References · Atkins, P.W., “Physical Chemistry”, Oxford University Press

References     Atkins, P.W., “Physical Chemistry”, Oxford University Press     Castellan, G.W., “Physical Chemistry”, Addison Wesley     Levine, I.R., “Physical Chemistry”, McGraw-Hill     Laidler & Meiser, “Physical Chemistry”, Houghton Mifflin Co.     Alberty, R.A. and Silbey, R., “Physical Chemistry”, Wiley

THERMODYNAMICS

definitionsdescriptions

no proofs

no violations

Events,Experiments,Observations

Laws of Thermodynamics:0th, 1st, 2nd, 3rd 

Applications, Verifications

abstract generalize

dynamics – changesthermo – heat

Note:      does not worry about rate of changes (kinetics) but the states before and after the change     not dealing with time

Classical Thermodynamics

Marcoscopic observables T, P, V, …

Statistical Thermodynamics

Microscopic details dipole moment, molecular size, shape

Joule’s experiment

T mgh adiabatic wall

(adiabatic process)

U (energy change) = W (work) = mgh

wh

Thermometer

w

T time interval of heatingU = q (heat)

Conclusion: work and heat has the same effect to system (internal energy change)

FIRST LAW: U = q + W*

Page 59

U: internal energy is a state function [ = Kinetic Energy (K.E.) + Potential Energy (P.E.) ] q: energy transfer by temp gradient W: force distance E-potential charge surface tension distance pressure volume

First Law: The internal energy of an isolated system is constant

Positive: heat flows into system work done onto systemNegative: heat flows out of system work done by system

Convention:

Pressure-volume Work

P1 = P2

V2V1

MM

d work F dlext M gdhpiston

weight

AAdh

P Adhext

P dVext work P V Vext 2 1

Mg h

If weight unknown, but only properties of systemare measured, how can we evaluate work?

PV1V2P

MM

Assume the process is slow and steady,

Pint = Pext

Free Expansion:

Free expansion occurs when the external pressure is zero, i.e. there is no opposing force

Reversible change: a change that can be reversed by an infinitesimal modification of a variable. Quasi equilibrium process: Pint = Pext + dP (takes a long time to complete)

infinitesimal at any time

quasi equilibrium process

ò dVPW extò dVPint

1 2P

V

Example: P1 = 200kPa = P2

V1 = 0.04m3 V2 = 0.1m3

W PdV1 21

2

P V V2 1

200 01 0 04 3kPa m. . 12kJ

what we have consider was isobaric expansion(constant pressure) other types of reversible expansion of a gas: isothermal, adiabatic

*

Page 64-66

Isothermal expansionremove sand slowlyat the same time maintain temperature by heating slowly

W PdVrev 1

2

nRTV

dV1

2

nRTdVV1

2

nRT

VV

ln 2

1

T

VV1 V2

P

area under curve

*

Page 65-66

Ex. V1 = 0.04m3 P1 = 200kPa V2 = 0.1m3

W nRTVVrev 1

2

1

ln

PV

VV1 1

2

1ln

200 0 04

010 04

3kPa m. ln.

. 7 33. kJ

PPVV

kPa21 1

2

80

Adiabatic Reversible Expansion

For this processPV = constant for ideal gas(proved later)

V1

P1 T1

V2

P2 T2

(slightly larger than 1)

CC

p

v

W PdVPV dV

V 1 1

1

2

1

2

PV V V1 1 21

11

1

P V PV2 2 1 1

1

Ex. V1 = 0.04m3, P1 = 200kPa, V2 = 0.1m3, = 1.3

P kPa2

1 3

2000 04010

60 77

.

..

.

W kJ

60 77 01 200 0 04

1 136 41

. . ..

