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I need help with exercise #2. Your help will be really appreciated and rated.
MAXWELLS EQUATION I. Maxwells Equation: Our first (of mony) distribution functions. Very important A. The Maxwell-Boltzman speed distribution gives the speed distribution, fiv), of particles confined to NN()d, which a volume, V, and in thermal equilibrium at a temperature, T. () is the number of particles moving within dv of a speed, v Distributions of this type can be considered as the product of three terms 1. The probabilitydensity η v) or η(energy is the probability that a particle will have a certain speed or energy. It can usually be interpreted as the probability that a microstate (a possible speed, v, or energy, in this case) will be occupied It is usualty expressed as ele 2. The number of ways or microstates, g, that a particle might have a certain speed or energy commonly described in a form mathematically analogous to a volume. (See E xercise #2, below) . dN N , or f(v)dv 이 = 3. A normalizing constant, K, such that B. As an example of how this might work, we consider a simple model of the atmosphere: ic pressure at an altitude, yi, is due to the weight of the air above yi ly > yal, we find that the pressure, P, decreases as the altitude, y, increases. That is: (1) dP=-pgdy. Also, the air density, p(=-), decreases with y. Now, if N is the number of molecules and m the mass of each molecule, then M·N m and ρ = (N/V) mn. So: (2) dP nmgdy For an ideal gas, PV = NkT so, if we assume T is constant, then (N/V) = η = P/(kT) and (3) d η = dP/(kT). is the probability density, see above.) and so:-=-mady Combining (2) and (4),dP=-nm g dyskTd, Exercise #2: Show that η/ η o-efngvan where y 0.) (Assume that T is uniform and let η,-the probability density
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Answer #1

Start with the differential equation

\small \frac{d\eta}{\eta}=-\frac{mg}{kT}dy

Integrate both sides

\small \int \frac{d\eta}{\eta}=-\frac{mg}{kT}\int dy

This gives

\small \ln \eta +c=-\frac{mg}{kT}y

where c is the integration constant. Now use the boundary condition that at

\small y=0,\,\eta=\eta_0

Therefore the above equation becomes

\small \ln \eta_0 +c=0

This gives

\small c=-\ln \eta_0

Therefore

\small \ln \eta -\ln \eta_0=-\frac{mg}{kT}y

which can be written as

\small \ln (\eta/\eta_0)=-\frac{mg}{kT}y

and hence

\eta/\eta_0=e^{-\frac{mg}{kT}y}

which is the equation we set out to prove.

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