2 the first law of thermodynamic
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2. The First Law
Thermodynamics
equilibrium thermodynamics
• characterize equilibrium states
• which equilibrium states can be reached from agiven initial state (i or 1): allowed processes
• which processes occur spontaneously
system
PChem I 2.1
isolated: no interaction at all with surroundings
example: Dewar flask with absolutely rigid walls in aFaraday cage on the space station
idealization, but useful
closed: exchange of energy, no exchange of matter
example: photochemical batch reactor
open: exchange of matter and energy
example: living systems
[Figure: Types of system; Atkins 9th ed., Fig. 2.1]
PChem I 2.2
work: one way to interact with system
Work is done if the process could be used to bringabout a change in the height of a weight somewherein the surroundings.
work is done by the system if the weight is raised
work is done on the system if the weight is lowered
PChem I 2.3
sign convention (IUPAC): viewpoint of the system
w positive, if work is done on the system
w negative, if work is done by the system
mechanics: w = ~F · ~d ; if ~F ||~d : w = F d
expansion/compression of a fluid
Pex Pex
P , V P −dP ,V +dV
}dx
P = F
A
expansion: external force F = PexA
dw =−PexA ·dx =−PexdV
minus sign: according to sign convention dw must
PChem I 2.4
be negative, since work is done by the system duringexpansion
compression: the same formula applies
dw =−PexdV
since dV < 0, we have dw > 0, as it should be, sincework is done on the system during compression
expansion/compression work (or volume-pressurework):
w =−∫ V2
V1
PexdV general
other types of work:
electrical work: dw = φdQ, φ = electric potential, Q= charge
surface expansion: dw = γdσ, γ = surface tension, σ= surface area
extension: dw = f dl , f = tension, l = length
PChem I 2.5
energy: capacity to do work
energy increased if w > 0, decreased if w < 0
experiments show that energy can also be increasedby heating
diathermic wall: heat flow possible; adiabatic wall: nochange in energy if no work can be done; no heat flowpossible
First Law of Thermodynamics
The work needed to change an adiabatic system fromone specified state to another specified state is thesame however the work is done.
for example: mechanical work, electrical work, mag-netic work
PChem I 2.6
wad,1 = wad,2 = wad,3 = wad,4 = . . . = wad(i , f )
Adiabatic work does not depend on the path, but onlyon the initial state i and the final state f .
U f −Ui ≡ wad(i , f )
U = internal energy, state function
can measure only ∆U =U f −Ui
replace adiabatic walls by diathermic walls
work is now path-dependent
mechanical definition of heat:
q ≡∆U −w = wad−w
First Law:
(1) There exists a state function U , the internal ener-gy, for the system.
(2) If the state of the system changes, then∆U = q +w
PChem I 2.7
∆U = q +w conservation of energy6 6 6
state function path function
? ?
what crosses the boundary
first law =⇒ U = const for an isolated system
to make the first law operational, consider infinitesi-mal changes
dU = dq +dw
PV T systems:
w =−∫ V f
Vi
PexdV
1. expansion into vacuum = free expansion, Pex = 0,no weight raised or lowered
w = 0 J
PChem I 2.8
2. expansion against constant external pressure:
w =−∫ V f
Vi
PexdV =−Pex
∫ V f
Vi
dV =−Pex(V f −Vi )
w =−Pex∆V
3. reversible expansion:
reversible change (process): change that can be re-versed, i.e., undone, by an infinitesimal modificationof a variable
system is in equilibrium with its surroundings
reversible paths lie entirely in the surface of equilibri-um states
reversible processes are quasistatic
expansion Pex < P : reversible Pex = P −dP
compression Pex > P : reversible Pex = P +dP
change from expansion to compression or vice versa:|2dddP |
dwrev =−PexdV =−(P ∓dP )dV
PChem I 2.9
dwrev =−PdV ±dPdV
dddwrev=−PdddV reversible process
wrev =−∫ V f
Vi
PdV
need equation of state P = P (V ,T )!
