Download (a) At equilibrium, the temperature in the larger (L) side and smaller (S)

Problem 3.1 Reif
(a) At equilibrium, the temperature in the larger (L) side and smaller (S)
side of the box will be equal, TL = TS . With a hole in the partition, it should
be intuitively clear that the pressures, and, therefore, the number densities of
particles should be equal on each side of the partition. Thus we have
NS
NL
=
,
VS
VL
(1)
for each species of particle. Since we are given that VL = 3VS , it follows
immediately that NL = 3NS and, thus, that 3/4 of the particles of each species
are in the large side and 1/4 in the small side. In terms of numbers, we have
750 Ne particles and 75 He particles on the large side and 250 Ne and 25 He
particles on the small side.
Because intuition in physics is a developed skill, the above solution may be
demonstrated in a systematic approach based on our knowledge of effusion. At
steady state, the two-way fluxes of each species through the hole will be equal
and given by the expression nv/4 for the total flux. Thus we have
(nv)L = (nv)S ,
(2)
for each species. Since TL = TS , we see that (v)L = (v)S for each species, and
thus we have
(n)L = (n)S ,
(3)
which is equivalent to Eq.(1). The only problem with this approach is that
knowledge of effusion and the total molecular flux is covered in Ch. 7, and
would not be available for use in Ch. 3 if we were covering the chapters in
numerical order. So there should be a way to solve this problem using only
information covered in Ch. 3. This is called "making an easy problem into a
difficult one."
The Ch. 3 method is to find the value of NS (or NL ) that maximizes the
total number of accessible states for the composite system, Ω(0) , where
Ω(0) = ΩL ΩS ,
(4)
and ΩL and ΩS are the numbers of accessible states for the large and small
subsystems, respectively. Assuming the gases are ideal, we can write (for each
species)
3N /2
B 3Nα /2 VαNα Eα α
Ωα =
, α = L, S
(5)
Nα !(3Nα /2)!
where B is the constant,
B=
2πm
.
h2
(6)
The total energy is constant
E (0) = EL + ES = N ε,
1
(7)
where ε is the average energy per particle and
ES = NS ε,
EL = NL ε.
(8)
After substituting the above expressions and using Stirling’s approximation for
ln(n!), we find
ln Ω(0)
3NL
3NL 3NL
ln ε − NL ln NL + NL +
+
ln B +
2
2
2
3NS
3NS
3NS
NS ln VS +
ln ε − NS ln NS + NS +
+
ln B. (9)
2
2
2
= NL ln VL +
Remembering that N = NL + NS , we can further simplify this expression to
read
3N
5N 3N
ln ε +
+
ln B.
2
2
2
(10)
Now we differentiate Eq.(10) with respect to NS , while holding N , ε, VS , and
VL constant and realizing that ∂NL /∂NS = −1, to find
ln Ω(0) = NL ln VL + NS ln VS − NL ln NL − NS ln NS +
VS NL
∂ ln Ω(0)
= − ln VL + ln VS + ln NL + 1 − ln NS − 1 = ln
.
∂NS
VL NS
(11)
The maximum value of Ω(0) occurs when ∂ ln Ω(0) /∂NS = 0, and thus when
VS NL
= 1,
VL NS
(12)
which is equivalent to the intuitive result found above in Eq.(1).
(b) Think of this problem as two simultaneous random walks, one with
N = 1000 and the other with N = 100. For each walk the probability of a
"step" into the small side is
VS
1
p=
= ,
(13)
V
4
and the probability for a "step" into the large side is
q =1−p=
VL
3
= .
V
4
(14)
The probability of simultaneously finding 100 He atoms in the small side
and 1000 Ne atoms in the large side is
P = W100 (100)W1000 (0),
(15)
where WN (n) is the binomial probability distribution for achieving n successes
in N trials. Using results from Ch. 1, we have
µ ¶100 µ ¶1000
1
3
P =
≈ 10−185 .
4
4
This is a vanishingly small number even for such relatively small values of N .
2