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Thermodynamic Definitions
Basic Physical Chemistry - Thermodynamics 1
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■ A thermodynamic system is defined as any part
of the Universe under consideration.
Thermodynamic System
2Basic Physical Chemistry - Thermodynamics
■ It may be something as simple as a beaker ofwater or as complicated as an entire galaxy!
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■ Thermodynamic surroundings are defined as
everything other than the thermodynamic
system. In other words, the entire rest of the
Universe.
Thermodynamic Surroundings
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■ The Universe is therefore the system plus its
surroundings.
The Universe
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The observable Universe is a sphere with a radius of ca. 4.66 × 1010 light years.
A light year is 9.46 × 1015 m. The observable Universe is thus ca. 880 Ym or
880,000,000,000,000,000,000,000,000 m across.
The Vastness of the Universe
5Basic Physical Chemistry - Thermodynamics
Image: NASA
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Thermodynamic Surroundings
6Basic Physical Chemistry - Thermodynamics
Fe2O3 + 2Al 2Fe + Al2O3
Image: Nikthestunned
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Thermodynamic Surroundings
7Basic Physical Chemistry - Thermodynamics
Image: NASA / ESA
super nova
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Thermodynamic Surroundings
8Basic Physical Chemistry - Thermodynamics
■ Surroundings are assumed infinite and remain at
constant temperature and pressure.
■ The vast size of the Universe validates thisassumption.
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■ The boundary may be actual or notional.
Boundary Conditions
9Basic Physical Chemistry - Thermodynamics
■ It controls transfer of work, heat and matter fromthe system to the surroundings and vice-versa.
■ The boundary may or may not impose
restrictions on such transfers.
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■ An open system may exchange both energy and
matter with its surroundings.
Open, Closed and Isolated
10Basic Physical Chemistry - Thermodynamics
Open
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■ A closed system may exchange energy but not
matter with its surroundings.
Open, Closed and Isolated
11Basic Physical Chemistry - Thermodynamics
Closed
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■ An isolated system may exchange neither
energy nor matter with its surroundings.
Open, Closed and Isolated
12Basic Physical Chemistry - Thermodynamics
Isolated
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■ A diathermic system allows heat flow into or out
of the system.
Diathermic and Adiabatic
13Basic Physical Chemistry - Thermodynamics
■ An adiabatic system prevents heat flow into orout of the system.
Hot Cold
Diathermic Adiabatic
Hot Cold
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Diathermic and Adiabatic Walls
14Basic Physical Chemistry - Thermodynamics
Both containers are attempts
to create adiabatic walls. Theflask is a good approximation
to an adiabatic calorimeter –
the cardboard cup isn’t…
Which will keep coffee
warmer for longer?
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■ Isothermal implies constant temperature, T .
■ Isobaric implies constant pressure, p.
■ Isochoric implies constant volume, V .
Isothermal, Isobaric and Isochoric
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■ A state function describes the state of a system.
State functions
16Basic Physical Chemistry - Thermodynamics
Pressure, p
Volume, V
Temperature, T
Mass, m
Quantity, n
Internal Energy, U
Enthalpy, H
Entropy, S
Gibbs Energy, G
■ The following are all state functions:
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■ A state function describes the current state of a
system.
Path functions
17Basic Physical Chemistry - Thermodynamics
■ How the system came to be in that particularstate is of no consequence.
■ Functions governing transition between states
are called path functions.
Heat, q Work, w
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■ The state of a system is changed by the supply
or removal of energy in the form of heat or work.
Heat and Work
18Basic Physical Chemistry - Thermodynamics
Work energy transfer is
uniform molecular motion
Heat energy transfer is
random molecular motion
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■ Path functions are positive when energy enters
the system.
Heat and Work Conventions
19Basic Physical Chemistry - Thermodynamics
■ Path functions are negative when energy leavesthe system.
■ They are often defined as:
heat supplied to the system, qin mechanical work done on the system, w on
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0th Law of ThermodynamicsTemperature
Basic Physical Chemistry - Thermodynamics 1
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■ Developed around 1850 by Rudolf Clausius.
2Basic Physical Chemistry - Thermodynamics
The First Law of Thermodynamics
■ Adaptation of the Law of Conservation of Energy
for thermodynamic systems.
■ Defined thermodynamic energy or internal
energy for a thermodynamic system.