.

volume change

|Wa|>|Wb|>|Wc|>|Wd|

W PdV 0

a

db

V

P200kPa

const

isothermal

adibatic

State Function Vs Path Function State function: depends only on position in

the x,y plane e.g.: height (elevation)

300200

100mB

AX

Y

1 2

Path Function: depends on which path is taken to reach destination

from 1 2, difference of 300m (state function)but path A will require more effort.Internal energy is a state function, heat and work are path functions

3

2481.13K

192.45K

P

200kPa

V/m30.04 0.1

1

5 moles of monoatomic gas

C R

U nC T R kJ

W kJQ U kJ kJW kJ kJQ kJ kJ kJ

V

V

32

32 289 5 18

1212 30

7 33 0 7 337 33 18 25 33

1 2

1 2

1 3 2

1 3 2

. .

. .

f x y z w, , , ,

Consider a 2 variable functionor f x y, z f x y , a surface in three

dimensional plot

C

C’

z or

yx

ycxc

Multivariable Calculus

At point C, there are two slopes orthogonal to each other in constant x or constant y direction

y

z z

x x=xC y=yc

C’C C

C’

the change in can be calculated as a sum of two parts: change in x direction and change in y direction

z z dx

dxy

dyC Cy xc c

partial derivative partial derivative w.r.t. x w.r.t. y

Joule’s second experiment

Energy UU(V,T) ???

V1V2

thermometer

At time zero, open valve

adiabatic wall

After time zero, V1 V1+V2

T=0 Q=0 W=0 no Pext U=0

thermometer

No temperature change

adiabatic wall

dTCdVVUdT

TUdU

VU

VTV

T

0

0 (for ideal gas)CV is constant volume heat capacity

U=U(T) Energy is only a function of temperature for ideal gas*

Page 99

Kinetic Model for GasesQualitatively:• Gases consists of

spheres of negligible size, far apart from one other.

• Particles in ceaseless random motion; no interactions except collisions

U

V

T

U

V

ConstantTemperature

VV T

UC

T

U

Energy is a function of Volume and temperature for real gases

Interaction among molecules

*

*Enthalpy

Define H = U + PV   state function intensive variables locating the state

Enthalpy is also a state function

H = U + PV + VP

Page 101

PPTP

CCdPPHdT

THdH

PTHH

0

,

for ideal gases C C RCC

P V

P

V

(proved later)

At constant pressureH = U+PV = U - W = QH = QP constant pressure heating

H is expressed as a functional of T and P

Thermochemistry Heat transferred at constant volume qV = U

Heat transferred at constant pessure qP = H

Exothermic H = -ve

Endothermic H = +ve

Standard states, standard conditions do not measure energies and enthalpies absolutely but only the differences, U or H

The choice of standard state is purely a matter of convenience

Analogy – differences in altitudes between 100 points and their elevation with respect to sea level

The standard states of a substance at a specified temperature is its pure form at 1 bar

What is the standard state ?*

25oC, 1 bar: the most stable forms of elements assign “zero enthalpy”

Ho298 = 0 used for chemical reactions

Standard enthalpy of formation

Standard enthalpy change for the formationof the compound from its elements in their reference states.Reference state of an element is its most stablestate at the specified temperature & 1 bar

C (s) + 2H2 (g) CH4 (g) Hfo = -75 kJ

289K, 1 atm

From the definition, Hfo for elements 0

Hess’s Law

The standard enthalpy of an overall reaction is the sum of thestandard enthalpies of the individual reactions into which areaction may be divided.

Standard reaction enthalpy is the change in enthalpy when the reactants in their standard states change to products in their standard states.

Hess’ Law

R X Y PH H H 2 3 4

H1

H1 = H2 + H3 + H4 state function

Hess’s law is a simple application of the first law of thermodynamics

e.g. C (s) + 2H2 (g) CH4 (g) H1o = ?

298K, 1 atm

C (s,graphite) + O2 (g) CO2 (g) Ho = -393.7 kJ

H2 (g) + ½O2(g) H2O (l) Ho = -285.8 kJ

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) Ho = -890.4 kJ

H1o = -393.7 + 2(-285.8) - (-890.4)

= -75 kJ/mole

Heat of Reaction (Enthalpy of Reaction) Enthalpy change in a reaction, which may be obtained from Hf

o of products and reactantsReactants Products

products tsreaac

oreactionfR

oprodfp

or HnHnH

tan,,

i fo

iproductsreac ts

Htan

I stoichiometric coefficient, + ve products, - ve reactantsE.g. CH4 (g) + Cl2 (g) CH3Cl (g) + HCl (g)

