WEBVTT
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let
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me start the law of equipartition of energy
as i have already told you for a given available
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energy of the system how the energy have been
distributed among its different degrees of
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freedom since we have already come to know
what are the different degrees of freedom
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for the various systems now let me see let
me see how the energy is distributed among
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the different degrees of freedom this is known
as the law of equipartition of energy okay
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let us deduce the law of equipartition of
energy which states that the energy of a dynamical
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system in thermal equilibrium at temperature
t is distributed equally among all the degrees
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of freedom and it is equal to half kt for
each degrees of freedom so what does it mean
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by equipartition equi means equal that means
energy u will be distributed equally among
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all its degrees of freedom and it can be shown
that energy for each degrees of freedom is
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half kt ok
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this law was first derived by maxwell and
later generalized by boltzmann as we have
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said earlier the particle distribution law
is given by dn = n times a a e to the power
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e by kt dx1 dx2 dx3 dq where e of xq is nothing
but it consists of kinetic energy plus potential
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energy in terms of new coordinates half alpha
1 x1 square + half alpha2 x2 square + eu of
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kt
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and a is a constant which is determined by
the normalization condition a entry integral
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e to the power of e by kt integrated over
all coordinates dx1 dx2 dx3 and dq = to 1
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if you will substitute these normalization
condition then you will get a in the above
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equations in the above equation what you will
get it you will get a integrating from minus
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infinity to plus infinity e to the alpha 1
x1 square by 2 kt dx1 minus infinity to plus
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infinity e to the alpha2 x2 square by 2 kt
dx2
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and up to nth term and last term is e to the
power of u by kt dq = 1 this is the normalization
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condition so we have got basically n + 1 integration
x1 x2 x3 xn and another integration over dq
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so in order to integrate our dx1 dx2 dx3 we
let us substitute alpha1 x1 square by 2 kt
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is y1 alpha2 x2 square by 2 kt is y2 etcetera
if i will substitute it then i can rewrite
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these equations as minus infinity to plus
infinity e to the alpha1 x1 square by 2 kt
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dx1
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after substitution it will look like 2 kt
by alpha1 to the power of half 0 to infinity
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e to the power minus y1 y1 to the power minus
of dy 1 which is nothing but the 2 pi kt by
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alpha 1 to the power half so you will get
in terms you will get terms of this type n
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number of terms you will get so if you so
because of the integration over x1 x2 x3 xn
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so finally what we will get a 2 pi kt by alpha1
to the power of half 2 pi kt by alpha2 to
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the power of half into last term e to the
power e u by kt dq = 1 so finally we can get
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a e to the power of e u by kt dq = 2 pi kt
by alpha1 to the power of half 2 pi k t by
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alpha 2 to the power of half we will get we
will be having n number of term so it is a
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product of 2 pi kt by alpha j to the power
half where j runs from 1 to n ok
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so then the mean energy now i will calculate
different quantities then mean energy corresponding
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to the moment towards xi is as we know energy
is some half alpha xi square so average mean
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of half alpha xi square is interest integration
half alpha xi square into probability f of
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xq dx dq so then you will substitute f of
xq so what we will get it half alpha xi square
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a e to the power by kt dx1 dx2 dx3 dxi into
dx cube so we know we know that average values
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of half alpha xi square is we know it is a
product of it is a so it is product of 2 pi
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kt by alpha j to the power – half so if
you put the values of e = half of alpha xi
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square summed over +u of kt so that part we
know what is it e to the power u by kt
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let me show you ae to the power e u by kt
dq is is a product of n terms 2 pi kt by alpha
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j to the power of half so just we have substituted
that thing product of 2 pi kt by alpha j to
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the power half which is nothing but a integral
e to the power u by kt now i have to do rest
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of the terms i have to integrate the race
term which is minus infinity to plus infinity
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half alpha1 x1 square e to the power –alpha1
x1 square by 2 kt dx1 half alpha2 x2 square
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e to the power alpha2 x2 square by 2 kt dx
2
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and finally half of alpha nth ith term is
alpha xi square e to the power of alpha xi
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square by 2 kt dxi so this term we also know
so apart from half kt so finally what we will
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get the product of 2 pi kt by alpha z to the
power of minus half into product of n terms
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of inverse thing means 2 ppi kt by alpha1
to the were plus half apart from half kt so
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obviously this term will cancel except half
kt
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so we will get the mean of we will finally
we will go we got the result the mean energy
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that means average values of half alpha xi
square is nothing but half kt
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so thus the mean kinetic energy for the momentoid
xi is half kt who are i =1 to n it is the
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equipartition of energy between the different
momentoid this result is applicable for translational
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rotational and vibrational motions and it