■ Concerns energy changes and lead to thedefinition of a new state function called enthalpy.
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■ Also developed around 1850 by Rudolf Clausius
following initial work by Sadi Carnot in 1824.
■ Describes the direction in which all processesspontaneously occur, e.g. a hot object loses heat
to its surroundings.
■ Lead to the definition of a new state function
called entropy which fundamentally measuresdisorder.
The Second Law of Thermodynamics
3Basic Physical Chemistry - Thermodynamics
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■ Developed in 1912 by Walter Nernst.
■ Considers the entropy of perfect crystals.
■ Lead to the definition of zero on the entropyscale thus enabling determination of absolute
entropy.
3rd Law of Thermodynamics
4Basic Physical Chemistry - Thermodynamics
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■ Temperature, which lay at the heart of this new
science of thermodynamics, was routinely
measured by thermometer.
■ Think how a thermometer works…
A Crisis of Temperature
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■ “When two objects are separately in
thermodynamic equilibrium with a third object,
they are in thermodynamic equilibrium with each
other.”
■ Whenever two objects are in contact with one
another energy will flow between them until they
reach a state of thermodynamic equilibrium.Upon reaching this state we say that the two
objects are at the same temperature.
The Zeroth Law of Thermodynamics
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■ “When two objects are separately at the same
temperature as a third object, they are at the
same temperature as each other.”
■ This sounds obvious but it has vital ramifications
for the measurement of temperature by
thermometer.
■ The obvious nature of this statement explainswhy it did not become a law of thermodynamics
until well after the other three laws.
The Zeroth Law of Thermodynamics
7Basic Physical Chemistry - Thermodynamics
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You are free:• to Share - to copy, distribute and transmit the work• to Remix - to adapt the work
Under the following conditions:• Attribution
You must attribute the work in the manner specifed by the author or licensor (but not in any waythat suggests that they endorse you or your use of the work).
• Noncommercial You may not use this work for commercial purposes.
•
Share Alike If you alter, transform, or build upon this work, you may distribute the resulting work only under thesame or similar license to this one.
For any reuse or distribution, you must make clear to others the license terms of this work. The best wayto do this is with a link to this web page: http://creativecommons.org/licenses/by-nc-sa/4.0/
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Nothing in this license impairs or restricts the author’s moral rights.
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1st Law of ThermodynamicsEnthalpy
Basic Physical Chemistry - Thermodynamics 1
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■ Stated by Rudolph Clausius in 1850.
2Basic Physical Chemistry - Thermodynamics
The First Law of Thermodynamics
■ “Energy can neither be created nor destroyed,
merely changed from one form into another.”
■ Every system possesses an internal energy, U .
DU = qin + w on [1]
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3Basic Physical Chemistry - Thermodynamics
Evolution of Gas from a Reaction
Zn (s) + 2HCl (aq) ZnCl2 (aq) + H2 (g)
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The Work Done by an Expanding Gas
4Basic Physical Chemistry - Thermodynamics
p = F / A
w = Fd
\ F = pA
\ w = pAd
w = pDV
ExternalPressure, p
Cross-section, A
ExpandingGas
Frictionless piston moves through a distance d
w on = – pDV [2]
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5Basic Physical Chemistry - Thermodynamics
The First Law of Thermodynamics
DU = qin + w on [1]
w on = – pDV [2]
DU = qin – pDV
■ Expansion work clearly depends on p and DV .
Image: NASA
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6Basic Physical Chemistry - Thermodynamics
Free Expansion
DU = qin – pDV
But in space p = 0
DU = qin
However, relatively veryfew experiments arecarried out in space!
Image: NYNAS
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7Basic Physical Chemistry - Thermodynamics
Reactions at Constant Volume
DU = qin – pDV
But at constant volume DV = 0
D
U = qv [3]
Constant volume reactors must withstand massivepressure changes – heavy chemical industry, £ € $.
Image: Lilly M
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8Basic Physical Chemistry - Thermodynamics
Reactions at Constant Pressure
■ Now we define a new state function, enthalpy:
DU = qin – pDV
\ DU + pDV = qin at contant pressure
\ DH = DU + D( pV )
H = U + pV
DH = DU + pDV + V D p
But, at constant pressure, D p = 0 hence V D p = 0\ DH = DU + pDV
DH = qp [4]
Product rule:d uv
d x = u
dvd x
+ v dudx
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You are free:• to Share - to copy, distribute and transmit the work
• to Remix - to adapt the work
Under the following conditions:• Attribution
You must attribute the work in the manner specifed by the author or licensor (but not in any waythat suggests that they endorse you or your use of the work).
