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thinkiit hcverma hindi
Consider two Boltzmann gases A and B, at pressures P A and P B and temperatures T A and T B , respectively, contained in two regions of space that communicate through a very narrow opening in the partitioning wall,
[I.1] Show that the dynamic equilibrium resulting from the mutual effusion of the two kinds of molecules, rather than ???
P A
=
1
PB , satisfies the condition,
That the two effusion rates are equal means, given that a Boltzmann gas is an ideal gas for which we have equipartition of energy, R A
=
RB
→
1 4
=
nA u A
1 4
n B uB
→
nA
kT A m A
=
nB
kTB
n =N / V ∝ P /T ; 2 1 mv = 12 kT 2
→
mB
PA
k TA
TA
mA
=
PB
kTB
TB
mB
→
PA
=
PB
TA m A TBm B
[I.2]
Appendix: derivation of the effusion rate from Pathria
[I.3] Consider velocities over the range [u, u + d u ] . Normalization condition,
∫
all velocities
[I.4]
f ( u ) ⋅ d u ≡ 1
How many particles will strike dA in time dt? Vinc = dA • u ⋅ dt → dninc = Vinc × d ρ = ( dA • u ⋅ dt ) ⋅ (n ⋅ f ( u) ⋅ du ) ;
d ρ ∆p z
= ∆p z ⋅ u zd ρ =
p zu z
=n
2 p zu z (nf ( u) ⋅ d u ) [I.5]
Integrating the density [I.5] over a volume gives pressure,
∫
P = n ⋅ d ρ ∆ p z
=
∫ 2np u z
z
∫
f ( u) ⋅ d 3u = n ( p zu z ) ⋅ f ( u) ⋅ d 3u
=n
pu cos 2 θ
=
1 3
n pu
[I.6]
Effusion
We are integrating from θ = [0, π 2 ] , half the range you might expect, since we are considering only u z is, only the particles that make it through the hole. Using the normalization R
=
∫
3
n u z f (u )d u
∞
=
n
∫ ∫ 0
2π
0
×
∫
π /2
0
2
u cos θ f (u ) ⋅ u sin θ ⋅ du ⋅ d θ ⋅ d φ
For more info: see SMT 01, p. 14, pr 22.
1
Which would be the case if the equilibrium had resulted from a h ydrodynamic flow