is appropriate to emphasize that these law
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does not refer to the potential energy of
particles in the external field the law of
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equipartition of energy is of great importance
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for example due to this law the mixture of
ideal gases obeys the diet dalton law of partial
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pressures okay so the microscopic origin of
dalton law partial pressure is nothing but
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the equipartition energy law of equipartition
of energy and as a result we got the daltons
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law of partial pressure who are total if that
means if the system has n number of components
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the total pressure exerted by the system is
the summed of partial pressures exerted by
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the individual components respectively
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so next i will tell atomicity of gases the
law equipartition of energy atomicity of gas
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means i will next couple of minutes i will
tell the result of equipartition of energy
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law of equipartition of energy in predicting
in predicting the ratio of the specific
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8 cp cv etcetera okay
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so the law of equipartition of energy enables
us to calculate the energy of an ideal system
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obviously we will will constraint ourselves
to an ideal system let us consider a gram
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molecule of a perfect gas at absolute temperature
the system contains any molecules where na
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is the avogadro number because we have considered
one gram molecule so obviously it contains
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avogadro number of molecules
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if n be the number of degrees of freedom of
each molecule then gas will have small n times
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capital na degrees of freedom thus the energy
of a gram molecule of a polyatomic gas is
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number of degrees of freedom into half kt
because per degree of freedom is half kt total
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degrees of freedom n times capital na so these
times half kt so total energy is half small
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n is the number of degrees of freedom
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total molecule is na times kt so the molar
specific heat at constant volume is cv is
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del e by del t at constant v so half small
n into capital na into k which is nothing
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but the half times n into r where r is nothing
but the universal gas constant which can be
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written as avogadro number times the boltzmann
constant so finally the molar specific heat
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at constant volume we have got is nr by 2
and exactly which matches with the experiment
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that at room temperature a molar specific
heat at constant volume is independent of
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the temperature
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since cp by cv = r for perfect all perfect
gases we have cp = cv + r since we have got
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cv is n by 2 times r so if you will put this
value so cp is n by 2 + 1 into r and hence
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the ratio of two specific heat is gamma equal
to cp by cv n by 2 + 1 so let us consider
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some particular cases let us start for the
mono atomic gas in this case as you know let
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us try to derive this ratio gamma equal to
cp by cc for three cases
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first one is the mono atomic gases second
one is the diatomic gases third one is the
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triatomic gases because we know monatomic
we know the degrees of freedom for mono atomic
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diatomic triatomic gases and just we will
put the degrees of freedom for three gases
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individually and calculate the ratio gamma
and let us see whether it matches with the
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experiment or not for the mono atomic gases
each molecule has only three translational
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degrees of freedom and hence n = 3 so cv = 3
by 2 r and cp = 5 by 2 r
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so the ratio equal to gamma 5 by 3 which is
167 which matches with the experiment beautifully
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now let us come to the diatomic gas we know
for the diatomic gases it has 5 degrees of
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freedom where three are the translational
degrees of freedom and two are the rotational
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degrees of freedom provided their inter particle
separation are held constant so each molecule
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of a diatomic gas has 5 degrees of freedom
where three translational and two rotational
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thus n = 5 now if we will put gamma = 5 by
2 r cv = 7 by 2 r they our ratio is 7 by 5
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which is which comes out to be 14 it also
matches with the experiment beautifully try
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to make gases in this case each molecule has
6 degrees of freedom who are 3 are due to
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translational motion and 3 are due to rotational
motion so we will put n = 6 we will get cp
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= 4r where cv = 3r where gamma equal to 4
by 3 = 133
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so here i should say something suppose if
you want to calculate the ratio for the polyatomic
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gases so they are the the observed value observed
value from this derivation and did not match
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with the experimental result so that is there
is in this form of derivation is not good
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for a multi for a polyatomic gas where you
need to look for some new theory may be statistical
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mechanics will be able to handle polyatomic
cases much better than this kinetic theory
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of gases
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atmosphere of planets so that is the one of
the most beautiful example through this we
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will see how the density of molecules are
distributed in our atmosphere
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planets including the earth are surrounded
by envelope of air known as atmosphere the
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molecules of the atmospheres are in thermal
motion and also experience a force of gravitational
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attraction of the planet therefore a distribution
of the molecules of the atmosphere varies