Hfo/ kJ

CH3Cl -83.7

HCl -92.0

Cl2 0

CH4 -75.3Hr

o = (-83.7-92.0) - (-75.3+0) = -100.4 kJ

Reactants Products

elements elements n HR f R

o , n Hp f po ,

jfo

jo

r HH ,

2A + B 3C + D

0 = 3C + D - 2A - B

Generally,0 = J J J

J denotes substances, J are the stoichiometric numbers

*

Page 83

Bond energy (enthalpy)Assumption – the strength of the bond is independent of the molecular environment in which the atom pair may occur.

C (s,graphite) + 2H2 (g) CH4(g)Ho = -75.4 kJH2 (g) 2H (g) Ho = 435.3 kJC (s,graphite) C (g) Ho= 715.8 kJ C (s,graphite) + 2H2 (g) C (g) + 4H (g) Ho = 2(435.3)+715.8 = 1586.5 kJ C (g) + 4H (g) CH4 (g) H = -75.4-1586.5

= -1661.9 kJCH Bond enthalpy = 1661.9/4 kJ = 415.5 kJ

Temperature dependence of Hr

productsreactantsreactants

reaction,products

products, ipiRippipP

T

Tp

or

Tr

CnCnCC

dTCHHo

CP,R

CP,P

Hro

P

R Hr

T

298 T

What is the enthalpy change for vaporization (enthalpy of vaporization) of water at 0oC?

H2O (l) H2O (g) Ho = -241.93 - (-286.1) = 44.01 kJmol-1 H2 = Ho + CP(T2-T1) assume CP,i

constant wrt T

H (273) = Ho(298) + CP(H2O,g) - CP (H2O,l)(273-298) = 44.10 - (33.59-75.33)(-25) = 43.0 kJ/mole

For ideal Gases:

HP

HT

C C RT P

P V

0,

dUUV

dVUT

dT

dHHP

dPHT

dT

T V

T P

dH dU PdV VdP

UV

dVUT

dT VdP PVP

dPVT

dT

UV

dVUV

VP

dPVT

dT

dHUT

UV

V

T V T P

T T T P

V T

,

TP

VT

dT V PVP

UV

VP

dP

HT

UT

PUV

VT

C R

P P T T T

P V T PV

0 = R/P for ideal gas

HP

V PUV

VP

V PRTP

V VT T T

2 0

0 for ideal gas

Prove PV = constant for adiabatic reversible expansion of an ideal gas

dVVUdTCdU

TV

0 for ideal gas

VCR

V

V

V

V

VV

TT

VV

CR

TT

VRdTdCVdVR

TdTC

dVVRTPdVdTC

QWQdU

1

2

1

2

1

2

1

2

0

lnln

lnln

,Adiabatic expansion

1122

2

1

1

2

1

1

2

1

2

11

22

VPVP

VV

VV

PP

VV

VPVP

V

V

CR

CR

Reversible vs IrreversibleNon-spontaneous changes vs Spontaneous

changesReversibility vs Spontaneity

First law does not predict direction of changes,cannot tell which process is spontaneous. Only U = Q + W

Second Law of Thermodynamics

•Origin of the driving force of physical and chemical change

•The driving force: Entropy

•Application of Entropy: • Heat Engines & Refrigerators• Spontaneous Chemical Reactions

•Free Energy

Second Law of Thermodynamics

No process is possible in which the sole result is the absorptionof heat from a reservoir and its complete conversion into work

Hot Reservoirq

w

Engine

* Page 120

Direction of Spontaneous Change

More Chaotic !!!

Entropy (S) is a measurement of the randomness of the system, and is a state function!

