Lecture 25
Goals:
• Chapters 18, micro-macro connection
• Third test on Thursday at 7:15 pm.
Physics 207: Lecture 25, Pg 1
Nitrogen molecules
near room temperature
Percentage of
molecules
15
10
100-1200
000-1100
00-1000
800-900
700-800
600-700
500-600
400-500
300-400
200-300
100-200
0-100
5
(m/s)
Physics 207: Lecture 25, Pg 2
Atomic scale

What is the typical size of an atom or a small
molecule?
A) 10-6 m
B) 10-10 m
C) 10-15 m
r
r ≈1 angstrom=10-10 m
Physics 207: Lecture 25, Pg 3
Mean free path

Average distance particle moves between collisions:
1
l=
4 2p ( N /V )r 2
N/V: particles per unit volume

The mean free path at atmospheric pressure is:
λ=68 nm
Physics 207: Lecture 25, Pg 4
Pressure of a gas
m
vx
vx
Physics 207: Lecture 25, Pg 5
Consider a gas with all
molecules traveling at
a speed vx hitting a
wall.

If (N/V) increases by a factor of 2, the pressure would:
A) decrease

B) increase x2
C) increase x4
If m increases by a factor of 2, the pressure would:
A) decrease

vx
B) increase x2
C) increase x4
If vx increases by a factor of 2, the pressure would:
A) decrease
B) increase x2
C) increase x4
Physics 207: Lecture 25, Pg 6
P=(N/V)mvx2

Because we have a distribution of speeds:
P=(N/V)m(vx2)avg

For a uniform, isotropic system:
(vx2)avg= (vy2)avg= (vz2)avg

Root-mean-square speed:
(v2)avg=(vx2)avg+(vy2)avg+(vz2)avg=Vrms2
Physics 207: Lecture 25, Pg 7
Microscopic calculation of pressure
P=(N/V)m(vx2)avg
=(1/3) (N/V)mvrms2
PV = (1/3) Nmvrms2
Physics 207: Lecture 25, Pg 8
Micro-macro connection
PV = (1/3) Nmvrms2
PV = NkBT (ideal gas law)
kBT =(1/3) mvrms2

The average translational kinetic energy is:
εavg=(1/2) mvrms2
εavg=(3/2) kBT
Physics 207: Lecture 25, Pg 9
 The average kinetic energy of the molecules of an ideal gas at 10°C
has the value K1. At what temperature T1 (in degrees Celsius) will the
average kinetic energy of the same gas be twice this value, 2K1?
(A) T1 = 20°C
(B) T1 = 293°C
(C) T1 = 100°C
 Suppose that at some temperature we have oxygen molecules
moving around at an average speed of 500 m/s. What would be the
average speed of hydrogen molecules at the same temperature?
(A) 100 m/s
(B) 250 m/s
(C) 500 m/s
(D) 1000 m/s
(E) 2000 m/s
Physics 207: Lecture 25, Pg 10
Equipartition theorem
 Things are more complicated when energy can be stored in other
degrees of freedom of the system.
monatomic gas: translation
solids: translation+potential energy
diatomic molecules: translation+vibrations+rotations
Physics 207: Lecture 25, Pg 11
Equipartition theorem
 The thermal energy is equally divided among all possible energy
modes (degrees of freedom). The average thermal energy is (1/2)kBT
for each degree of freedom.
εavg=(3/2) kBT (monatomic gas)
εavg=(6/2) kBT (solids)
εavg=(5/2) kBT (diatomic molecules)
 Note that if we have N particles:
Eth=(3/2)N kBT =(3/2)nRT (monatomic gas)
Eth=(6/2)N kBT =(6/2)nRT (solids)
Eth=(5/2)N kBT =(5/2)nRT (diatomic molecules)
Physics 207: Lecture 25, Pg 12
Specific heat
 Molar specific heats can be directly inferred from the thermal energy.
Eth=(6/2)N kBT =(6/2)nRT (solid)
ΔEth=(6/2)nRΔT=nCΔT
C=3R (solid)
 The specific heat for a diatomic gas will be larger than the specific
heat of a monatomic gas:
Cdiatomic=Cmonatomic+R
Physics 207: Lecture 25, Pg 13
Entropy
 A perfume bottle breaks in the corner of a room. After some time,
what would you expect?
A)
B)
Physics 207: Lecture 25, Pg 14
very unlikely
 The probability for each particle to be on the left half is ½.
probability=(1/2)N
Physics 207: Lecture 25, Pg 15
Second Law of thermodynamics
 The entropy of an isolated system never decreases. It can only
increase, or in equilibrium, remain constant.
 The laws of probability dictate that a system will evolve towards the
most probable and most random macroscopic state
 Thermal energy is spontaneously transferred from a hotter system to
a colder system.
Physics 207: Lecture 25, Pg 16