# root mean squared velocity (Urms) vs velocity in KE equation

#### JKMBC

What's the difference between the velocity described in the equation KE=1/2mv^2=3/2*Kb*T and the root mean squared velocity Urms=sqrt(3RT/M)?

Can you use both to find the most like velocity of a given gas particle?

I thought about this while working on Kaplan FL #5, it was in the Chem/Phys section, question #44.

Thanks!!

#### basophilic

5+ Year Member
KE=1/2mv^2 refers to a single particle - so this will give a specific velocity of a single particle

KE = 1/2mv(rms)^2 is average velocity of a population of particles - this is used in conjunction with equipartition theorem (which pretty much says that the avg. kinetic energy of all particles in a population is approximately equal in all translational directions - 1/2Kb*T in one direction, so 3 times that for all 3 dimensions); i.e. 1/2mv(rms)^2=3/2*Kb*T

When you solve this equality you'll get the "Urms" eqn you posted above. Doing this requires a fundamental phys chem relationship: R = N*Kb; here N is Avogadro's number (i.e. number of particles per mole).
Derivation:
1/2mv(rms)^2=3/2*Kb*T --> v(rms)^2 = 3KbT/m
Hence, Kb = R/N --> v(rms)^2 = 3(R/N)T/m = 3RT/Nm
m = mass of single particle; hence Nm = mass of a mole of particles --> i.e. molecular mass (which you designated as 'M')
Hence: v(rms) = sqrt(3RT/Nm) = sqrt(3RT/M)

In short, v is velocity of a particle; v(rms) is average velocity in a population of particles.

The "most like" velocity you refer to is I'm guessing the most probably velocity which is DIFFERENT from the v and v(rms). This is found by calculating the maximum of the Boltzmann distribution; in a standard Gaussian bell curve, you'd expect average and most probable to be identical b/c a Gaussian curve is symmetric; because Boltzmann distribution is asymmetric, they are not the same - v(rms) always greater than v(probable)

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#### JKMBC

Thanks for the derivation! I meant most likely.. I'm referring to this question below, where in the solutions they mention 1/2mv^2=3/2*Kb*T. To do this did they assume the 'most likely velocity' would be equal to v(rms)? I don't understand how they equated those two if they were just referring to a single gas particle. If they were in fact referring to a population of gas particles, then I should have gotten the correct answer by using v(rms)=sqrt(RT/M)...right?

(The relevant passage excerpt is below the question)..thanks!!!  • Get it done