S Q S 1 / T

Spontaneous change is usually accompanied by a dispersal of energy into a disorder form, and its consequence is equivalent to heating

Expansion into a vacuum

irreversible reversible

Vac

V1

V2V1

V2

W P dV

nRTVV

q W

rev ext

rev rev

ln 2

1

W P dV

U Q

V V W nRTVV

irr ext

irr

0

0

2 12

1

, lnmin

qrev > qirr

-Wrev > -Wirr Wirr = 0 = qirr = U

TdqdS rev

Entropy S

For a reversible process, the change of entropy is defined as

Another expression of the Second Law:

The entropy of an isolated systems increases in the course of aspontaneous change:

Stot > 0

where Stot is the total entropy of the isolated system

(thermodynamic definition of the entropy)*

Page 122

Entropy S

The entropy of an isolated systems increases in the course of areversible change:

Stot = 0

where Stot is the total entropy of the isolated system

V1

V2

revrev

extrev

revrev

WqVVnRT

dVPW

VVnR

TqdqS T

1

2

1

21

ln

ln

Entropy Change for an isothermal expansion of a perfect gas

Entropy is a state function

Depend only on V

Carnot’s Theoretical Heat Engine

Heat flows from a high temperature reservoirto a low temperature body. The heat can be utilized to generate work.

e.g. steam engine.

For the cycle,-Wnet = -(W1+W2+W3+W4)

= qnet = q1+q3

Ucycle=0

q1 positiveq3 negativeqnet positive

|q3| < |q1|

Consider the sequence of reversible processes

Net effect of the gas going through a cycle

|q1| qH

|q3| qL

|Wnet| W

Efficiency of theoretical heat engine

1

3

1

3

1

31

1

11qq

qq

qqq

qW

qW

heatavailableW net

H

netnet

Carnot theorem: Engines operating between two temperature TH, TL have the same efficiency

H

Lth

A

B

C

D

A

B

D

C

A

B

A

D

B

C

C

D

A

B

TT

TT

VVnRT

VVnRT

VVnRT

qqq

VV

VV

VV

TT

VV

TT

UqVVnRTW

UqVVnRTW

1

1

0

0

1

3

1

31

1

31

1

1

3

1

1

3

333

111

ln

lnln

,

,ln

,ln

limiting thermal efficiency of a heat engineIn actual cases, heat engine have much lowerefficient. irreversible processes, friction losses, etc.. Kelvin: It is impossible, by means of a cycle to

take heat from q reservoir and convent it to work without at the same time transferring heat from a hot to cold reservoir. (We cannot have a 100% efficient heat engine)

Show that 2nd law and Kelvin’s principle areequivalent

Examine an adiabatic irreversible process. We want to evaluate the entropy change for the process by an reversible path BCDA

D C B C adiabatic

compression C D isothermal A compression B D A adiabatic Q=0 expansion

Wnet

irreversible

1st law for the cycle, U = 0 0-(WAB + WBA) = QBA + QAB

-Wnet = QBA

By Kelvin’s principle, -Wnet must be negative,otherwise a 100% efficient heat engine QBA = -Wnet must be negative QBA 0

S

QT

rev

QT

A BA B

B A

H

0

Evaluation of Entropy Changesisothermal expansion:TdS dq dU PdV

Sdq

TPdV

TnRdV

VnR

VV

nRPP

rev

rev

ln ln2

1

1

2

Tm: melting pt., Tb: boiling pt.,If CP = constant,

S CTTPln 2

1

H/S VL

S

Tm Tb T2

isobaric heating:

2

1

T

T

gasp

b

vT

T

liqP

m

mT

T

solidP

b

b

m

m

dTT

CTHdT

TC

THdT

TC

S ,,,

T1

change of temperature and volume (pressure)

