1
Homework 1 (due September 24)
Problem 1-1 (3 pts.)
A volume V contains a large number N of molecules. Find the probability that a given region of volume v contains
exactly n molecules, for any n. Express the result in terms of average particle concentration, C = N/V (take into
D E
account that N is very large, but n can be arbitrary, even 0) . Calculate hni, and (n − hni)2 . How does the width
of the distribution behave in the limit when hni is large? Assume v << V and n << N .
Solution:
Probability for each molecule to be in the volume v is v/V . Therefore, the probability of a particular state with
given n molecules in the volume v and the rest outside of it is,
³ v ´n ³ v ´N −n
1−
V V
The total number of such "microstates" for a given value of n is equal to the number of ways to pick n molecules
out of N , therefore
N! ³ v ´n ³ v ´N −n
P (n) = 1−
n! (N − n)! V V
We can now simplify this expression by considering log (P ):, and using the Stirling approximation ( log(m!) ≈
m log m − m)
³ v´ 1 ³ v ´n
log P (n) ≈ N [log N − 1] − (N − n) [log (N − n) − 1] + (N − n) log 1 − + log ≈
V n! V
µ ¶n
1 Nv Nv
≈ log − + O (n/N )
n! V V
Hence
n
(Cv)
P (n) ≈ exp (−Cv)
n!
One can check that normalization has not been lost as a result of our approximations.
We can now calculate the average and mean square deviation for this distribution:
∞
X ∞
X (Cv)n
hni = nP (n) = exp (−Cv) = cV.
n=0 n=1
(n − 1)!
∞ ∞
2® X X (Cv)n
n = (n (n − 1) + n) P (n) = exp (−Cv) + cV = cV (cV + 1)
n=0 n=2
(n − 2)!
D E ® ®
2 2 2
(n − hni) = n2 − 2 hni hni + hni = n2 − hni = cV = hni
1/2
As expected, the width of the distribution behaves as hni (surprisingly, even for small hni).
Problem 1-2 (3 pts.)
You toss the a dice N times in a row. What is the probability of getting the "6" (i.e. ":::") exactly N6 times out
of N ?
Similarly to the previous problem,
2
µ ¶N6 µ ¶N −N6
N! 1 5
P (N6 ) =
N6 ! (N − N6 )! 6 6
Problem 1-3 (3 pts.)
Find the minimum volume V of a helium balloon capable of carrying a person of mass m. Obtain the general
formal expression, and then calculate V for m = 70kg. Assume the pressure inside the balloon to be approximately
the same as outside (Patm ≈ 105 P a). Mass composition of air is roughly 80% N2 and 20% O2 . Note that the total
pressure of a mixture of gases is sum of their partial pressures (i.e. Pair ≈ PN2 + PO2 ).
Solution:
m
[m + V (ρHe − ρair )] g = 0, therefore V =
ρair − ρHe
µ ¶ µ ¶−1
.2 .8 Patm .2 .8
Patm = ρair + RT ; ρair = +
μO2 μN2 RT μO2 μN2
μHe Patm
ρHe = ;
RT
We obtain:
£ J ¤
m mRT 70 · 103 [g] · 8 mole·K · 300K
V = = h i= £ J ¤¡ £ g ¤ £ g ¤¢ ' 70m3
ρair − ρHe −1
Patm (0.2/μO2 + 0.8/μN2 ) − μHe 105 m3 29 mole − 4 mole
Problem 1-4 (3 pts.)
Ideal gas of particles of mass m in the presence of gravity g has a non-uniform density distribution along the vertical
coordinate z:
µ ¶
mgz
n (z) = n0 exp −
kB T
D E
2
Find the average vertical position hzi and its mean square deviation (z − hzi) for such a gas in a very high
cylindrical container. Choose z = 0 to be the bottom level.
Solution:
R
∞ ³ ´
mgz
z exp − kB T dz
kB T
hzi = 0R
∞ ³ ´ =
mgz mg
exp − kB T dz
0
R
∞ ³ ´
mgz
z 2 exp − kB T dz µ ¶2
2® 0 kB T
z = R
∞ ³ ´ =2
mgz mg
exp − kB T dz
0
3
D E ® µ ¶2
2 2 2 2® 2 kB T
(z − hzi) = z − 2 hz hzii + hzi = z − hzi =
mg
Problem 1-5 (3 pts.)
Velocity distribution in an ideal gas has the following general form:
µ 2¶
v
f (v) = α exp −
2σ
Find the constants α and σ, as functions of temperature T and pressure P . Take the mass of a gas molecule to be m.
Solution:
Z ∙Z µ 2¶ ¸3
P v 3/2
n= = f (v) d3 v = α exp − x dvx = α (2πσ)
kB T 2σ
R ¡ 2 ¢
2® v2 exp −vx /2σ dvx
vx = Rx =σ
2
exp (−vx /2σ) dvx
Therefore,
2 ® kB T n P ³ m ´3/2
σ = vx = ; α= 3/2
= 5/2
m (2πσ) (kB T ) 2π
³ m ´3/2 µ ¶
P mv 2
f (v) = 5/2
exp −
(kB T ) 2π 2kB T