Equipartition Theorem Question

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The discussion revolves around calculating the temperature at which the root-mean-square speed of a helium atom equals that of an air molecule at standard temperature and pressure (STP). The formula used is vrms = Sqrt[3 kT/m], where T is temperature and m is the mass of the molecule. The calculations involve comparing the speeds of helium and oxygen, using molecular weights to derive the temperature. The final result is determined to be approximately 307 K, which rounds to 34.1°C when converted. The conclusion confirms the accuracy of the calculation and the significance of using absolute temperature in Kelvin.
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Homework Statement


Assuming that pressure remains constant, at what temperature is the root-mean square speed of a helium atom equal to the root-mean-square speed of an air molecule at STP? Express your answer to 3 significant figures in Kelvin.


The Attempt at a Solution



the root mean square speed is given by vrms = Sqrt[3 kT/m]

where T is the temperature and m is the mass of one molecule (k is the Boltzmann constant)

so if we take ratios, we get

vrms(He)/vrms(O2) = Sqrt[T(he)/T(O2)] Sqrt[m(O2)/m(He)]

I am using oxygen since air is a mixture and there is no such thing as a molecule of air. If you need to compare an ensemble of He atoms to an ensemble of air molecules, use molecular weight = 28.97

so we have

vrms(He)/vrms(O2) = Sqrt[T(He)/273][Sqrt[32/4]

for vrms(He) = vrms()2), then

1 = (T/273)(8) => T=273/8 solve for T

So you get 34.125 C? Then going to Kevlin it would be 307.125 and rounding to 3 significant figures is 307?

Is this right??
 
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T is the absolute temperature in K.

ehild
 
So it is 34.1 rounding to 3 significant figures?
 
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