TdS dU PdVUT

dT PdV nC dT PdV

S nCT

dTPT

dV

nCTT

nRVV

nCTT

nRPP

VV

V

V

P

1

2

1

2

2

1

2

1

2

1

1

2

ln ln

ln ln 1

2P

V

The efficiencies of heat engines

Hot Reservoirq

w

Engine

S = - |q|/Th < 0 not possible! contrary to the second law*

Page 142

The efficiencies of heat engines

Hot Reservoirqh

w

Cold Reservoir

qc

Sh = - |qh|/Th

Sc = + |qc|/Tc

S = - |qh|/Th + |qc|/Tc 0

|wmax| = |qh|- |qc,min| = (1- Tc /Th) |qh|

Maximum efficiency:rev= |wmax|/|qh|= 1- Tc /Th

rev 1 as Tc 0 or Th

Page 130

Energetics of Refrigeration

Hot Reservoir

Cold Reservoir

Sh = + |qc|/Th

Sc = - |qc|/Tc

S = - |qc|/Tc + |qc|/Th < 0

not possible!|qc|

Page 144

The energetics of refrigeration

Hot Reservoirqh

w

Cold Reservoir

qc

Sh = + |qh|/Th

Sc = - |qc|/Tc

S = |qh|/Th - |qc|/Tc 0

|wmin| = |qh,min|- |qc,| = (Th /Tc-1) |qc|

Maximum efficiency of performance:crev= |qc|/ |wmin| = Tc /(Th- Tc)

qh

Page 144

The energetics of refrigeration

Hot Reservoirqh

w

Cold Reservoir

qc

Sh = + |qh|/Th

Sc = - |qc|/Tc

How to keep it cool?dqc/dt = A(Th -Tc)

d|w|/dt = (1/ crev) dqc/dt = A (Th -Tc)2 / Tc

qh

Page 145

The Nernst Theorem

The entropy change accompanying any physical or chemical transformation approaches zero as the temperature approaches zero.

S 0 as T0

Third Law of Thermodynamics

If the entropy of each element in its most state is taken as zero at the absolute zero of temperature, every substance has a positive entropy. But at 0K, the entropy of substance may equals to 0, and does become zero in perfect crystalline solids.

Crystalline form: complete ordered, minimum entropy

Implication: all perfect materials have the same entropy (S=0) at absolute zero temperature

* Page 147

Statistical Interpretation of S

S = 0 at 0K for perfect crystals S = k ln

  Boltzmann number of arrangements postulate of entropy Boltzmann constant

Entropy Change of Mixing

one distinguish arrangement

S k S kA B ln!!

, ln!!

44

044

0

A B

A B

A B

A B

N NN N

S k

!! !

ln

8 7 6 54 3 2 1

70

70

number of arrangement increased

In general mixing NA, NB

!!

!lnBA

BAmix NN

NNkS

Entropy change of mixingStirling’s approximation: ln N! N ln N + 0(N)

for large N

0

BBAABA

BBAABABA

BBAABABAmix

XXXXNNkNXNXNNNNk

NNNNNNNNkS

lnlnlnlnln

lnlnln

From classical thermodynamics, isothermal reversible expansion of gases A & B

0

BBAAT

A

BAB

A

BAAmix

B

BAB

B

rev

A

BAA

A

BAA

A

rev

A

rev

XXXXRn

VVVRn

VVVRnS

VVVRn

Tq

VVVRn

VVV

TRTn

TW

Tq

lnln

lnln

ln

ln

ln

Assignment (due on 06/12/1999)

2.4, 2.5, 2.37, 3.5, 3.23, 4.10, 4.29

Extensions of 2nd Law 

TdS dq Clausius Inequality

For adiabatic process,

TdS or dS 0 0

Entropy will always attain maximum in adiabatic processes.

A similar function for other processes?

*Page 133

Define Helmholtz free energyA = U - TS Thermodynamic

State Function dA = dU - TdS - SdTSubstitute into Clausius Inequality

000

dq TdSdq dA dU SdTPdV dA SdT

for isothermal, isochoric (constant volume)process, 0dA

*

*

Page 149

A will tend to a minimum value

isothermal, isochoric process,dA 0dA 0

equilibriumspontaneous

If only isothermal, 0

dA PdVPdV dA

W dAisothermal reversible expansion

W RTVV

A ln 2

1

*

change in Helmholtz free energy = maximum isothermal work

Example of isothermal, isochoric process: combustion in a bomb calorimeter

O2 +fuel

Temp. bath

O2, CO2,H2O

Higher P heat givenout

Gibbs Free Energy

Define Gibbs free energy G = H - TS Thermodynamic

state function = U + PV -TS dG = dU +PdV + VdP -TdS -SdT

substitute into Clausius inequality

SdTVdPdGSdTVdPdGTdSPdVdU

TdSdq

000

* Page 149-155

constant pressure, constant temperature

dG0

G will tend to a minimum value equilibrium spontaneous change

dG 0dG 0

More applications since most processes areisothermal, isobaric

chemical reactions at constant T, PReactants Productsendothermic H is positiveexothermic H is negative

time

G

*

*Page 154

H is not a criteria for spontaneityS only isolated system G = H - TS

RP

G

rxn. Co-ord.

