[2810713]

Why does root mean square velocity has importance in  Kinetic theory?
Whats its specific importance? I mean why not just mean of magnitudes or cube-root mean cube velocity etc.

I don't know more abut statistics, so please help...

hrushikesh

[FeLiXe]

the significance of the root mean squared velocity is that by it the kinetic energy is defined (E = mv^2 / 2). a few more kinetic thoughts also relate it to temperature.

with every hit between two particles the kinetic energy stays the same. if the particles are the same therefore also squared velocity stays the same. the vectorial sum of velocities also stays the same but not necessarily the absolute value.

the mean velocity is only a statistical value and might differ over time if you only look at a few particles. the root mean square stays exactly the same due to energy conservation.

maybe that helped a little bit but I don't know all the proper English terms.

[2810713]

Thanks, now i understand that it remains the same in a given state so it may be used as a state-co-ordinate. Moreover it helps in getting mean K.E. Thanks again.

Now all molecules in gas don't have same K.E., there would be a curve formed if we plot no. of molecs on Y-axis and velocit on X axis.
Still the whole system has one value of temperature!
I think this is because- A system has gas with total volume V=nv , v are small volumes and n is their no. If you keep thermometer in some volume v the the no. of molecuales collided [Y-axis] to it versus the velocity [X-axis] of those will give similar curve as no. of molecules in such a generally accesible volume is fairly large. And the total momentum transfered [ heat] would be the same in each unit volume as we are using the same thermometer in similar position.
If we are able to just measure the momentum transfer in very small volume, we may get different reading for temperature as total momentum transfered won't be the same due to less no. of molecules under consideration.
Is my explanation correct?


hrushikesh

[Donaldson Tan]

RMS Velocity is not the equal to average velocity

eg. we have 2 molecules in a tiny box, each moving at different speed.

average velocity c = (c₁ + c₂)/2

rms velocity rms-c = square-root[[ (c₁² + c₂²)/2 ]]

we can express the velocity of eachh molecule as:
c₁ = c + ?c₁
c₂ = c + ?c₂

c₁² = c² + 2.c.?c₁ + (?c₁)²
c₂² = c² + 2.c.?c₂ + (?c₂)²

c₁² + c₂² = 2.c² + 2.c.(?c₁+?c₂) + (?c₁)² + (?c₂)²

unless all the ?c terms equal to zero (which is not possible), then c₁² + c₂² ? 2.c², thus c ? rms-c

[FeLiXe]

you have to differentiate between momentum transfer which macroscopically leads to pressure
and kinetic energy which leads to temperature

average kinetic energy might statistically differ if you look at different parts of your system. the important point is though that in an ideal enclosed system the kinetic energy (and therefore temperature and rms velocity) stays exactly the same. average velocity will only stay statistically close to the value.

for pretty much any practical purpose I guess this does not matter at all though because statistical deviations don't matter if you watch 10^23 molecules. actually the only reason I thought about these things is that I wrote a program which simulates gas particles. since my computer can't handle much more than I hundred I notice the mean velocity changing and the rms velocity staying exactly the same.

--

I thought of something else why rms velocity is important. You need it for deriving the Maxwell-Boltzmann distribution. You can start assuming that velocity in each direction is normal distributed with the mean to be 0. Than you still have to determine the parameter sigma. You can do this when you insert the rms velocity that you can derive with different thoughts.

[2810713]

Thanks both of you,that ?C thing gave a clear mathematical idea.


Felixe-
"you have to differentiate between momentum transfer which macroscopically leads to pressure
and kinetic energy which leads to temperature".
But the thermometer can measure the momentum transferred. M I correct?

KIndly explain that Mxwell-Boltzman distribution thing...
How can RMS velocity give us 'sigma'?

hrushikesh

[FeLiXe]

A barometer measueres momentum. A thermometer measures energy. If you want the exact explanation: A thermometer is a closed system which is in thermodynmic equilibrium with the system you want to measure. According to the zeroth law of thermodynamics then the temperature inside the thermometer and the other system are the same.

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now let's go to Maxwell-Boltzmann

first we need the kinetic temperature
http://hyperphysics.phy-astr.gsu.edu/hbase/kinetic/kintem.html#c1

you start out assuming the speeds in each direction are normal distributed. then we find out that the mean is zero. and the standard deviation is the same as the rms velocity. I'll write the rest with Mathematica:
http://stud4.tuwien.ac.at/~e0425252/mbdist/

the first part is not mathematical mostly thinking a little bit. The second part goes easier with a little bit of advanced calculus but it's the same idea as in the radial probability function of electron orbitals for example.

---

actually I'd like to add something to geodome. I'm kind of a part time mathematician ...

it's a general theorem of mathematics that with any set of numbers the root mean sqare is always larger or equal to the arithmetic average. (equal only if all the values are the same). same for arithmetic and geometric Sqrt(a*b). even smaller the harmonic average: 2/(1/a+1/b)

[2810713]

Thanks for that density-function derivation... i sort of enjoyed it.

 Barometer measures momentum and thermometer measures energy. But the energy transfer is by means of momentum transfer or say K.E. transfer. So, my arguments containing momentum can now be converted into K.E.[ a scalar] in that direction. Or the momentum arguments can be converted into 'energy'  i.e. temperature arguments by using appropriate factors. right?

hrushikesh

[FeLiXe]

well you can convert it but you would need the exact mass and the gas has to behave ideal

[2810713]

 Yeah,  we may get exact mass but ideal behaviour is not in our hands , so better i trust thermometers for practical purposes.

But using the improved  relation between P, V , n and T we may get closer to the answer.
hrushikesh