ideal gas P = nRT /V
wrev =−∫ V f
Vi
nRT
VdV
Can not evaluate the integral; the integrand dependsnot only on V , but also on n and T
Need to specify a path! Work w is a path function!
closed system n = const and isothermal T = const
wrev =−∫ V f
Vi
PdV =−∫ V f
Vi
nRT
VdV =−nRT
∫ V f
Vi
1
VdV
PChem I 2.10
=−nRT [lnV ]V f
Vi=−nRT (lnV f − lnVi )
wrev =−nRT ln
(V f
Vi
)ideal gas, isothermal, reversible
since P f V f = PiVi for ideal gas and isothermal, wehave the equivalent form
wrev =−nRT ln
(Pi
P f
)ideal gas, isothermal, reversible
indicator diagram
isothermal, reversible: w = ///
constant external pressure: w = \\\
PChem I 2.11
How do we measure ∆U for general systems?
dU = dq +dwexp+dwe
where dwexp is expansion work and dwe is “extra”work, i.e., work in addition to expansion work, e.g.,electrical work.
Consider a constant-volume process (isochoric pro-cess). Then dwexp = 0. Assume further that no otherkind of work is possible, i.e., dwe = 0. Then
dU = dqV
or
∆U = qV
In general
dq =C (T )dT
where C (T ) is the heat capacity, an extensive quan-tity. Since q is a path function, the heat capacity C
PChem I 2.12
depends on the type of process or change that oc-curs. If C (T ) = const is a good approximation for therange of temperatures involved in the process, then
q =C∆T
Caution: C can be infinite, C =∞, for certain paths!
For an isochoric process,
dqV =CV dT
where CV (T ) is the heat capacity at constant volume.
For a constant-pressure process (isobaric process)
dqP =CP (T )dT
where CP (T ) is the heat capacity at constant pres-sure. Generally CP > CV , except if the fluid contractsupon heating, e.g., water in the range 0◦C – 4◦C at1 atm.
PChem I 2.13
relation between CV and CP (for a derivation, see Further
Information 1 on page 2.43)
CP −CV = α2V T
κT
where
α≡ 1
V
(∂V
∂T
)P
isobaric thermal expansivity
κT ≡− 1
V
(∂V
∂P
)T
isothermal compressibility
For ideal gases, the relation reduces to
CP −CV = nR
For a constant-volume process,
dU =CV dT, V = const, (general)
Considering U to be a function of T and V , U =
PChem I 2.14
U (T,V ), then
CV = dU
dTfor V = const
More rigorously,
CV =(∂U
∂T
)V
, general definition of CV
Use constant-volume bomb calorimeter to measure∆U :
∆U =∫ f
idqV =
∫ f
iCV (T )dT
If CV (T ) =CV = const is a good approximation for therange of temperatures involved in the process, then
∆U = qV =CV∆T for isochoric processes
reactions typically occur at P = const
dU = dq +dwexp if dwe = 0
PChem I 2.15
since dwexp < 0, we have
dU < dqP
some heat was used by the system to do expansionwork!
new state function
H =U +PV enthalpy
dH = dU +d(PV )
= dU +V dP +PdV
= dq +dw +V dP +PdV (dw = dwexp+dwe)
= dq −PdV +dwe+V dP +PdV
= dq +V dP +dwe
(Since H is a state function, it is no restriction of gen-erality that we have used dwexp = dwrev =−PdV .)
For a constant-pressure process and no extra work
dH = dqP
PChem I 2.16
Considering H to be a function of T and P , H =H(T,P ), then
CP = dH
dTfor P = const, (general)
More rigorously,
CP =(∂H
∂T
)P
, general definition of CP
∆H =∫ f
iCP (T )dT for P = const
If CP (T ) = CP = const is a good approximation for therange of temperatures involved in the process, then
∆H = qP =CP∆T for isobaric processes
If it is necessary to take into account the variationof the heat capacity with temperature, the following
PChem I 2.17
approximate empirical expression is commonly used:
C P (T ) = a +bT + c
T 2
where a, b and c are empirical parameters that areindependent of T and are tabulated for various sub-stances.
