Question

The Principles of Statistical Mechanics :-
The exponential atmosphere.

If the temp. is same at all heights, the problem is to discover by what law the atmosphere becomes tenuous as we go up. If N is no. of molecules in a volume V of gas at pressure P, then we know PV = NkT, or P = nkT, where n =N/V is the no. of molecules per unit volume, if we know the no. of molecules per unit volume, we know pressure and vice-versa : they are proportional to each other, since the temperature is constant. But the pressure is not constant, it must increase as the altitude is reduced.
If we take a unit a

Answer 1

at height h, then the vertical force from below is pressure P. The vertical force pushing down at height h+dh would be same, in absence of gravity.
so, dn/dh = -mgn/kt
(tells how density goes down as we go up in energy)

=> n = n0. e^ ( -mgh/kt)

I want you to explain this to me.

Answer 2

if you are reading from Feynmen lectures then

you know dP=-mgndh

also we have P=nkT so, dP=kTdn

using both equation we have

kTdn=-mgndh

or dn/n=-mgdh/(kT)

on integrating we get ln(n)=-mgh/(kT) + constant

or n=exp(-mgn/(kT))* exp(constant)

here exp(constant)=n0

Answer 3

@ramkrishna Yes sir, this is an excerpt from Feynmen lectures.
I was looking for the physical significance of Boltzman factor, e ^ {-( E2 - E1 ) / kT }. Can you please explain that ..

Answer 4

For a system in equilibrium at temperature T, the probability that the
system is in a particular quantum state i, a particular microstate, with energy \[E_i\]
is proportional to e^{-Ei/kT}

The ratio of probabilities of the particles in the state 1 and 2 is
n1/n2=e ^ {-( E2 - E1 ) / kT }