First Law of Thermodynamics
1.1 Work and Heat
Changes in the state
spontaneous non-spontaneous
Isothermal Isobaric isochoric
Adiabatic Diabatic
irreversible reversible
Processes
Work
Energy transfer through the mechanical link between the system and the surroundings
V P r A A r f
f
w=− • =− • =− ∆
J 325 , 101 Nm 325 , atm 101
1
N/m 325 , 101 L
10 m atm 1
L 1
2 3
3 × = =
×
Change in State Accompanying Work
ext
d
dw = − P V
21 ext
d
w = − ∫ P V w = − ∫12P T V ( , )d V
path dependant quasistatic process
path A–C–B w = – 2 X (30 – 10) + 0 = –40 atm L path A–D–B w = 0 – 1 X (30 – 10) = –20 atm L
Joule’s Experiment
James Prescott Joule (1818-1889)
Joule “Mechanical Equivalence of Heat”
Work and heat are equivalent.
4.184 J = 1.0 calorie
The heat required to raise the
temperature of a gram of water 1℃, from 14.5℃ to 15.5℃
Thermal energy transfer in thermal contact through the temperature gradient
Heat
2.2 First Law of Thermodynamics
Forms of energy crossing a boundary
in an adiabatic process
∆ = U w
Express the change in state of a system in an adiabatic process in terms of work The internal energy
U is a state function.
with no work done
∆ = U q
The heat absorbed by a closed system in a process in which no work is done is equal to the increase in internal energy of the system.
The First Law of Thermodynamics
U q w
∆ = + d U = 0
∫ U is a state function
2.3 Exact and Inexact Differentials
The differential of a state function is exact b
d
b aa
U = U − U = ∆ U
∫
The differential of a path function is inexact b b a
a
dw = ≠ w w − w
∫
Euler’s criterion ( , ) z = f x y
d d d
y x
z z
z x y
x y
∂ ∂
= ∂ + ∂
y x x y
z z
y x x y
∂ ∂ = ∂ ∂
∂ ∂ ∂ ∂
d z = M x y x ( , )d + N x y y ( , )d
x y
M N
y x
∂ ∂
=
∂ ∂
AnIntegrating factor multiplies an inexact differential to yield an exact differential.
2.4 Isothermal Compression of a Gas
2
p mg
= A w = mgh = − P V
2(
2− V
1)
21
d
Vd
w = ∫ w = − ∫V P V
The minimum amount of work is
required in the limit of an infinite number of steps – the pressure is changed by an infinitesimal amount for each
infinitesimal step.
reversible process
2 1 2 2
1 2 1 1
cycle V
d
Vd
Vd
Vd 0
V V V V
w = − ∫ P V − ∫ P V = − ∫ P V + ∫ P V =
consists of a series of successive equilibria.
Isothermal Expansion of a Gas
2
p mg
= A w = mgh = − P V
2(
2− V
1)
21
d
Vd
w = ∫ w = − ∫V P V
The largest amount of work in the surroundings and the largest negative work done on the system are obtained in the limit of an infinite number of steps.