With this form
∆H = H(T2)−H(T1) = a(T2−T1)+ 1
2b(T 2
2 −T 21 )
− c
(1
T2− 1
T1
)P = const
Note:
∆H =∆U +∆(PV )
=∆U +P f V f −PiVi
For an ideal gas, PV = nRT : ∆H =∆U +∆(nRT ), andif n = const, i.e., closed system, ∆H =∆U +nR∆T
PChem I 2.18
thermochemistry
endothermic process – exothermic process
(a) container with adiabatic walls: endothermic: T de-creases; exothermic: T increases
(b) container with diathermic walls; isothermal pro-cess: endothermic: q positive; exothermic: q nega-tive
[Figure: endo- and exothermic processes; Atkins 9th ed., Fig. 2.2]
PChem I 2.19
most chemical reactions take place at constant pres-sure: enthalpy changes ∆H
frequently reported for a set of standard conditions
standard enthalpy changes ∆H−◦
standard state
The standard state of a substance at a specified tem-perature is its pure form at that temperature and apressure of 1 bar. (P−◦ = 1 bar)
Remarks: (1) older values (pre-1982) use P−◦ = 1 atm
(2) for real gases the nonideality is removed, discussed later
conventional temperature: 298.15 K
phase = a specific state of matter that is uniformthroughout in composition and physical state
phase transition: conversion of one phase of a sub-stance, say a, to another phase, say b, at constantpressure
∆trsH = Hb−Ha = qP enthalpy of transition
PChem I 2.20
examples: enthalpy of vaporization, enthalpy of fu-sion, enthalpy of sublimation, . . .
∆trsH > 0, from more ordered to less ordered phases
standard enthalpy of vaporization; example
H2O(l) −→ H2O(g) ∆vapH−◦(373 K) =+40.66 kJ mol−1
H2O(l) −→ H2O(g) ∆vapH−◦(298 K) =+44 kJ mol−1
standard enthalpy of fusion; example
H2O(s) −→ H2O(l) ∆fusH−◦(273 K) =+6.01 kJ mol−1
sublimation: solid → vapor; reverse: vapor deposition
s → g = s → l + l → g
∆subH−◦ =∆fusH−◦ +∆vapH−◦
standard enthalpy of solution, enthalpy of ionization,. . . , bond dissociation enthalpy
PChem I 2.21
standard reaction enthalpy; example
CH4(g)+2O2(g) −→ CO2(g)+2H2O(l) ∆rH−◦ =−890kJmol−1
thermochemical equation
can be written as
∆rH−◦ = H
−◦(CO2(g))+2H
−◦(H2O(l))−H
−◦(CH4(g))−2H
−◦(O2(g))
or in general
∆rH−◦ = ∑
products
νH−◦ − ∑
reactantsνH
−◦
∆rH > 0 endothermic reaction, ∆rH < 0 exothermic re-action
enthalpy is a state function =⇒ Hess’s Law:
If a process occurs in stages or steps (all at the sametemperature and pressure), even if only hypothetical-ly, the enthalpy change for the overall reaction (netprocess) is the sum of the enthalpy changes for theindividual steps.
PChem I 2.22
example:
2C(s)+2O2(g) −→ 2CO2(g) ∆H =?
