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stunner5000pt
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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.
[tex] \mu_{tot} = \mu_{int} + \mu_{ext} [/tex]
right?
[tex] \mu_{tot} = \tau\log\frac{n}{n_{Q}} + \mu_{ext} [/tex]
where n is te concentration and nQ is the quantum concentration
[tex] n_{Q} = \frac{M\tau}{2\pi\hbar^2} [/tex]
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
[tex] \mu_{ext} = \frac{1}{2} M\omega^2 r^2 [/tex]
so then
[tex] \mu = \tau\log\frac{n}{n_{Q}} + \frac{1}{2} M\omega^2 r^2[/tex]
[tex] n(r) = n_{Q} \exp\left(\frac{1}{\tau}\left(\mu-\frac{1}{2}M\omega^2 r^2\right)\right) [/tex]
[tex] n(0) = n_{Q} \exp\left(\frac{\mu}{\tau}\right) [/tex]
[tex] n(r) = n(0) \exp\left(-\frac{1}{2\tau} M\omega^2 r^2\right) [/tex]
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 teh surface, show that at constant temperature the total numbers of molecules in the atmosphere is
[tex] N = 4\pi n(R) \exp\left(-MgR/\tau\right) \int_{R}^{\infty} dr r^2 \exp\left(MgR^2/r\tau\right) [/tex]
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!
[tex] \mu_{tot} = \mu_{int} + \mu_{ext} [/tex]
right?
[tex] \mu_{tot} = \tau\log\frac{n}{n_{Q}} + \mu_{ext} [/tex]
where n is te concentration and nQ is the quantum concentration
[tex] n_{Q} = \frac{M\tau}{2\pi\hbar^2} [/tex]
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
[tex] \mu_{ext} = \frac{1}{2} M\omega^2 r^2 [/tex]
so then
[tex] \mu = \tau\log\frac{n}{n_{Q}} + \frac{1}{2} M\omega^2 r^2[/tex]
[tex] n(r) = n_{Q} \exp\left(\frac{1}{\tau}\left(\mu-\frac{1}{2}M\omega^2 r^2\right)\right) [/tex]
[tex] n(0) = n_{Q} \exp\left(\frac{\mu}{\tau}\right) [/tex]
[tex] n(r) = n(0) \exp\left(-\frac{1}{2\tau} M\omega^2 r^2\right) [/tex]
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 teh surface, show that at constant temperature the total numbers of molecules in the atmosphere is
[tex] N = 4\pi n(R) \exp\left(-MgR/\tau\right) \int_{R}^{\infty} dr r^2 \exp\left(MgR^2/r\tau\right) [/tex]
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!
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