reversible process
Isothermal Reversible Expansion of an Ideal Gas
2 2
1 1
2 rev
1
d d ln
V V
V V
V
w P V nRT V nRT
V V
= − ∫ = − ∫ = −
P nRT
= V
1 1 2 2
PV = P V
rev 21
ln P w nRT
= P
1 1
2 rev
1
1 bar
(a) ln (1 mol)(8.3145 J K mol )(298K) ln = 3988 J 5 bar
w nRT P P
− −
= = −
irrev 2 2 1 2
2 1
1 1
(b) ( )
1 1
= (1 bar)(1 mol)(8.3145 J K mol )(298K) = 1982 J 1 bar 5 bar
nRT nRT
w P V V P
P P
− −
= − − = − −
− − −
Example 2.3
One mole of an ideal gas expands from 5 to 1 bar at 298K. Calculate (a) wrev and (b) wirrev.2.5 Various Kinds of Work
Surface work
Elongation work
Electrical work
d w = γ d A
Sd w = f L d d w = φ d Q
Work Type Intensive Variable
Extensive Variable
Differential Work
Hydrostatic P V –PdV
Surface γ AS γdAS
Elongation f L fdL
Electrical φ Q φdQ
2
Sw = ∆ = f x L x ∆ = ∆ γ γ A 2
f = L γ
2.6 Change in State at Constant Volume
( , )
U T V d d d
V T
U U
U T V
T V
∂ ∂
= ∂ + ∂
d U = d q + d w = d q − P
extd V
d d
extd d
V T V
U U U
q T P V T
T V T
∂ ∂ ∂
= ∂ + + ∂ = ∂
at constant volume,d V = 0
The heat capacity
C
V at constant volume,d d
V V
V
q U
C T T
∂
≡ = ∂
2 1
T
d
V T V V
U C T q
∆ = ∫ =
2 1
( )
V V V
U C T T C T
∆ = − = ∆
Joule’s Experiment – Internal Pressure
Joule detected no discernable change in
temperature upon expansion.
d q = 0
No work is done, since
P
ext= 0 d w = 0
At constant temperature, d
T = 0
d U = d q + d w = 0
d d 0
T
U U V
V
∂
= ∂ =
0
T
U V
∂ =
∂
Note that the heat capacity of the bottles is large relative to the gas. The internal pressure changes with the volume because as the volume is increased, the average intermolecular distances increase and the average intermolecular potential energy changes.
Kinetic Theory of Gases
Theory for gaseous physical properties by analyzing constituent molecular motions.
In a state of a pure gas, an enormous number of identical molecules fills a volume of space homogeneously. The average distance among gas molecules is very large compared to the size of the molecule.
Gas molecules move randomly and their speed makes a distribution.
Gas molecules do not interact and exert no force unless they collide.
Molecular collision is exactly elastic.
Molecular average kinetic energy is proportional to the absolute temperature.
Ideal Gas Assumptions
1 1 2
rev
1
1 bar
(a) ln (1 mol)(8.3145 J K mol )(298K) ln = 3988 J 5 bar
w nRT P P
− −
= = −
irrev 2 2 1 2
2 1
1 1
(b) ( )
1 1
= (1 bar)(1 mol)(8.3145 J K mol )(298K) = 1982 J 1 bar 5 bar
nRT nRT
w P V V P
P P
− −
= − − = − −
− − −
Example 2.3
One mole of an ideal gas expands from 5 to 1 bar at 298K. Calculate (a) wrev and (b) wirrev.Expansion of an Ideal Gas
Example 2.5
Calculate (a) qrev and (b) qirrev.rev rev
(a) q = ∆ − U w = − − 0 ( 3988 J) = 3988 J
irrev irrev
(b) q = ∆ − U w = − − 0 ( 1982 J) = 1982 J
2.7 Enthalpy and Change of State at Constant Pressure
P P
U q w q P V
∆ = + = − ∆
2 1 P
(
2 1)
U − U = q − P V − V
2 2 1 1
( ) ( )
q
P= U + PV − U + PV
H ≡ + U PV
Define a new state function, the
enthalpy H
2 1
q
P= H − H
andd q
P= d H ( , )
H T P d d d
P T
H H
H T P
T P
∂ ∂
= ∂ + ∂
d
Pd
P
q H T
T
∂
= ∂
at constant pressure,d P = 0
The heat capacity
C
P at constant pressure,d d
P P
P
q H
C T T
∂
≡ = ∂
2.8 Heat Capacities
CH4
NH3 CO2 H2O N2 He
2 3
C
P= + α β T + γ T + δ T
JANAF(Joint-Army-Navy-Air Force) Thermochemical Tables (1959 – )
Heat Capacities
d d
extd d
V T V
U U U
q T P V T
T V T
∂ ∂ ∂
= ∂ + + ∂ = ∂
d
P Vd d
T
q C T P U V
V
∂
= + + ∂
At constant pressure, Pext= P
P V
T P
U V
C C P
V T
∂ ∂
− = + ∂ ∂
P
P V T
∂
∂
T P
U V
V T
∂ ∂
∂ ∂
the work produced per unit increase in temperature at constant pressure
the energy per unit temperature required to separate the molecules against intermolecular attraction
For an ideal gas,
U 0 V
∂ =
∂
V nR
T P
∂ =
∂
and
C
P− C
V= nR
2.9 Joule-Thomson Expansion
Push one mole of gas through the porous plate,
P
1> P
2, in an insulated pipe,q = 0
2 1 1 1 2 2
U − U = = w PV − P V
2 2 2 1 1 1
U + P V = U + PV
2 1
H = H
At low temperature. As a gas expands, the average distance between molecules and the potential energy of the gas increases. Due to the enthalpy conservation, the increase in potential energy thus implies a decrease in kinetic energy and therefore in temperature.