• Noncommercial You may not use this work for commercial purposes.
• Share Alike If you alter, transform, or build upon this work, you may distribute the resulting work only under thesame or similar license to this one.
For any reuse or distribution, you must make clear to others the license terms of this work. The best wayto do this is with a link to this web page: http://creativecommons.org/licenses/by-nc-sa/4.0/
Any of the above conditions can be waived if you get permission from the copyright holder.
Nothing in this license impairs or restricts the author’s moral rights.
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Reversible IsothermalExpansion of an Ideal Gas
Basic Physical Chemistry - Thermodynamics 1
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2Basic Physical Chemistry - Thermodynamics
The Ideal Gas Equation
pV = nRT
p is pressure
V is volume
n is quantity
R is the ideal gas constant, 8.314 J K –1 mol –1
T is absolute temperature
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3Basic Physical Chemistry - Thermodynamics
The Ideal Gas Equation
p = nRT / V
Consider 1 mol of an ideal gas at 298 K
p = 1 mol × 8.314 J K –1 mol –1 × 298 K / V
p = 2477.572 J / V
Using the above equation, calculate the pressure
of the gas for volumes of 1, 3, 5, 7, 9 and 11 m3.
Answer: 2478, 826, 496, 354, 275 and 225 Pa
respectively.
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4Basic Physical Chemistry - Thermodynamics
Isothermal Expansion
External
Pressure, p
One mole of
Gas
External pressure, p = 2478 Pa volume = 1m3
Now imagine the external pressure suddenly drops
to 826 Pa. What happens to the gas?
Ambient temperature = 298 K
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0
500
1000
1500
2000
2500
1.00 2.00 3.00 4.00 5.00 6.00 7.00 8.00 9.00 10.00 11.00
5Basic Physical Chemistry - Thermodynamics
p /
P a
V / m3
w = pDV826 Pa × 2 m3
w = 1652 J
w = 450 Jw = 550 Jw = 708 J
w = pDV
496 Pa × 2m3
w = 982 J
w total = 1652 + 982 + 708 + 550 + 450 J = 4342 J
Isothermal Expansion
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0
500
1000
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2500
1.00 2.00 3.00 4.00 5.00 6.00 7.00 8.00 9.00 10.00 11.00
6Basic Physical Chemistry - Thermodynamics
p /
P a
V / m3
Isothermal Expansion
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0
500
1000
1500
2000
2500
1.00 2.00 3.00 4.00 5.00 6.00 7.00 8.00 9.00 10.00 11.00
7Basic Physical Chemistry - Thermodynamics
p /
P a
V / m3
p = nRT / V
Isothermal Expansion
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8Basic Physical Chemistry - Thermodynamics
0
500
1000
1500
2000
2500
1.00 2.00 3.00 4.00 5.00 6.00 7.00 8.00 9.00 10.00 11.00
p /
P a
V / m3
w = 1652 J
w = 450 Jw = 550 Jw = 708 J
w = 982 J
w on = –4342 J
Isothermal Expansion
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9Basic Physical Chemistry - Thermodynamics
0
500
1000
1500
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2500
1.00 2.00 3.00 4.00 5.00 6.00 7.00 8.00 9.00 10.00 11.00
p /
P a
V / m3
1239 J826 J
619 J496 J
413 J 354 J 310 J 275 J 248 J 225 J
w on = –5004 J
Isothermal Expansion
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10Basic Physical Chemistry - Thermodynamics
0
500
1000
1500
2000
2500
1.00 2.00 3.00 4.00 5.00 6.00 7.00 8.00 9.00 10.00 11.00
p /
P a
V / m3
w on = –5428 J
Isothermal Expansion
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11Basic Physical Chemistry - Thermodynamics
0
500
1000
1500
2000
2500
1.00 2.00 3.00 4.00 5.00 6.00 7.00 8.00 9.00 10.00 11.00
p /
P a
V / m3
w on = –5672 J
Isothermal Expansion
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12Basic Physical Chemistry - Thermodynamics
0
500
1000
1500
2000
2500
1.00 2.00 3.00 4.00 5.00 6.00 7.00 8.00 9.00 10.00 11.00
p /
P a
V / m3
w on = –5803 J
Isothermal Expansion
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13Basic Physical Chemistry - Thermodynamics
Reversible Isothermal Expansion
0
500
1000
1500
2000
2500
1.00 2.00 3.00 4.00 5.00 6.00 7.00 8.00 9.00 10.00 11.00
p /
P a
V / m3
w on = –5941 Jw on = –5941 J
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14
w on for Reversible Isothermal Expansion
w on = – pdV
But, pV = nRT
So, p = nRT / V
w on = –(nRT / V) dV
Now consider an expansion from V i to V f
dV V
w on = –nRT
V f
V i
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15
w on for Reversible Isothermal Expansion
dV V
w on = –nRT
V f
V i
w on = –nRT V f
V i
lnV
w on = –nRT (lnV f – lnV i)
w on = –nRT ln V
V f
i
[5]
d x x
= ln x + c
ln A
−lnB
=ln
A
B
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16Basic Physical Chemistry - Thermodynamics
w on = –nRT ln V
V f
i
w on = –5941 Jw on = –1 mol × 8.314 J K
–1
mol –1
× 298 K × ln(11)
w on for Reversible Isothermal Expansion
■ Reversible change is a theoretical construct.
