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