At high temperature. During gas molecular collisions, kinetic energy is temporarily converted into potential energy. As the average intermolecular distance increases, the number of collisions per time decreases, and the average potential energy decrases. The enthalpy is conserved, so this leads to an increase in kinetic energy and in temperature)
Joule-Thomson Effect
Zero for an ideal gas, positive at low temperatures and negative at high temperatures for real gases.
The inversion temperature:
N 607K, H 204K, and He 43K The Joule-Thomson coefficient,
µ
JT2 1
JT 0 2 1
lim
P HT T T
P P P
µ
∆ →− ∂
= − = ∂
2.10 Adiabatic Processes
No heat is gained or lost,
q = 0 d U = d w = − P
extd V
2 2
1
d
2 1 1 extd
U V
U V
U U U U w P V
∆ = ∫ = − = = − ∫
For an ideal gas,
dU = C
VdT
2 21
d
2 1 1d (
2 1)
U T
V V
U T
U U U U C T C T T
∆ = ∫ = − = ∫ = −
Reversible adiabatic expansion of one mole of an ideal gas,
P
ext= P
d
Vd d RT d
U C T P V V
= = − = − V
d d
V
T V
C R
T = − V
2 2 1 1
d d
T V
V T V
T V
C R
T = − V
∫ ∫
2 1 1 2
ln ln
V
T V
C R
T = V
We assumed that the temperature range is small enough so that CVdoes not change very much.
Isothermal and Reversible Adiabatic Expansions
Isothermal and reversible adiabatic expansions of one mole of an ideal monatomic gas
P V
C − C = R
PV
C γ = C
2 1 1 2
ln ln
V
T V
C R
T = V
1 2 1
1 2
T V
T V
γ−
=
1
2 2
1 1
T P
T P
γ γ
−=
1 1 2 2
PV
γ= P V
γAdiabatic expansion
results in a smaller volume than isothermal expansion
2.11 Thermochemistry
Enthalpy is an extensive state function.
1
d d
Ns
i i i
H H n
=
= ∑
A single chemical reaction can be represented as
1
0 B
NS
i i i
ν
=
= ∑
Define the
extent of reaction
such thatn
i0 is the initial amount of substancei
0
i i i
n = n + ν ξ
1
d d d
Ns
P i i
i
H q ν H ξ
=
= = ∑
r
, 1
d d
Ns
P
i i T P i
q
H H ν H
ξ ξ
=∂
∆ = ∂ = = ∑
The reaction enthalpy
(heat of reaction at constant temperature and pressure)
Thermodynamic Standard States
r
1 Ns
i i i
H ν H
=
∆
= ∑
A pure gaseous substance (g) at a given temperature – the (hypothetical) ideal gas at 1 bar
A pure liquid substance (l) at a given temperature – the pure liquid at 1 bar
A pure crystalline substance (s) at a given temperature – the pure crystalline substance at 1 bar pressure
A substance in solution at a given temperature – the ideal solution of standard state molality (1 m) at 1 bar pressure
A strong electrolyte completely dissociated in water (ai)
An undissociated in water (ao) including ions
2.12 Enthalpy of Formation
∆
fH ° Standard enthalpy of formation
Enthalpy change accompanying the formation reaction for 1 mol of a compound from its constituent elements in the standard state
The enthalpy of formation of an element in its standard state is defined to be 0 at every temperature.
