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Lecture 14
Maxwell-Bolztmann distribution.
Heat capacities.
Phase diagrams.
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
Maxwell-Boltzmann distribution

higher T
higher speeds are
more probable

3/2

2
 m 
f (v ) = 4π 
v 2e −mv /(2kT )
÷
2π kT 


Maxwell-Boltzmann
distribution
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
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
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
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!
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
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.
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
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
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)
pT diagram for water
DEMO:
Boiling water
with ice

p
solid

Critical point

liquid
demo

1 atm

gas

0°C

100°C

T
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.
Melting (at ~ -50°C)

solid

liquid

Boiling (at ~ -5°C)

For a boiling transition,
pressure must be lower:.
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
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
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
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
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
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
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
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.
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.
pVT diagram: Ideal gas

States are points on
this surface.
pVT diagram: Water

Phase transitions
appear as angles.

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Lecture 14 maxwell-boltzmann distribution. heat capacities

  • 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
  • 3. Maxwell-Boltzmann distribution higher T higher speeds are more probable 3/2 2  m  f (v ) = 4π  v 2e −mv /(2kT ) ÷ 2π kT   Maxwell-Boltzmann distribution
  • 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.
  • 26. pVT diagram: Ideal gas States are points on this surface.
  • 27. pVT diagram: Water Phase transitions appear as angles.