S H G    

+ ve - ve - ve spontaneous dissociation of unstable compounds

- ve + ve + ve non-spontaneous

forming unstable compounds

+ ve + ve ? ? dissociation of a strong compound

- ve - ve ? ? recombination reactione.g. H + H H2

Change of Gibbs free energy with temperature(constant pressure)

STG

SdTGG

VdPSdTdG

P

12

0

oo

oo

PfRTGG

PPRTGG

ffRT

PPRT

VdPGG

SdTVdPdG

ln

ln

ln

ln

1

2

1

2

12

0

Ideal gas

Real gas fugacity

Change of Gibbs free energy with pressure(constant temperature)

Ideal gas

Real gas

Gibbs free energy

oio

oio

i

PfRTG

PpRTGG

ln

ln Ideal Gases

Real Gases

Total Gibbs free energy of a mixture of gases

BBAAiBBiAA

oB

BoBBo

AA

oAA

BBAA

XRTXXRTXGXGXn

PpRTnGn

PpRTnGn

GnGnG

lnln

lnln

,,

÷øöçèæ

÷øöçèæ

*

Page 168-174

0

mix

iiimixmix

iii

oii

H

XRTXSTG

XRTXGXn

ln

ln

for ideal gases

Chemical Equilibria

aA + bB cC + dDconsider G for a pass of the reaction at constant T & P

G = cGC + dGD - aGA - bGB

molar Gibbs free energy

ioii fRTGG ln

pure state fugacity

ii

oii

bB

aA

dD

cCo

BADCoB

oA

oD

oC

fRTG

ffffRTG

fbRTfaRTfdRTfcRTbGaGdGcGG

ln

ln

lnlnlnln

At equilibrium, G = 0

Kffff

RTG

bB

aA

dD

cC

o

lnln

equilibrium constant

fi in the unit of bar

0 = JJ J

*

for ideal gaseous mixture fi = pi

nxb

Ta

T

cT

dT

bB

aA

cC

dD

p

T

ii

nc

na

Ab

B

cC

dD

p

ii

i

p

p

o

bB

aA

dD

cC

PKPPPP

XXXXK

ppX

RTKRTccccK

RTcVRTnp

constK

constK

RTG

pppp

÷÷øöççèæ.

.ln

ln

ci = concentration

varies withtotal pressure

n = c+d-a-b

For liquids, use activity

GRT

a aa a

Ko

Cc

Dd

Aa

Bb eqln ln

Gibbs - Helmholtz Equation

G VdP SdTGT

SP

Also, G H TS H TGT P

GT

HT

GT

GTT

GT T

GT

P

P

P

2

1

GTT

HT

orG

T

TH

P

P

2

1

Gibbs - Helmholtz Equation

For a reaction

GT

TH

T

GT

TH

P

P

2

1

Similarly,

AT

TU

P1

1/T

G/TSlope=H

G2

H

G1

1/T

G/T

Change of equilibrium constant with temperature

G RT K

GT

R K

To

To

ln

ln

Gibbs Helmholtz equation

GT

THT

RK

THT

O

To

To

2

2

ln

ln KT

HRT

To

2Van’t Hoff Equation

If constant (i.e. HP -HR is constant)Hrxno

ln tanKHRT

cons to

endothermic, H + ve, K with Texothermic, H - ve, K with T

Relation between Thermodynamic functions

dU = TdS-PdV 1st lawdH = TdS+VdP, H = U+PVdA = -SdT-PdV, A = U-TSdG = -SdT+VdP, G = H-TS

From multivariable differential calculus dz = M dx + N dy total differential, i.e.

