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First Law of Thermodynamics

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(1)

First Law of Thermodynamics

(2)

1.1 Work and Heat

Changes in the state

spontaneous non-spontaneous

Isothermal Isobaric isochoric

Adiabatic Diabatic

irreversible reversible

Processes

(3)

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 × = =

×

(4)

Change in State Accompanying Work

ext

d

dw = − P V

2

1 ext

d

w = − ∫ P V w = − ∫

12

P 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

(5)

Joule’s Experiment

James Prescott Joule (1818-1889)

(6)

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

(7)

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

(8)

2.3 Exact and Inexact Differentials

The differential of a state function is exact b

d

b a

a

U = UU = ∆ U

The differential of a path function is inexact b b a

a

dw = ≠ w ww

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.

(9)

2.4 Isothermal Compression of a Gas

2

p mg

= A w = mgh = − P V

2

(

2

V

1

)

2

1

d

V

d

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

V

d

V

d

V

d 0

V V V V

w = − ∫ P V − ∫ P V = − ∫ P V + ∫ P V =

consists of a series of successive equilibria.

(10)

Isothermal Expansion of a Gas

2

p mg

= A w = mgh = − P V

2

(

2

V

1

)

2

1

d

V

d

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

(11)

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 2

1

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.

(12)

2.5 Various Kinds of Work

Surface work

Elongation work

Electrical work

d w = γ d A

S

d 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

S

w = ∆ = f x L x ∆ = ∆ γ γ A 2

f = L γ

(13)

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 qP

ext

d V

d d

ext

d 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

∆ = − = ∆

(14)

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.

(15)

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

(16)

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

(17)

2.7 Enthalpy and Change of State at Constant Pressure

P P

U q w q P V

∆ = + = − ∆

2 1 P

(

2 1

)

UU = qP VV

2 2 1 1

( ) ( )

q

P

= U + PVU + PV

H ≡ + U PV

Define a new state function, the

enthalpy H

2 1

q

P

= HH

and

d q

P

= d H ( , )

H T P d d d

P T

H H

H T P

T P

∂ ∂

   

=   ∂   +   ∂  

d

P

d

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

 ∂ 

≡ =   ∂  

(18)

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 – )

(19)

Heat Capacities

d d

ext

d d

V T V

U U U

q T P V T

T V T

 

∂ ∂ ∂

     

=   ∂   +   +   ∂     =   ∂  

d

P V

d 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

(20)

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

UU = = w PVP 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)

(21)

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

JT

2 1

JT 0 2 1

lim

P H

T T T

P P P

µ

∆ →

−  ∂ 

= − =   ∂  

(22)

2.10 Adiabatic Processes

No heat is gained or lost,

q = 0 d U = d w = − P

ext

d V

2 2

1

d

2 1 1 ext

d

U V

U V

U U U U w P V

∆ = ∫ = − = = − ∫

For an ideal gas,

dU = C

V

dT

2 2

1

d

2 1 1

d (

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

V

d 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.

(23)

Isothermal and Reversible Adiabatic Expansions

Isothermal and reversible adiabatic expansions of one mole of an ideal monatomic gas

P V

CC = R

P

V

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

γ

Adiabatic expansion

results in a smaller volume than isothermal expansion

(24)

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 that

n

i0 is the initial amount of substance

i

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)

(25)

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

(26)

2.12 Enthalpy of Formation

f

H ° 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= 2fH (CO) –fH (CO2) = 2(– 110.527 kJ/mol) – ( –393.533 kJ/mol) = 172.468 kJ/mol

(27)

Enthalpy of Formation for Carbon Allotropes

C(graphite) → C(diamond)H

fo

= 1.9 kJ·mol

–1

C(graphite) → C(buckminsterfulleren)H

fo

= 34 kJ·mol

–1

(28)

Enthalpy of Reactions

reactants

reactants

products

products

r

H

T

r

H

298

298

,reactants

d

T

C

P

T

∫ ∫

298T

C

P,procucts

d T

298

r T P,reactants

d

r 298 298T P,products

d

H

T

C T H C T

= ∫ + ∆

+ ∫

r

H

T r

H

298 298T

( C

P,products

C

P,reactants

)d T

r

H

298 298T r

C T

P

d

= ∆

+ ∫

= ∆

+ ∫ ∆

r P i P i,

i

C ν C

= ∑

298 298T

d

T P

H

H

= ∫ C

T

(29)

2.13 Calorimetry

A

( )

1 P

0

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

∆ = −∆ = − + −

(30)

Adiabatic Calorimeters

jacket water electrodes sample bomb

(31)

Calorimetry

Adiabatic bomb calorimeter for carrying out combustion at constant volume

U q

V

∆ =

r

H

r

U 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

ν

(32)

The 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) + efH = 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)

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