How Does Angular Velocity Affect Gas Concentration in a Rotating Cylinder?

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SUMMARY

The discussion focuses on the relationship between angular velocity and gas concentration in a rotating cylinder, specifically deriving the concentration formula n(r) based on the ideal gas model. The total chemical potential is expressed as μ_tot = μ_int + μ_ext, where μ_ext is defined as μ_ext = (1/2) Mω²r², leading to the concentration equation n(r) = n(0) exp(-Mω²r²/(2τ)). The participants also explore the implications of these equations for atmospheric molecular concentration, referencing the textbook "Thermal Physics" by Kittel and Kroemer for further context.

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  • Understanding of ideal gas laws and chemical potential
  • Familiarity with angular velocity and its effects on gas behavior
  • Basic knowledge of quantum concentration and its calculation
  • Proficiency in calculus for evaluating integrals involving exponential functions
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A circular cylinder of radius R rotates about hte long axis with angular velocity omega. The cylinder contains an ideal gas of atoms of mass M at temperature tau. Find the expression for the dependence of the concentration n(r) on the radial distance r from the axis in terms of n(0) on the axis. Take the chemical potnetial u to be that of an ideal gas.

\mu_{tot} = \mu_{int} + \mu_{ext}

right?

\mu_{tot} = \tau\log\frac{n}{n_{Q}} + \mu_{ext}
where n is te concentration and nQ is the quantum concentration
n_{Q} = \frac{M\tau}{2\pi\hbar^2}
now I am wondering what external potentisl should be...
should it be the energy of the particle while it is rotation at some radiual distance r with angular velocity omega?
Wold that mean taht
\mu_{ext} = \frac{1}{2} M\omega^2 r^2

so then
\mu = \tau\log\frac{n}{n_{Q}} + \frac{1}{2} M\omega^2 r^2

n(r) = n_{Q} \exp\left(\frac{1}{\tau}\left(\mu-\frac{1}{2}M\omega^2 r^2\right)\right)
n(0) = n_{Q} \exp\left(\frac{\mu}{\tau}\right)

n(r) = n(0) \exp\left(-\frac{1}{2\tau} M\omega^2 r^2\right)


If n is the concentration of molectules at the surface of the Earth, M the mass of a molecule and g the gravitational acceleration at the surface, show that at constant temperature the total numbers of molecules in the atmosphere is

N = 4\pi n(R) \exp\left(-MgR/\tau\right) \int_{R}^{\infty} dr r^2 \exp\left(MgR^2/r\tau\right)

i don't understand why the number of molecules at the bottom is multiplied by the concentration at the above layers... shouldn't it be added?

why they change the exponential's argument to what it is in the integrand...? IM a bit


thank you for the input!
 
Last edited:
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Just asking: what does the setting of the first problem have to do with any quantum corrections ?

Daniel.
 
dextercioby said:
Just asking: what does the setting of the first problem have to do with any quantum corrections ?

Daniel.

well it's a question that appears in the Textbook...
the textbook is Thermal Physics by Kittel and Kroemer

i tihnk they wnat us to solve it in that way
 

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