Section 3: Thermal Properties:
Topic 3.1 Thermodynamics: states and phases
Topic 3.2 Critical temperature and latent heat
Topic 3.3 Thermal expansion in solids
3. Thermal Properties > 3.1 Thermodynamics: states and phases
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Topic 3.1 Thermodynamic aspects of stability
•
•
•
Solid: - …………
volume and shape
Gas: - volume and shape ……………………………..
Liquid:- ……………volume, shape ……………………………
This behaviour relates to:
COMPRESSIBILITY (C) : response to an attempt to change the volume
VISCOSITY (V) and RIGIDITY (R): response* to an attempt to change
the shape
(*N.B. the measurement time-scale is important)
•
•
•
Solid: - ....
Gas: - ….
Liquid:- ….
These properties relate to the PACKING and ORDER of the atoms:• Solid: …………… packing, …………………. order
• Gas: …………….. packing, ……….………… order
• Liquid: ………….. packing, …………………. order
3. Thermal Properties > 3.1 Thermodynamics: states and phases
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States of Matter
• The ……… of a substance depends on the values of P, V and T
(pressure, volume and thermodynamic (i.e. Kelvin) temperature).
• For a particular amount (e.g. 1 mole) of substance, in a
particular state, these quantities are linked by an ………….
……………………………………. (e.g. PV = RT for ideal gas)
• Possible values of P,V and T form a surface in PVT space.
Different regions correspond to different STATES
• Boundaries between these regions
correspond to …………………….
……………….
• The links between PVT and STATE
is usually displayed on P-T or P-V
PHASE DIAGRAMS.
3. Thermal Properties > 3.1 Thermodynamics: states and phases
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Experimental determination
• Fixed amount (1 mole) of
substance in a closed cylinder, P
and T can be varied.
• Each pair of P,T values leads to
value(s) of V and STATE
• Only some [P,V,T, state]
combinations possible.
• SYSTEM must be at
EQUILIBRIUM
3. Thermal Properties > 3.1 Thermodynamics: states and phases
pressure p
constant
temperature
enclosure
T
system
under
investigation
4
P-T Phase Diagram
•
•
•
•
•
The range of variables where the solid,
liquid and gas phases exist are shown as
areas.
Notice diagram doesn’t show regions where
phases coexist.
Reason: while a mix of phases exists P is
constant at constant T.
States of mixed phases are lines not regions
in the PT diagram
Lines are boundaries and represent
conditions where phase transitions take
place
TP: Triple Point : ………………………….
…………………………
CP: Critical Point : ………………………..
………………………………………………
3. Thermal Properties > 3.1 Thermodynamics: states and phases
P
SOLID
CP
TP
LIQUID
GAS
T
Brown: sublimation curve
Blue: melting/fusion curve
Red: vaporization/condensation curve
(also vapour pressure)
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P-T Phase Diagram (2)
α
Changes at constant P: …………..
β
γ
P
SOLID
Changes at constant T: ……………
CP
TP
LIQUID
GAS
Examples: 3 isotherm line α, β and γ
α
GAS -> SOLID (below triple point temperature, sublimation)
β
GAS -> LIQUID -> SOLID
γ
GAS -> SOLID (above critical temperature, no liquid phase)
3. Thermal Properties > 3.1 Thermodynamics: states and phases
T
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P-V Phase Diagram
Separates regions of P and V where substance is in different STATES
In PV diagram: regions of single phase and regions of mixed phases
S+G
Critical point curve (fixed T)
SOLID
CP
S+L
GAS
L
L+G
TP-line
SOLID+GAS
3. Thermal Properties > 3.1 Thermodynamics: states and phases
Triple point line (fixed T)
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P-V Phase Diagram (2)
T information can be included by ISOTHERMS
ISOTHERMS:
α:
compression of gas (ideal gas: P ∝ 1/V)
⇒ solidification (……..)
⇒ compression of solid
β:
compression of gas
⇒ liquefaction/condensation (…….)
⇒ compression of liquid
⇒ solidification/fusion (……..)
⇒ compression of solid
γ:
compression of gas
⇒ solidification/sublimation (……..)
⇒ compression of solid
α
β
Notice in all (phase transition) regions of mixed phase, ………………………………...
3. Thermal Properties > 3.1 Thermodynamics: states and phases
γ
8
The PVT surface
If the physically allowed sets of PVT
values are plotted as points along 3 axes
=> PVT surface with regions for each
PHASE
PT and PV PHASE DIAGRAMS are
projections of the 3D PVT surface onto
the 2D PT and PV planes.
