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 2 years ago
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 viceversa : 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
 2 years ago
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 viceversa : 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

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shubham
 2 years ago
Best ResponseYou've already chosen the best response.0at 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.

ramkrishna
 2 years ago
Best ResponseYou've already chosen the best response.0if 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

shubham
 2 years ago
Best ResponseYou've already chosen the best response.0@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 ..

ramkrishna
 2 years ago
Best ResponseYou've already chosen the best response.0For 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 }
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