Understanding Kinetic Molecular Theory and Graham's Law

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Discussion Overview

The discussion revolves around the Kinetic Molecular Theory and Graham's Law, focusing on the kinetic energy of ideal gases, the derivation of average kinetic energy, and the relationship between diffusion rates and molar mass. Participants explore theoretical concepts, mathematical formulations, and clarify misunderstandings related to these topics.

Discussion Character

  • Homework-related
  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant expresses confusion about the transition from the kinetic energy formula E = 1/2 mv^2 to the average molar kinetic energy expression Eaverage, molar = 3/2 RT, seeking clarification on the role of Avogadro's constant.
  • Another participant introduces the root-mean-square speed formula v_{rms} = sqrt(3RT/M_m) and suggests substituting it into the kinetic energy equation to find the average kinetic energy of an ideal gas, while advising caution in trusting their explanation.
  • A participant corrects their earlier misunderstanding regarding the notation in the average kinetic energy equation, indicating it should be v with a bar instead of v^-2.
  • Discussion arises about the gas constant R, with one participant asserting it is always 8.314, while another notes that the choice of R depends on the units used for pressure and volume.
  • One participant challenges the characterization of root-mean-square speed as a property of the electron, asserting it applies to any set of moving objects, particularly gas molecules.

Areas of Agreement / Disagreement

Participants exhibit some agreement on the formulas related to kinetic energy and root-mean-square speed, but there is disagreement regarding the characterization of these concepts and the application of the gas constant in different unit systems. The discussion remains unresolved on certain points, particularly the interpretation of the root-mean-square speed.

Contextual Notes

Some participants express uncertainty about the derivations and applications of the formulas discussed, highlighting potential limitations in their understanding of the concepts involved.

Who May Find This Useful

Students studying thermodynamics, kinetic theory, or gas laws, as well as those interested in the mathematical relationships between temperature, kinetic energy, and molecular behavior in gases.

revacious
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Homework Statement



Ok, firstly, I apologise for posting something which is probably trivial to any physics student, but my understanding of physics is pretty poor, so baby steps would be appreciated!

Homework Equations



PV=nRT
E = 1/2 mv2

The Attempt at a Solution



My understanding of it so far:

At the molecular level, a point particle of an ideal gas (ignoring rotational/vibrational components and forces between particles) has kinetic energy equal to:
E = 1/2 mv2

But particles are constantly colliding with each other as well as the walls of the container. Hence an expression for average kinetic energy tells us more information. My course notes give this as:
Ebar = 1/2 mv(bar)-2

an explanation as to what that means and how they got there would be nice :S

And then there's another jump to average molar kinetic energy of an ideal gas, which is given in a different form by different sources. If I have reasoned correctly, the form in my notes for Eaverage, molar = 3/2 RT = 3/2 PV when n=1 in the ideal gas equation.

but how did they jump from 1/2 mv2 to this? No doubt avogadro's constant comes into play, but its clearly not as simple as taking the second expression and multiplying.

And finally, the jump to graham's law. I understand and accept that kinetic energy is only dependent on the temperature; hence two gases at equal temperatures have equal kinetic energy.

If the rate of diffusion/effusion is a velocity, then how did we get to ratex = constant/sqrt(molar massx)?

thanks!
 
Last edited:
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There's a property of the electron called the Root-mean-square speed. The formula for root-mean-square speed is something like this:

v_{rms} = \sqrt{\frac{3RT}{M_m}}

Where v is the rms speed, M_m is the molar mass of the gas, T is temperature, R is 8.314, and 3 is 3.

Plug the rms speed into the kinetic energy of a particle equation and you should get the average kinetic energy of an ideal gas. But don't trust me. (seriously, don't.) Try it.
 
Ok i figured half of it out, it was meant to be a v with a bar instead of v^-2 for the second equation. gah, i feel stupid now.

And with the rms equation, is the constant always 8.314? Would we ever need to use the other versions of the gas constant?
 
Well, the constant is always 8.314, and here's the reason...

The units of the radicand are ...

(\frac{J}{mol-K})(K)(\frac{mol}{g}) = <br /> \frac{J-mol-K}{mol-K-g} = \frac{J}{g}

And J/g is near m^2/s^2, the square of speed, and then the square root takes care of everything.

But, you ask, Joules aren't g-m^2/s^2, they're kg-m^2/s^2. Well, I'm pretty sure that you can ignore this for the same reason that you can throw a 3 in the equation... it's all about proportionality.
 
Char. Limit said:
There's a property of the electron called the Root-mean-square speed

This is not property of the electron :bugeye:

revacious said:
And with the rms equation, is the constant always 8.314? Would we ever need to use the other versions of the gas constant?

Depends on what units are other values expressed in. If you are given pressure in PSI and volume in cubic feet you may prefer other R value. It is all about convenience. But it is different just because it is expressed in different units, physical sense it still the same.

--
 
I meant to say gas...
 
Last edited:
okay, thanks for the input, i think i get this concept better now :)
 

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