The surface is described by the
Equation(s) of State of the material and
describes all equilibrium states of the
material.
3. Thermal Properties > 3.1 Thermodynamics: states and phases
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What determines the state of a system?
(Microscopically)
The balance between the interatomic (intermolecular) potential energy*
and the kinetic energy+ (thermal energy) of the atoms (molecules)
*depends on separation (i.e. P or V)
+ depends on temperature (T)
• Interatomic p.e. dominant (…………………. …..)
=> SOLID
• thermal energy dominant (……………………….)
=> GAS
• both important (………………………….)
=> LIQUID
3. Thermal Properties > 3.1 Thermodynamics: states and phases
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Phase transitions
•
The distinctions between the three states outlined above may not
always be clear cut. The transitions (phase transitions) may not always
be sharp or well-defined.
•
Certain materials or materials under certain conditions can exhibit
intermediate properties.
IN PHYSICS:
• LIQUIDS are sometimes associated with solids as CONDENSED
MATTER (emphasising the close packing)
• LIQUIDS are sometimes associated with GASES as FLUIDS
(emphasizing the low viscosity)
3. Thermal Properties > 3.1 Thermodynamics: states and phases
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3 Thermal Properties
Topic 3.2 Critical temperature and latent heat
We will look at
Collision cross-section
(i) critical temperature (the
temperature above which no
liquefaction occurs)
(ii) latent heat for the van der Waals
solid
(iii) latent heat for the ionic solid
3 Thermal Properties > 3.2 Critical temperature and latent heat
V(r)
density
a0
r
-ε
Critical temperature
Latent heat
Surface energy
thermal expansion,
compressibility
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Critical temperature
(For the L-J 6-12 van der Waals potential)
•
-ε is the potential energy between two molecules at their equilibrium separation
∴ +ε is the BINDING ENERGY of the pair
≡ ……………………………………………………
•
THERMAL ENERGY ≡ kBT
(Boltzmann’s constant: kB = 1.38×10–23 J/deg)
when THERMAL ENERGY > BINDING
ENERGY molecules do not stay together
∴ substance does not liquefy irrespective
of pressure applied
3 Thermal Properties > 3.2 Critical temperature and latent heat
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Critical temperature
(For the L-J 6-12 van der Waals potential)
∴ at CRITICAL TEMPERATURE
For example: argon TC=151 K
=> ε = …………... (measured ~2.2x10-21 J)
= ……………
3 Thermal Properties > 3.2 Critical temperature and latent heat
(1 eV = 1.6×10–19 J)
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Latent Heat (van der Waals solid/liquid)
• The MOLAR LATENT HEAT OF SUBLIMATION (or
VAPORISATION) is the energy required to change one mole (NA
molecules) of a substance from solid (or liquid) to a gas.
[Note: SPECIFIC LATENT HEAT is for 1 kg]
To change separation from:
r = a0
(value for …………….)
(value for ……….)
requires approximately an energy ………………………….
3 Thermal Properties > 3.2 Critical temperature and latent heat
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Latent Heat (van der Waals solid/liquid) cont.
•
In 1 mole there are 1/2 n NA pairs.
– factor of 1/2 ensures pairs are not counted twice,
– n is coordination number (lecture 2.1).
For sublimation use n for solid ; For vaporisation use n for liquid
MOLAR LATENT HEAT
∴ using assumption kinetic energy is small compared to potential energy (i.e.
low temperatures)
E.g.
Heats of sublimation
Calculated (kJ/mol)
Measured (kJ/mol)
He
0.33
0.08
Ne
1.7
1.3
Ar
5.9
7.3
N2
4.7
5.4
3 Thermal Properties > 3.2 Critical temperature and latent heat
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Latent Heat: temperature variation
•
The derivation above ignores the kinetic energy of the atom/molecules
and is valid only at low temperatures.
•
Experimentally latent heats vary strongly with temperature (as do
density, compressibility etc.)
•
e.g. Argon (the best L-J 6-12 atomic system)
sublimation
8
6
L0
evaporation
ρ
4
1
liquid
10 kgm
3
103Jmol-1
2
0
solid
2
-3
melting
0
70
100
150 T/K
0
50
100
150 T/K
(At higher T part of the increase in separation happens in between phase transitions
(decompression). i.e. it isn’t counted in the latent heats)
3 Thermal Properties > 3.2 Critical temperature and latent heat
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Latent heat of ionic crystal
m

a0 
αe
1  a0 
V (r ) =
   − 
r 
4πε 0 a0  m  r 
2
Recall the interatomic potential is:
and at r=a0, V(a0)= -ε, therefore
 1  αe
ε = 1 − 
 m  4πε 0 a0
2
This is the binding energy per pair of ions in a crystal.
