# Root mean square speed equation

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#1
How do you go from:
vr.m.s = (3kT/m)^1/2 to vr.m.s = (3RT/M)^1/2
where M is the molar mass and m is the mass
0
3 years ago
#2
R is some random gas constant, there is a formula for it somewhere in the book but a quick calculation based on guess shows R = kNa, Na being Avocado.. Avogadro's, constant.

So. 3kT/m to 3RT/M
3k/m to 3R/M

m is just the atomic mass, and M is the atomic mass * avocados constant, i.e. mNa
And we know that R = kNa

So kNa divided by mNa will give k/m, the initial value.

I hope I helped?

If you just know that R = kNa (Na being the number of molecules in a mole of something) and k being Maxy-Boltzy' (Maxwell-Boltzmann) constant that should help you derive the right equation
0
3 years ago
#3
Hi!

if you consider the kinetic energy to the entire sample of the gas, replacing the mass m by molecular mass M of the sample, plus add the number of moles n, the equation of kinetic energy Ek for gas molecules results in:

Ek = 1/2mv² = 1/2nMv²

For constant temperature T, the kinetic Energy can be written as formula for ideal gases in the one hand, but also as average kinetic energy of the particles in the other hand.

For ideal gases, it is

Ek = 3/2nRT, where R is the molecular gas constant for ideal gases.

For average kinetic energy of particles, it is

Ek = 3/2NkT, where k is the Boltzmann-constant and N the number of gas molecules.

This leads to Ek = 1/2nMv² = 3/2nRT = 3/2NkT.

So you get v = (3RT/M)^1/2 by converting the equation 1/2nMv² = 3/2nRT.

Replacing M by m (mass of one molecule of the gas) and n by N it is Ek = 1/2nMv² = 1/2Nmv² now.

By converting 1/2Nmv² = 3/2NkT you get v = (3kT/m)^1/2 and therefore finally

v = (3kT/m)^1/2 = (3RT/M)^1/2

quod erat demonstrandum!
2
#4
(Original post by Kallisto)
Hi!

if you consider the kinetic energy to the entire sample of the gas, replacing the mass m by molecular mass M of the sample, plus add the number of moles n, the equation of kinetic energy Ek for gas molecules results in:

Ek = 1/2mv² = 1/2nMv²

For constant temperature T, the kinetic Energy can be written as formula for ideal gases in the one hand, but also as average kinetic energy of the particles in the other hand.

For ideal gases, it is

Ek = 3/2nRT, where R is the molecular gas constant for ideal gases.

For average kinetic energy of particles, it is

Ek = 3/2NkT, where k is the Boltzmann-constant and N the number of gas molecules.

This leads to Ek = 1/2nMv² = 3/2nRT = 3/2NkT.

So you get v = (3RT/M)^1/2 by converting the equation 1/2nMv² = 3/2nRT.

Replacing M by m (mass of one molecule of the gas) and n by N it is Ek = 1/2nMv² = 1/2Nmv² now.

By converting 1/2Nmv² = 3/2NkT you get v = (3kT/m)^1/2 and therefore finally

v = (3kT/m)^1/2 = (3RT/M)^1/2

quod erat demonstrandum!
Thank you so much
0
3 years ago
#5
(Original post by znx)
Thank you so much
Your thank appreciates me! I was so endeavoured to come to the solution by writing the equations and relations on papers. Was not an easy-going!
0
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