THERMODYNAMICS

dynamics – changesthermo – heat

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

Note: Thermodynamics ……..

     is not concerned about rate of changes (kinetics) but the states before and after the change

     does not deal 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

w

h

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 expansion

remove sand slowlyat the same time maintain temperature by heating slowly

W PdVrev 1

2

nRT

VdV

1

2

nRTdV

V1

2

nRT

V

Vln 2

1

T

VV1 V2

P

area under curve

*

Page 65-66

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

W nRTV

Vrev 1

2

1

ln

PV

V

V1 12

1

ln

200 0 04

01

0 043kPa m. ln

.

.kJ 7.33

PPV

VkPa2

1 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)

C

Cp

v

W PdVPV dV

V 1 1

1

2

1

2

PV V V1 1 21

11

1

PV PV2 2 1 1

1

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

= 1.3

P kPa2

1 3

2000 04

01060 77

.

..

.

W kJ

60 77 01 200 0 04

1 136 41

. . .

..

volume change

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

W PdV 0

a

db

V

P

200kPaconst

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 kJ

Q U kJ kJ

W kJ kJ

Q kJ kJ kJ

V

V

3

23

2 289 5 18

12

12 30

7 33 0 7 33

7 33 18 25 33

1 2

1 2

1 3 2

1 3 2

. .

. .

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

dTCdVV

UdT

T

UdU

V

U

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*

*Enthalpy

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

Enthalpy is also a state function

H = U + PV + VP

PPTP

CCdPP

HdT

T

HdH

PTHH

0

,

for ideal gases C C R

C

C

P V

P

V

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 = -

Endothermic H = +

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

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

What is the standard state ?

25oC, 1 atmosphere: 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 atmosphere

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 H1

o = -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 reactants

Reactants Products

products tsreaac

oreactionfR

oprodfp

or HnHnH

tan,,

i foi

productsreac ts

Htan

I stoichiometric coefficient, + products,

- reactantsE.g. CH4 (g) + Cl2 (g) CH3Cl (g) + HCl (g)

Hfo/ kJ

CH3Cl -83.7

HCl -92.0

Cl2 0

CH4 -75.3

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

Reactants Products

elements elements

n HR f Ro , n Hp f p

o ,

jfo

jor 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 kJ

H2 (g) 2H (g) Ho = 435.3 kJ

C (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

T

por

Tr

CnCnCC

dTCHHo

CP,R

CP,P

Hro

P

R

HrT

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

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 Reservoir

q

w

Engine

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

T

dqdS 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

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.

The efficiencies of heat engines

Hot Reservoir

q

w

Engine

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

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

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

!

!

4

40

4

40

A B

A B

A B

A B

N N

N N

S k

!

! !

ln

8 7 6 5

4 3 2 170

70

number of arrangement increased

In general mixing NA, NB

!!

!ln

BA

BAmix NN

NNkS

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

for large N

0

BBAABA

BBAABABA

BBAABABAmix

XXXXNNk

NXNXNNNNk

NNNNNNNNkS

lnln

lnlnln

lnlnln

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?

Define Helmholtz free energyA = U - TS Thermodynamic

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

0

0

0

dq TdS

dq dA dU SdT

PdV dA SdTfor isothermal, isochoric (constant volume)process,

0dA

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

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

Change of Gibbs free energy with temperature(constant pressure)

ST

G

SdTGG

VdPSdTdG

P

12

0

dP

dT

H

T V

H

TV

P H

RT

dP

Pd P

H

RTdT

PH

RTconst

P

P

H T T

RT T

vap

vap

vap

vap

vap

vap

sat vap

sat

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

dP

dT

H

T V V

calmol l atmcal

K lmol

atmK

torr K

dT

dPK torr

sat

sat

vap

v l

9725 0 04129

37315 30180

0 03566

2710

0 0369

1 1

1

1

1

1

.

. .

.

.

.