2. Molecular speeds
Not all molecules have the same speed.
If we have N molecules, the number of molecules with
speeds between v and v + dv is:
dN = Nf (v )dv
f (v ) = distribution function
f (v )dv = probability of finding a molecule with speed
between v and v + dv
4. Distribution = probability density
f (v )dv
v2
∫v
1
= probability of finding a molecule with speed between v and v + dv
f (v )dv = probability of finding molecule with speeds between v1 and v2
= area under the curve
Normalization:
∫
∞
0
f (v )dv = 1
5. Most probable speed, average speed,
rms speed
2kT
m
v mp =
Most probable speed
(where f(v) is maximum)
∞
∫ vf (v )dv =
=
∫
∫ f (v )dv
v
0
∞
vf (v )dv = ( ... ) =
∞
0
0
v
2
3kT
= ∫ v f (v )dv = ( ... ) =
0
m
vrms =
∞
2
v
2
3kT
=
m
8kT
πm
Average speed
Average squared speed
rms speed
6. Molar heat capacity
How much heat is needed to change by ΔT the
temperature of n moles of a certain substance?
m = nM
n = number of moles
M = mass of one mole (molar mass)
Q = mc ∆T = nMc ∆T
Q = nC ∆T
C = Mc
C = molar heat capacity
7. Constant volume or constant pressure?
The tables of data for specific heats (or molar capacities) come
from some experiment.
For gases, the system is usually
kept at constant volume
CV
For liquids and solids, the
system is usually kept at
constant pressure (1 atm)
CP
You can define both for any system.
You need to know what’s in the table!
8. Heat capacity for a monoatomic ideal gas
Average total kinetic energy K total =
3
3
NkT = nRT
2
2
3
d K total = nRdT
2
From the macroscopic point of view, this is
the heat entering or leaving the system:
3
nRdT = nCV dT
2
3
CV = R
2
dQ = nCV dT
Molar heat capacity at
constant volume for
monoatomic ideal gas
Point-like particles
9. Beyond the monoatomic ideal gas
Until now, this microscopic model is only valid for monoatomic
molecules.
Monoatomic molecules (points) have 3 degrees of freedom
(translational)
Diatomic molecules (points) have 5 degrees of freedom: 3
translational + 2 rotational)
Principle of equipartition of energy: each velocity
component (radial or angular) has, on average,
associated energy of ½kT
The equipartition principle is very general.
10. Diatomic ideal gas
K = K tr + Krot
3
2
= kT + kT
2
2
5
Average energy
= kT
per molecule
2
Average total energy
K total =
The same temperature involves more
energy per molecule for a diatomic
gas than for a monoatomic gas..
5
5
NkT = nRT
2
2
dQ = nCV dT
DEMO:
Mono and diatomic
“molecules”
5
CV = R
2
Including rotation
Molar heat capacity at
constant volume for
diatomic ideal gas
11. Monoatomic solid
Simple model of a solid crystal: atoms held together by springs.
K1 =
3
kT
2
Vibrations in 3 directions
But we also have potential energy!
PE1 = K1
For N atoms:
Molar heat capacity at
constant volume for
monoatomic solid
UN = N KE1 + N PE1
= 3NkT
= 3nRT
(for any harmonic oscillator)
d U = 3nRdT
dQ = nCV dT
CV = 3R
12. pT diagram
Melting curve (solid/liquid transition)
solid
Triple
point
Sublimation curve
(gas/solid transition)
liquid
Critical
point
gas
Vapor pressure curve
(gas/liquid transition)
13. pT diagram for water
DEMO:
Boiling water
with ice
p
solid
Critical point
liquid
demo
1 atm
gas
0°C
100°C
T
14. In-class example: pT diagram for CO2
Which of the following states is NOT possible for CO 2 at 100 atm?
A. Liquid
B. Boiling
liquid
C. Melting
solid
D. Solid
E. All of the
above are
possible.
15.
16. Melting (at ~ -50°C)
solid
liquid
Boiling (at ~ -5°C)
For a boiling transition,
pressure must be lower:.
17. Triple point for CO2 has a pressure > 1 atm.
At normal atmospheric pressure (1 atm), CO 2 can only be solid or gas.
Sublimation at T = -78.51°C
18. pT diagram for N2
At 1 atm, Tboiling = 77 K Tmelting = 63 K
Triple point for N2: p = 0.011 atm, T = 63 K
p
1 atm
solid
liquid
Triple
point
DEMO:
N2 snow
63 K
demo
gas
77 K
T
19. pV diagrams
Convenient tool to represent states and transitions from one
state to another.
Expansion at constant
pressure
p
states
A
B
process
(isobaric process)
VB
VA
V
Example: helium in balloon expanding in the room and warming up
If we treat the helium in the
balloon is an ideal gas, we can
predict T for each state:
TA/B =
pVA/B
nR
DEMO:
Helium balloon
20. ACT: Constant volume
This pV diagram can describe:
p
A. A tightly closed container cooling
down.
pB
B. A pump slowly creates a vacuum
inside a closed container.
C. Either of the two processes.
pA
B
A
(isochoric process)
In either case, volume is constant and
pressure is decreasing.
In case A, becauseT decreases.
In case B, because n decreases.
V
21. Isothermal curves
For an ideal gas, p =
nRT
V
T1 <T2 <T3 <T4
(For constant n, a hyperbola for each T )
Each point in a pV
diagram is a possible
state (p, V, T )
Isothermal curve = all states with the same T
22. ACT: Free expansion
A container is divided in two by a thin wall. One side contains an ideal gas, the
other has vacuum. The thin wall is punctured and disintegrates. Which of the
following is the correct pV diagram for this process?
A
1
Initial state
B
2
Initial state
Final state
Final state
3
C
Initial state
Final state
4
D
Initial state
Final state
23. Final state has larger V, lower p
During the rapid expansion, the gas does NOT uniformly fillV at a
uniform p
⇒ hence it is not in a thermal state.
⇒ hence no “states” during process
⇒ hence this process is not represented by line
A
1
Initial state
B
2
Initial state
Final state
Final state
3
C
Initial state
Final state
4
D
Initial state
Final state
24. Beyond the ideal gas
When a real gas is compressed, it eventually becomes a liquid…
Decrease volume at constant
temperatureT2:
• At point “a”, vapor begins to condense
into liquid.
• Between a and b: Pressure and T
remain constant as volume decreases,
more of vapor converted into liquid.
• At point “b”, all is liquid. A further
decrease in volume will required large
increase in p.
25. The critical temperature
Critical point for water: 647K and 218 atm
For T >> Tc, ideal gas.
p
Supercritical fluid
solid
Triple
point
liquid
Critical point
gas
T
critical temperature = highest temperature where a
phase transition happens.