Kinetic Theory

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Question : 13
Total: 14
A gas in equilibrium has uniform density and pressure throughout its volume. This is strictly true only if there are no external influences. A gas column under gravity, for example, does not have uniform density (and pressure). As you might expect, its density decreases with height. The precise dependence is given by the so-called law of atmospheres
n2=n1exp[mg
(h2h1)
kBT
]

where n2,n1 refer to number density at heights h2 and h1 respectively. Use this relation to derive the equation for sedimentation equilibrium of a suspension in a liquid column : n2=n1exp[mgNA(ρρ)
(h2h1)
(ρRT)
]
.
where r is the density of the suspended particle, and ρ′ that of surrounding medium.
[NA is Avogadro’s number, and R is the universal gas constant.]
Solution:  
According to the law of atmospheres,
n2=n1exp.[
mg
kBT
(h2h1)
]
... (i)
where n2,n1 refer to number density of particles at heights h2andh1 respectively.
If we consider the sedimentation equilibrium of suspended particles in a liquid, then in place of mg, we will have to take effective weight of the suspended particles.
Let V = average volume of a suspended particle, r = density of suspended particle,
r′ = density of liquid, m = mass of equal volume of liquid displaced.
According to Archimede’s principle, effective weight of one suspended particle = actual weight – weight of liquid displaced =mgmg
=mgVρg=mg(
m
ρ
)
ρ
g
=mg(1
ρ
ρ
)
Also, Boltzman constant, kB=
R
NA

where, R is gas constant and NA is Avogadro’s number.
Putting, mg(1
ρ
ρ
)
in place of mg and value of kB in (i), we get
n2=n1exp.[
mgNA
RT
(1
ρ
ρ
)
(h2h1)
]
which is the required relation.
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