2C(s)+O2(g) −→ 2CO(g) ∆H =−221.0 kJ/mol
2CO(g)+O2(g) −→ 2CO2(g) ∆H =−566.0 kJ/mol
2C(s)+2O2(g) −→ 2CO2(g) ∆H =−787.0 kJ/mol
standard enthalpy of formation ∆fH−◦ = standard
reaction enthalpy for the formation of the compoundfrom its elements in their references states
reference state = most stable form at the specifiedtemperature and P−◦
standard enthalpies of formation are enthalpies permole of molecules or formula units of the compound
the standard enthalpy of formation of each elementin its reference state is zero
standard enthalpy of formation of ions: a solution is
PChem I 2.23
electrically neutral and always contains cations andanions; principle of electroneutrality
need an additional zero point: ∆fH−◦(H+,aq) = 0
∆rH−◦ = ∑
products
ν∆fH−◦ − ∑
reactantsν∆fH
−◦
Introduce stoichiometric numbers νJ:
νJ positive for products
νJ negative for reactants
∆rH−◦ =∑
J
νJ∆fH−◦(J)
How do we obtain ∆H at temperatures different fromthe conventional temperature?
H increases with increasing temperature
H(T2) = H(T1)+∫ T2
T1
CP (T )dT P = const
PChem I 2.24
applies to each species in a reaction, thus
∆rH−◦(T2) =∆rH
−◦(T1)+∫ T2
T1
∆CP (T )dT P = const
where
∆CP (T ) = ∑products
νC P −∑
reactantsνC P Kirchhoff’s Law
∆CP (T ) =∑J
νJC P (J)
for moderate changes in temperature C P (T ) ≈C P and∆CP (T ) ≈∆CP ; otherwise use C P (T ) = a +bT + c/T 2
state function: U , H , P , . . .
∆U =∫ f
idU =U f −Ui
path function: w , q
q =∫ f
i ;pathdq ( 6= q f −qi !!!!)
PChem I 2.25
dU exact differential; dq inexact differential
state function ⇐⇒ exact differential
PVT system
U (P,V ,T ), equation of state F (P,V ,T ) = 0
(example: ideal gas F (P,V ,T ) = PV −nRT = 0)
U (T,V ) or U (T,P ) or U (V ,P )
most convenient choice: U (T,V )
(T,V ) −→ (T +dT,V +dV ) :
U (T,V ) −→U (T +dT,V +dV ) =U +dU
dddU =(∂U
∂T
)V
dddT +(∂U
∂V
)T
dddV
general function f (x, y)
differential: d f =(∂ f
∂x
)y
dx +(∂ f
∂y
)x
dy
physical meaning of partial derivatives:
PChem I 2.26
dddU =CV dddT +πT dddV
πT ≡(∂U
∂V
)T
internal pressure
response of internal energy U to changes in temper-ature T for V = const: CV = (
∂U∂T
)V
response of internal energy U to changes in volumeV for T = const: πT = (
∂U∂V
)T
[Figure: Internal pressure; Atkins 9th ed., Fig. 2.23]
PChem I 2.27
what is πT ?
ideal gas: Joule experiment
[Figure: Joule experiment; Atkins 9th ed., Fig. 2.25]
i : gas in left compartment, Vi ; right compartmentvacuum; temperature Ti
f : gas in both compartments: V f , T f
experiment: T f = Ti
∆U = q+w ; w = 0, free expansion; q = 0, since T f = Ti
and C 6=∞ =⇒ ∆U = 0
PChem I 2.28
Conclusions:
for ideal gas U (T,V ) =U (T )
for ideal gas πT =(∂U
∂V
)T= 0
for ideal gas ∆U =CV∆T
This holds for all changes for an ideal gas; V neednot be constant!
ideal gas: isothermal ≡ isoergic !!
ideal gas in the strict sense: CV (T ) =CV
PChem I 2.29
πT for real gases
[Figure: Internal pressure; Atkins 9th ed., Fig. 2.24]
ideal gas: H =U +PV =U (T )+nRT =⇒ H = H(T )
∆H =∆U +∆(PV ) =∆U +∆(nRT )
n = const: ∆H =CV∆T +nR∆T = (CV +nR)∆T
for ideal gas ∆H =CP∆T
This holds for all changes for an ideal gas; P neednot be constant!
