Ex 6020: A divided box with a hole in one side
Submitted by: Yotam Sherf
The problem:
A cylinder of length L and cross section A is divided into two compartments by a piston. The
piston has mass M and it is free to move without friction. Its distance from the left basis of the
cylinder is denoted by x. In the left side of the piston there is an ideal Bose gas of Na particles with
mass ma . In the right side of the piston there is an ideal Bose gas of Nb particles with mass mb .
The temperature of the system is T . Assume that the left gas can be treated within the framework
of the Boltzmann approximation. Assume that the right gas is in condensation. In items (3-5)
consider separately two cases:
(a) A small hole is drilled in the left wall of the box.
(b) A small hole is drilled in the right wall of the box.
The area of the hole is δA.
(1) Find the equilibrium position of the piston.
(2) What is the frequency of small oscillations of the piston.
(3) What is the velocity distribution F (v) of the emitted particles?
(4) What is the flux (particles per unit time) of the emitted particles?
(5) Is the piston going to move? If yes write an expression for its velocity.
In item (3) use normalization that makes sense for the calculation in item (4). In item (5) assume
that the process is quasi-static, such that at any moment the system is at equilibrium. Express
your answers using L, A, δA, Na , Nb , ma , mb , T, M .
Z
0
∞
xdx
π2
=
ex − 1
6
The solution:
(1)
Equilibrium condition
Pa = Pb
(1)
Na T
Where for Boltzmann Pa =
and for the condense gas Li5/2 (1) = ζ(5/2) and
Ax
3
5
b 2
Pb = ζ 52 m
T 2.
2π
Define x0 as the equilibrium position we find from (1).
Na
2π 3/2
x0 =
ζ(5/2)A mb T
(2)
1
(2)
Looking at the total force on the piston and expanding it around x0 for small oscillations.
dF (x) dP (x) F (x0 + dx) = F (x0 ) +
dx = A P (x0 ) + dx dx
x0
| dx
{z } x0
(3)
−M w2
We find
A
Na T
w = − P 0 (x0 ) =
M
M x20
2
A
w = ζ(5/2) √
Na M
⇒
mb
2π
3/2
T2
(4)
(3)
(a) In the Boltzmann case we have the known velocity distribution
n(v) =
ma
2πT
3/2
4πv 2 e−
2
βmva
2
(5)
And for Na particles we have
Fa (v) = Na n(v) = Na
ma
2πT
3/2
4πv 2 e−
βma v 2
2
(6)
(b)condense Bose gas.
1
Fb (v)dv =
(2π)3
Z
d x m3B d3 v
1
3
βmB v 2
−
2
e
=V
−1
mb
2π
3
4πv 2
1
βm v 2
− 2b
e
dv
(7)
−1
Where V = A(L − x).
(4)
By definition J(v) = ρ(v)v, and ρ(v) =
(a) Boltzmann gas.
F (v)
, so the total flux.
V
Z
Z
Na ma 3/2 ∞ 3 − βmB v2
2Na π ma 3/2 2T 2 ∞ −x
2
Ja = 4π
v e
dv =
xe dx =
4V 2πT
Ax0 πT
ma
0
0
2 s 3
2 s 3 Z ∞
mb
mb
T
T
ζ(5/2)
xe−x dx = ζ(5/2)
2π
ma 0
2π
ma
|
{z
}
(8)
1
And the flux per unit time Ia = δAJa
(b) Bose gas.
Ib = (δA)2π
mb
2π
3 T
mb
2 Z
0
|
∞
x
mb
dx = (δA)
e−x − 1
6
{z
}
π 2 /6
2
2
T
2
(9)
with no dimensional dependence.
(5)
(a) In the Boltzmann case using the quasi-stationary assumption we can use equilibrium
condition (1) in each moment, so x0 → x0 (t).
x0 (t) =
2π
2π 3/2
Na (t)
Aζ(5/2) mb T
(10)
dNa (t)
= Ia , the piston velocity would be (ẋ0 (t) = vp ).
dt
r
δA 2πT
vp =
A
ma
where
(11)
3 5
b 2
(b) In the condense case we have Pb = ζ 25 m
T 2 with no dependence on Nb , so the pressure
2π
will remain the same and the equilibrium condition (1) still holds .
3