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with altitude consequently there is a change
in pressure of the gas with altitude let us
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derive let us derive how the pressure changes
with the altitude okay
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let the pressure at the altitude h be p on
the altitude changes by an hi bye bye d h
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the pressure changes by an amount d p where
d p is related to dh by dp = rho g times dh
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where rho is the density of the air and g
is the acceleration due to gravity of the
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planet if n be the number of molecules per
unit volume and m is the mass of the molecule
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then rho = m times n okay
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so according to the kinetic theory we know
p = n kt so n = p by kt consequently rho = small
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mp by kt and dp if you substitute values of
rho in the dp so we will get dp equal to mg
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by kt into pdh or dp by p = mg by kt into
dh so so here the acceleration due to gravity
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g is not constant but depends on the altitude
h in accordance with the law of gravitation
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in our school level physics we know how to
derive
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we feel go away from this surface if you go
towards the center then how the acceleration
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changes with the distance so the g = gm by
r square where g is m by r naught + h whole
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square where g is the gravitational constant
m is the mass of the planet r naught is its
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radius okay so that means if you will go away
from the surface and that means you are away
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from so r naught + h if you will go towards
the center by a distance h then it will be
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reduced r naught r naught h should be
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so substituting this value in the equation
we will write dp by p is mg into small m into
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capital m by kt dh by r naught + h whole square
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if we assume that temperature is the same
at all altitude for the sake of simplicity
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then we can take outside t we can take outside
of this integration then on integration we
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get log of p =g small m capital m by kt 1
by r naught + h + log of a where log of a
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is a integration constant hence p if we rewrite
this equation p = a times e to the power g
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small m capital m by kt 1 by r naught + h
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let p naught be the pressure of the air on
the surface of the planet where h = 0 just
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surface so then that means we want to say
how the pressure changes with the altitude
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as you will be away from this surface okay
let us say pressure is p = p naught at the
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surface that means h = 0 the constant a can
be calculated from the condition that p = p
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naught so the constant is a = p naught e to
the power – g small m capital m by kt into
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1 by r naught
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if we will put back the constant a so we will
get p as a function of h is the pressure of
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the earth is p of h = p naught e to the power
g small m capital m by into 1 by r naught
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1 by r naught + h so this equation represents
the law of decrease of pressure with altitude
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according to this pressure of a gas decreases
with altitude according to an exponential
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law and exactly it happens in our atmosphere
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if you will be away from the atmosphere if
you go above to our surface where we will
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feel less pressure and that is the that is
the reason we have seen less pressure in the
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air of atmosphere since the pressure of the
gas is proportional to the number molecules
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per unit volume so the above equation pressure
can be written in terms of number of molecules
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as a height so n of h = n naught n naught
means number of molecules per unit volume
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at the surface into e to the minus m g naught
r naught by kt into 1 1 by r naught + h
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where g naught is the acceleration due to
gravity on the surface of the planet of radius
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r naught n and n naught be the number of number
densities of molecules means number of molecules
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per unit volume at the altitude h and on the
surface of the planet for h assumed to be
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0 respectively this is the law of diminishing
of the density of the molecules with the latitude
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that means as we as we will go above to the
surface the density of molecules will be decreasing
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so that is expected and that it happened in
this expression also
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according to this formula the density tends
to a finite value as h tends to infinity which
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if you will substitute in the above equation
h equal to infinity then we will get a n at
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h tends to infinity = n naught e to the m
naught g naught r naught by kt this means
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that the density of gas should be distributed
over the entire infinite space this is physically
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impossible the atmosphere of the planet is
not in equilibrium state
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however any system ultimately tends to an
equilibrium state so dependent for different
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systems suppose if we allow the system for
the coulomb to come to equilibrium different
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system will take different time to come to
the equilibrium but asymptotically any system
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will come to the equilibrium so this leads
us to assume that the atmosphere of the planets
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gradually dissipate some celestial bodies
such as the moon completely lost they are
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atmospheres
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while the mars have has a very rarefied atmosphere
the atmosphere of the moon has already attained
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the equilibrium state the atmosphere of the
earth is dense and hence in the way towards
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the equilibrium state thank you very much