PChem I 2.30
ideal gas: isothermal ≡ isenthalpic !!
test for exact differential
d f =(∂ f
∂x
)y
dx +(∂ f
∂y
)x
dy
is exact, if and only if
∂2 f
∂y∂x= ∂2 f
∂x∂y
state function ⇐⇒ exact differential ⇐⇒mixed second derivatives are equal
U is a state function ⇒∂2U
∂V ∂T= ∂2U
∂T∂V
∂CV
∂V= ∂πT
∂T
for ideal gas: CV = const and πT = 0 =⇒∂CV
∂V= 0 = ∂πT
∂TX
PChem I 2.31
we know that(∂U∂T
)V=CV
what about(∂U∂T
)P?
changing the independent variables from (T,V ) to(T,P ), i.e., considering U (T,P ) instead of U (T,V ), oneobtains (for a derivation, see Further Information 2 on page
2.44)(∂U
∂T
)P=CV +πT
(∂V
∂T
)P(
∂U
∂T
)P=CV +απT V
thermodynamic relation, universally valid
special case: ideal gas πT = 0 ⇒ (∂U∂T
)P=CV
state function enthalpy H ; consider to be a functionof T and P
H = H(T,P )
dH =(∂H
∂T
)P
dT +(∂H
∂P
)T
dP
PChem I 2.32
dH =CP dT +(∂H
∂P
)T
dP
µT ≡(∂H
∂P
)T
isothermal Joule-Thomson coefficient
dH =CP dT +µT dP
relations for partial derivatives:(∂y
∂x
)z= 1(
∂x∂y
)z
(1)
(∂x
∂y
)z
=−(∂x
∂z
)y
(∂z
∂y
)x
(2)
Using relation (2):(∂H
∂P
)T=−
(∂H
∂T
)P
(∂T
∂P
)H(
∂H
∂P
)T=−CP
(∂T
∂P
)H
µ≡(∂T
∂P
)H
Joule-Thomson coefficient
PChem I 2.33
µT =−CPµ
dH =CP dT −CPµdP
(∂T
∂P
)H=µ=
{ < 0 if ∆T > 0 for ∆P < 0> 0 if ∆T < 0 for ∆P < 0
ideal gas: H(T,P ) = H(T ) =⇒ µT = (∂H∂P
)T= 0 =⇒ µ= 0
example of a constant enthalpy process: push gasin an adiabatic container through a throttle (porousplug); i = high pressure side, f = low pressure side;fixed amount of gas
[Figure: Throttling experiment; Atkins 9th ed., Fig. 2.26 and Fig. 2.27]
PChem I 2.34
∆U = q +w
∆U = w
w = w1+w2
w1 =−Pi (0−Vi ) = PiVi
w2 =−P f (V f −0) =−P f V f
∆U =U f −Ui = w = PiVi −P f V f
U f +P f V f =Ui +PiVi
H f = Hi
usually µ is measured by measuring µT : pump gasthrough a heat exchanger (ensure T = const), througha porous plug inside a thermally insulated container,then through a heater to offset the cooling; measure∆P and ∆H = heat provided by heater: µT =∆H/∆Pas ∆P → 0
PChem I 2.35
[Figure: isothermal Joule-Thomson coefficient; Atkins 9th ed., Fig. 2.29]
µ< 0 −−−−−−−−−−−−−−−−−−−−→inversion temperature TI
µ> 0
[Figure: Inversion temperature; Atkins 9th ed., Fig. 2.30]
PChem I 2.36
[Figure: Inversion temperatures fro nitrogen, hydrogen, and helium; Atkins 9th ed., Fig. 2.31]
changing the independent variables for the enthalpyH from (T,P ) to (T,V ), we obtain (see Further Information