z depends on x & y

My

Nx

x y

*

Page 149

Maxwell Relations

TV

PS

TP

VS

SV

PT

SP

VT

S V S P

T V T P

,

,

Phase equilibrium

Clapeyron equationmolar Gibbs free energy

G=0G=G-G

Page 163

G G

dG dG

V dP S dT V dP S dT

dPdT

S SV V

SV

HT V

P

T

liq

vaps

Clausius- Clapeyron equationFor vaporization and sublimation

dPdT

HT V

HTV

P HRT

dPP

d PH

RTdT

PH

RTconst

PP

H T TRT T

vap

vap

vap

vap

vap

vap

sat vapsat

vap

2

2

2

1

2 1

1 2

ln

ln .

ln1/T

In P

Example: What is the change in the boiling point of water at 100oC per torr change in atmospheric pressure?

 Hvap = 9725 cal mol-1

Vliq = 0.019 l mol-1

Vvap = 30.199 l mol-1

dPdT

H

T V V

cal mol l atmcal

K l mol

atm K

torr KdTdP

K torr

sat

satvap

v l

9725 0 04129

37315 30180

0 03566

27 10

0 0369

1 1

1

1

1

1

.

. .

.

.

.

Example: Calculate the change in pressurerequired to change the freezing point of water 1oC.

 At 0oC, the heat of the fusion of ice is 79.7cal g-1,the density of water is 0.9998 g cm-3 and that of ice is 0.9168 g

V V l g

PT

H

T V V

cal g l atmcal

K l gatm K

l s

sat

satfus

l s

10 9998

10 9168

9 06 10

79 7 0 04129

2731 9 06 10133

5 1

1 1

5 11

. ..

. .

. .

liq

vapsolid

P

T

50kg

skate blade

iceliq. H2O

Example 2.1A certain electric motor produced 15 kJ of energy each second as mechanical work and lost 2 kJ as heat to the surroundings. What was the change in the internal energy of the motor and its power supply each second?

Example 2.2Calculate the work done when 50 g of iron reacts with hydrochloricacid in: (a) a closed vessel of fixed volume; (b) an open beaker at25oC.

Example 2.3The internal energy change when 1.0 mole CaCO3 in the form ofcalcite converts to aragonite is 0.21 kJ. Calculate the difference between the enthalpy change and the change in internal energy.

Example 2.4The enthalpy change accompanying the formation of 1.00 mole NH3(g) from its elements at 298 K is -46.1 kJ. Estimate the change in internal energy.

Example 2.5Water is heated to boiling under pressure of 1.0 atm. When an electric current of 0.50 A from 12-V supply is passed for 300 sthrough a resistance in thermal contact with it, it is found that 0.798 g of water is vaporized. Calculate the molar internal energyand enthalpy changes at the boiling point (373.15 K).

Exercise 2.6At very low temperatures the heat capacity of a solid is proportionalto T3, and we can write Cv=aT3. What is the change in enthalpy ofsuch a substance when it is heated from 0 to T?

Example 3.6A sample of argon at 1.0 atm pressure and 25oC expands reversiblyand adiabatically from 0.50 L to 1.00 L. Calculate its final tempera-ture, the work done during the expansion, and the change in internalenergy. The molar heat capacity of argon at constant volume is 12.48 JK-1 mol-1.

Example 4.1Calculate the entropy change in the surroundings when 1.00 molH2O(l) is formed from its elements under standard conditions at298.15 K.

Example 4.4Calculate the entropy change when argon at 25oC and 1.00 atm in a container of volume 500 cm3 is allowed to expand to 1000 cm3

and is simultaneously heated to 100oC.

Example 5.4The pressure deep inside the Earth is probably greater than 3×103 kbar,and the temperature is around 4×103 oC. Estimate the change in Gon going from crust to core for a process in which V=1.0 cm3 mol-1

and S = 2.1 JK-1 mol-1.

Example 5.5Calculate the change in molar Gibbs energy when water vaporizes at1 bar and 25 oC. Note that the molar Gibbs energies of formation are-237.13 and -228.57 kJ mol-1 for water and its vapor, respectively.

Example 5.6Suppose that the attractive interactions between gas particles can be neglected and find an expression for the fugacity of a van der Waals gas in terms of the pressure.