethically because I like it better :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0we use Equipartition theorem to get E(k)=3KT/2, and use Boltzmann statistics to prove PV=nRT, and in this way we get Boltzmann Constant quantitatively ,namely K=AR, where A is Avergadro Constant.
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