3 on page 2.45)(∂H
∂T
)V=
(1−αµ
κT
)CP
PChem I 2.37
adiabatic expansion/compression
dU = dq +dw
adiabatic process dddq = 0: dU = dw
ideal gas for the remainder:
dU =CV dT (for all processes in an ideal gas)
wad =∫ f
iCV dT =CV∆T
(1) fixed external pressure
w =−Pex∆V (general)
−Pex∆V =CV∆T
∆T =−Pex∆V
CV
T f −Ti =−Pex(V f −Vi )
CV=−
Pex
(nRT f
P f− nRTi
Pi
)CV
PChem I 2.38
(2) reversible
dw =−PdV (general)
dwad =CV dT
CV dT =−PdV reversible, adiabatic, ideal gas
P = nRT
V
CV dT =−nRT
VdV
CV
TdT =−nR
VdV∫ T f
Ti
CV
TdT =−
∫ V f
Vi
nR
VdV
CV ln
(T f
Ti
)=−nR ln
(V f
Vi
)
ln
(T f
Ti
)=−nR
CVln
(V f
Vi
)= nR
CVln
(Vi
V f
)
c ≡ CV
nR
PChem I 2.39
T f
Ti=
(Vi
V f
)1/c
T f V 1/cf = TiV
1/ci or T V 1/c = const
1
c= nR
CV= CP −CV
CV= CP
CV−1
γ≡ CP
CV
T V γ−1 = const adiabats
equivalent, alternative form of ideal gas adiabats
PV = nRT, T = PV
nRPV
nRV γ−1 = const
PV γ= nR const
PV γ= const adiabats
compare to isotherms PV = const
PChem I 2.40
[Figure: Adiabats and isotherms; Atkins 9th ed., Fig. 2.18]
third form of the adiabats: T V γ−1 = const and theequation of state PV = nRT yield
T
(nRT
P
)γ−1
= const
PChem I 2.41
T · T γ−1
Pγ−1 =const
(nR)γ−1
T γ ·P−γ+1 = const
T ·P−1+ 1γ = const adiabats
Caution: the adiabats can only be used for re-versible adiabatic processes in an ideal gas!!
PChem I 2.42
Further Information 1:
CP −CV =(∂H
∂T
)P−
(∂U
∂T
)V
=(∂(U +PV )
∂T
)P−
(∂U
∂T
)V
=(∂U
∂T
)P+
(∂PV
∂T
)P−
(∂U
∂T
)V
= CV +απT V +P
(∂V
∂T
)P−CV
= απT V +PαV
CP −CV =αV (πT +P )
prove later (2nd law) that(∂U
∂V
)T=πT = T
(∂P
∂T
)V−P
so
CP −CV =αV T
(∂P
∂T
)V(
∂P
∂T
)V=−
(∂P
∂V
)T
(∂V
∂T
)P
Eq. (2)
PChem I 2.43
(∂P
∂T
)V=−
(∂P
∂V
)T·Vα(
∂P
∂T
)V=− Vα(
∂V∂P
)T
Eq. (1)
(∂P
∂T
)V= Vα
V κT= α
κT
Further Information 2:
dV =(∂V
∂T
)P
dT +(∂V
∂P
)T
dP
dU =CV dT +πT dV
dU =CV dT +πT
[(∂V
∂T
)P
dT +(∂V
∂P
)T
dP
]dU =
[CV +πT
(∂V
∂T
)P
]dT +πT
(∂V
∂P
)T
dP
dU =(∂U
∂T
)P
dT +(∂U
∂P
)T
dP
compare coefficients
PChem I 2.44
(∂U
∂T
)P=CV +πT
(∂V
∂T
)P(
∂U
∂T
)P=CV +απT V
Further Information 3
dH =(∂H
∂T
)P
dT +(∂H
∂P
)T
dP
=CP dT +(∂H
∂P
)T
dP
P = P (T,V )
dH =CpdT +(∂H
∂P
)T
[(∂P
∂T
)V
dT +(∂P
∂V
)T
dV
]dH =
[Cp +
(∂H
∂P
)T
(∂P
∂T
)V
]dT +
(∂H
∂P
)T
(∂P
∂V
)T
dV(∂H
∂T
)V=CP +
(∂H
∂P
)T
(∂P
∂T
)V(
∂H
∂T
)V=CP −µCP
α
κT
PChem I 2.45
(∂H
∂T
)V=
(1−αµ
κT
)CP
PChem I 2.46
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