■ Determines maximum possible expansion work.
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You are free:• to Share - to copy, distribute and transmit the work
• to Remix - to adapt the work
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You must attribute the work in the manner specifed by the author or licensor (but not in any waythat suggests that they endorse you or your use of the work).
• Noncommercial You may not use this work for commercial purposes.
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If you alter, transform, or build upon this work, you may distribute the resulting work only under thesame or similar license to this one.
For any reuse or distribution, you must make clear to others the license terms of this work. The best wayto do this is with a link to this web page: http://creativecommons.org/licenses/by-nc-sa/4.0/
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Nothing in this license impairs or restricts the author’s moral rights.
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Heat Capacity
Basic Physical Chemistry - Thermodynamics 1
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2Basic Physical Chemistry - Thermodynamics
Heating Objects
Heat Capacity, C , is the heat energy required toraise the temperature of an object by 1 °C or 1 K.
Low
Heat
Capacity
High
Heat
Capacity
C = q / DT [6]
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3Basic Physical Chemistry - Thermodynamics
Definitions of Heat Capacity
Heat capacity is an extensive property.
Specific Heat Capacity, C s = C / m units J K –1 kg –1
Molar Heat Capacity, C m = C / n units J K –1 mol –1
Specific Heat capacity and Molar Heat Capacity
are intensive properties.
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130
140 390
440
710
710
13402000
2010
4Basic Physical Chemistry - Thermodynamics
520
Specific heat capacities of substances / J K –1 kg –1
880
1670
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5Basic Physical Chemistry - Thermodynamics
Isochoric and Isobaric Heat Capacities
Isochoric heat capacity is defined as:
C V = qV / DT = DU / DT [7]
Isobaric heat capacity is defined as:
C p = qp / DT = DH / DT [8]
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H m = U m + pV m
H = U + pV
For an ideal gas: pV = nRT
pV m = RT
6Basic Physical Chemistry - Thermodynamics
RT = 8.314 J K –1 mol –1 × 298 K ~ 2.5 kJ mol –1
H m = U m + RT
This is not negligible for a gas.
C p,m & C V,m Relationship for an Ideal Gas
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For a change of temperature DT
H m = U m + RT
DH m = DU m + R DT
DH m
DT =
DU m
DT +R
7Basic Physical Chemistry - Thermodynamics
C p,m = C V,m + R [9]
C p,m & C V,m Relationship for an Ideal Gas
where R = 8.314 J K –1 mol –1
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• to Remix - to adapt the work
Under the following conditions:• Attribution
You must attribute the work in the manner specifed by the author or licensor (but not in any waythat suggests that they endorse you or your use of the work).
• Noncommercial You may not use this work for commercial purposes.
• Share Alike
If you alter, transform, or build upon this work, you may distribute the resulting work only under thesame or similar license to this one.
For any reuse or distribution, you must make clear to others the license terms of this work. The best wayto do this is with a link to this web page: http://creativecommons.org/licenses/by-nc-sa/4.0/
Any of the above conditions can be waived if you get permission from the copyright holder.
Nothing in this license impairs or restricts the author’s moral rights.