r f
1 Ns
i i
i
H ν H
=
∆
= ∑ ∆
Example 2.8
Calculate the standard enthalpy of reaction at 298.15K.CO2(g) + C(graphite) = 2CO(g)
∆rH◦= 2∆fH◦ (CO) – ∆fH◦ (CO2) = 2(– 110.527 kJ/mol) – ( –393.533 kJ/mol) = 172.468 kJ/mol
Enthalpy of Formation for Carbon Allotropes
C(graphite) → C(diamond) ∆ H
fo= 1.9 kJ·mol
–1C(graphite) → C(buckminsterfulleren) ∆ H
fo= 34 kJ·mol
–1Enthalpy of Reactions
reactants
reactants
products
products
r
H
T∆
r
H
298∆
298
,reactants
d
T
C
PT
∫ ∫
298TC
P,procuctsd T
298
r T P,reactants
d
r 298 298T P,productsd
H
TC T H C T
∆
= ∫ + ∆
+ ∫
r
H
T rH
298 298T( C
P,productsC
P,reactants)d T
rH
298 298T rC T
Pd
∆
= ∆
+ ∫ −
= ∆
+ ∫ ∆
r P i P i,
i
C ν C
∆
= ∑
298 298T
d
T P
H
− H
= ∫ C
T
2.13 Calorimetry
A
( )
1 P0
H H T H
∆ = ∆ + ∆ =
A R
(
2) 0
H H H T
∆ = ∆ + ∆ =
R(T2)+Cal(T2)
R(T1)+Cal(T1)
∆H(T2)
∆H(T1)
∆HR ∆HA = 0 ∆HP
P(T2)+Cal(T2)
P(T1)+Cal(T1)
1 P 2 1
( ) [
P(P)
P(Cal)]( )
H T H C C T T
∆ = −∆ = − + −
2 R 2 1
( ) [
P(R)
P(Cal)]( )
H T H C C T T
∆ = −∆ = − + −
Adiabatic Calorimeters
jacket water electrodes sample bomb
Calorimetry
Adiabatic bomb calorimeter for carrying out combustion at constant volume
U q
V∆ =
r
H
rU RT ν
g∆ = ∆ + ∑
Example 2.10
The combustion of ethanol in a constant-volume calorimeter produces 1364.34 kJ mol –1at 25.0℃. Calculate ∆rH◦ for the following combustion reaction.C2H5OH(l) + 3O2(g) = 2CO2(g) + 3H2O(l)
∆ ◦= ∆ ◦
Σ
νThe Enthalpy of Formation of Ions
The neutralization reaction of a dilute solution of a strong acid with a dilute solution of a strong base
OH–(ao) + H+(ao) = H2O(l)
Since for strong electrolytes in dilute solution the thermal properties of the ions are essentially independent of the accompanying ions, it is convenient to use enthalpies of formation of
individual ions.
∆fH◦(298 K) = –55.835 kJ mol –1
H2O(l) = OH–(ao) + H+(ao) ∆fH◦ = 55.835 kJ mol –1 H2(g) + ½ O2(g) = H2O(l) ∆fH◦ = –285.830 kJ mol –1 H2(g) + ½ O2(g) = OH–(ao) + H+(ao) ∆fH◦ = –229.995 kJ mol –1 By convention, with ∆fH◦(e–) = 0
½H2(g) = H+(ao) + e– ∆fH◦ = 0
∆fH◦ = –229.995 kJ mol –1
½ H2(g) + ½ O2(g) + e–= OH–(ao)
∆fH◦ = –167.159 kJ mol –1
½ H2(g) + ½ Cl2(g) = H+(ao) + Cl–(ao)
∆ ◦