∴ binding energy per mole (NA pairs of ions) ≡ molar latent heat
3 Thermal Properties > 3.2 Critical temperature and latent heat
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Latent heat of ionic crystals
•
Example: crystalline NaCl
Experimentally L0 = 7.63x105 J mol-1
∴ Binding energy per pair of ions ε = 1.27x10-18 J = 7.92 eV
Does this agree with our potential energy model?
Using:
a0 = 2.8x10-10 m, m ~ 10, α for NaCl = 1.75,
∴ ε = ……………………………….
L0 = N Aε
2
e
1
α


ε = 1 − 
 m  4πε 0 a0
∴ L0=……………..
Excellent agreement!
3 Thermal Properties > 3.2 Critical temperature and latent heat
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3 Thermal Properties
Topic 3.3 Thermal expansion in solids
Collision cross-section
V(r)
density
a0
r
We have already discussed:
(i) Density
(ii) Critical temperature
(iii) Latent heat
Next we discuss:
Thermal expansion
-ε
Critical temperature
Latent heat
Surface energy
thermal expansion,
compressibility
3. Thermal properties > 3.3 Thermal expansion
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Thermal expansion in solids
MACROSCOPICALLY (HRW)
Coefficient of linear expansion
Coefficient of volume expansion
MICROSCOPICALLY (QUALITATIVE)
Thermal expansion arises from
increasing atomic vibration through increasing temperature
in combination with
the fact that the V(r) curve is not symmetric
(i.e. two atoms are more easily pushed apart than pushed together).
3. Thermal properties > 3.3 Thermal expansion
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Thermal expansion
Consider two atoms only:
∴ increasing temperature
⇒
……………………………….
+ε
V(r)
0
a0
r
Increasing
T
-ε
T=0
3. Thermal properties > 3.3 Thermal expansion
If V(r) were symmetric
(parabolic ≡ SHM) there
would be no thermal
expansion.
∴ anharmonicity
⇒
……………………………..
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Thermal expansion cont.
• MICROSCOPICALLY (QUANTITATIVE)
How does V(r) change around the equilibrium
position (r= a0)?
∆V
Putting ∆V = V(r)-V(a0)
and x = r- a0,
∆V can be approximated by
∆V = Bx + Cx
2
parabola
(=SHM)
3
0
x
C(<0) and B are constants relating to the shape of the V(r) curve
3. Thermal properties > 3.3 Thermal expansion
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•
Thermal expansion cont.
If the total mean energy (k.e. + p.e.) of the vibrating system is E , then this
equals the p.e. at the end-points of the vibration x1 and –x2.
∴E =
∴
∴
x1 − x 2
Mean position ≡
2
Then for x1 ≈ x 2 (i.e. small anharmonicity),
x1 + x 2 ≈ 2 x; x13 + x 23 ≈ 2 x 3 ,
where x = amplitude of vibration
x1 − x 2
∴ Mean position :
=
2
This corresponds to the increased separation between two atoms
originally a0 apart (i.e. a0 → a0 – (C/2B)x2, where C<0)
3. Thermal properties > 3.3 Thermal expansion
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Thermal expansion cont.x
• The coefficient of linear expansion relates to how
(i.e. ∆L/L) varies with T (at constant pressure)
1
− x2
a0
2
Around the equilibrium separation the thermal energy of vibration k.e. = E (p.e.
at minimum): k.e. =Bx2 +Cx3 ≈Bx2 (neglecting the small 2nd term)
∴ the molar heat capacity for NA atoms vibrating in 3D is given by
Thus α is directly related
to the heat capacity CP.
This is known as the
Gruneisen relation
3. Thermal properties > 3.3 Thermal expansion
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Thermal expansion (Lennard Jones solid)
• For L-J 6-12,
 a0 12  a0  6 
V (r ) = ε   − 2  
 r  
 r 
Using Taylor’s theorem,
(see Flowers and Mendoza)
(This is a good approximation up to x = 0.1 a0, ~melting
point.)
Putting in values of C and B
3. Thermal properties > 3.3 Thermal expansion
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Example: Argon
For Ar, in the high temperature regime:
Molar heat capacity: Cp ~ 3NAkB (= 3R)
7 CP
1
≈
∴α =
kB
216ε N A 10ε
For Ar:
ε =1.7x10-21J
∴
so
3. Thermal properties > 3.3 Thermal expansion
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