Problem Sheet 2 solution - WebRing

PHYS1B28: Thermal Physics
Department of Physics and Astronomy, University College London.
Problem Sheet 2 (2005) – Model answers
1. Five gas molecules chosen at random are found to have speeds of 500, 600, 700, 800, 900 m/s.
Find the rms speed and the average speed. Is the rms spead the same as the average speed? Explain
the difference.
2 2 2 2 2
v1
+ v2
+ v3
+ v4
+ v5
ν rms =
=
5
500
2
+ 600
v1
+ v2
+ v3
+ v4
+ v5
v = = 700 m/s
5
The rms speed is larger than the average speed.
2
+ 700
5
2
+ 800
2
+ 900
2
= 714.14 m/s
Since the five molecules have been chosen at random, they should obey the Maxwell-Boltzmann
distribution of molecular speeds. According to this distribution v = 1.60 kB T / m and
v = 1.73 kB T / m , i.e. the rms speed should be larger.
rms
2. (a) Obtain expression for the mean free path of molecules in a gas assuming that their speeds are
much greater than the rms speed. Compare with the exact expression.
M
λ
Mean free path is equal to the average distance between collisions.
Assume that molecules are spheres of diameter = d
Since we consider the mean free path of molecules which are moving very fast – much faster than
the rms speed – we can assume that most of other molecules, which move slower, are stationary
between collisions.
1
[5]

On average, molecule M travels distance λ before colliding with any other molecule. Before
collision there should no other molecule except M in the cylinder of volume π ×d 2 × λ.
We assume that the number density of molecules nV = N/V is constant everywhere in the gas.
In the cylinder it is equal nV =
1
. Therefore in this approximation
d λ
d n
2
1
λ = .
π
π 2
This is slightly bigger than an exact value λ =
1
2π
d nV
2
(b) Derive expression for collision frequency of molecules in an ideal gas. Estimate the collision
frequency of hydrogen molecules in a container containing hydrogen gas at T= 1000 K and
P = 1 atm. (Assume that hydrogen molecules are spherical with the effective diameter equal to
twice the diameter of the 1s orbit in the hydrogen atom.)
Collision frequency is the number of collisions per unit time. Time between collisions can be
estimated as mean free path divided by the average velocity of molecules in the gas ν , i.e.
t = λ/ν . Therefore the collision frequency is f = 1/t = V n
2
2 πd ν .
From the ideal gas law
n
V
P
= =
k T
−20
k BT
1.
38×
10
v = 1 . 60 × = 1. 60
−27
m 3.
346×
10
d 2 = (4×0.0529×10 -9 ) 2 = 0.0448×10 -18 m 2
f = 2.96×10 9 s -1
B
1.
01
1.
38×
10
5
× 10
−23
Pa
= 0.
732 × 10
× 1000
× = 2.030×10 3 m/s
25
V
molecules/m 3
3. Calculate the most probable speeds of H2 and O2 molecules at 20 o C. On a single diagram sketch
the Maxwell-Boltzmann distribution of molecular speeds for H2 and O2 molecules at this
temperature.
vmp = kB T / m
2 = 1.41 kB T / m
vmp(H2) = 1560 m/s
vmp(O2) = 389 m/s
2
[10]

[5]
4. Calculate the probability that a molecule of oxygen in oxygen gas at 1000 K has a speed
between 1000 m/s and 1001 m/s.
f(v) = 4π
m 3/2 2 2 4 ⎛ m ⎞ 2 ⎛ m 2 ⎞
( ) v × exp(-mv /2kBT)dv = v
v dv
2πk
BT
⎜
k BT
⎟ × × exp ⎜
⎜−
k BT
⎟
π ⎝ 2 ⎠
⎝ 2 ⎠
m(O2) = 32 × 1.66×10 -27 kg = 5.312 ×10 -26 kg; dv = 1 m/s
2kBT = 2 × 1.38×10 -23 ×1000 = 2.76 × 10 -20 J
m
= 1.
92×
10
2k
T
B
f(v)
−6
O2
400 1600
−6
3 / 2 6
−3
f ( v)
= 2.
26 × ( 1.
92 × 10 ) × 10 × exp( −1.
92)
× 1 = 0.
88×
10
Some students can do this through integration of the distribution function.
5. The proportion of various gases in the earth’s atmosphere changes somewhat with altitude.
Would you expect the proportion of oxygen at high altitude to be greater or less than at sea level
compared to the proportion of nitrogen? Explain your answer.
P = nV
k BT
; nV = N / V = the number density of molecules
mgh
Law of atmospheres: nV
= n0
× exp( − ) , where n0 is the number density of molecules at the
k BT
sea level. m(O2) > m(N2). Therefore the number density and pressure of oxygen will decrease
faster with altitude and the proportion of oxygen will be less than nitrogen.
[5]
H2
3 / 2
[5]
v
3