05_01_The Second Law of Thermodynamics

7/29/2019 05_01_The Second Law of Thermodynamics

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1 2010 Imam Siswanto Kimia Fisika I

universitas airlangga

excellence with morality

KIMIA FISIKA ISecond Law of Thermodynamics


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Second Law of Thermodynamics

2010 Imam Siswanto Kimia Fisika I



H = -1000 kkal


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H = +700 kkal


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???

H = -300 kkal


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H = -3000000 kkal


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The first law of thermodynamics does not define the ease or extent ofconvertibility of one form of energy into another.

Formulation by Kelvin :

No process is possible in which the sole result is the absorption of

heat from a reservoir and its complete conversion into work


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Entropy

The second law of thermodynamics can also be expressed in term ofthe entropy :

The entropy of an isolated system increases in the course of a spontaneous

change :

0totS


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Entropy

The thermodynamic definition of entropy is based on theexpression :

For a measurable change between two states i andfthis

expression integrates to

revdqdST

f

rev

i

dqS

T


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Entropy Change in Isolated Systems

Isolated system :

a system and heat reservoir adiabatically insulated from theirsurrounding.

Suppose that the substance in the cylinder undergoes anisothermal and reversible expansion from volume V1 and V2.

The substance will absorb from the reservoir a quantity of heat qr,and the reservoir will lose a quantity of heat qr.

rs qS

T

rr qS

T

1 0s rS S S


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If we compress the substance now isothermally and reversibly fromV2 to V1

Further the total change in entropy for the complete cycle, S, is

' rsq

S

T

' rrq

S

T

2 ' ' 0s rS S S

1 2 0S S S


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If the process is irreversible

1 0rq qS

T T

2 ' ' 0s rS S S

0rq q

S T T


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The Second Law of Thermodynamics

If instead of finite changes we consider infinitesimal ones, it can besummarized that for any process occurring in an isolated system the entropychange dSi is given by

0idS

All process in nature tend to occur only with an increase in entropyand that the direction of change is always such as to lead to the

entropy increase


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Entropy Change for Systems Only

When we ignore the surroundings, the entropy of the system can be zero,positive, or negative.

The differential change in entropy :

Or

Form the first law : dq = dE + dW, where dWconsist inpdVas well as other typesof works, dw, then

0i rdS dS dS

/rdS dq T

0TdS dq


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Then we get

When only PVwork is involved dw = 0, we get

However, for reversible processp = P, dw = dw

m , we obtain :

And when dwm = 0,

'dq dE pdV dw

' 0TdS dE pdV dw

0TdS dE pdV

' 0mTdS dE PdV dw

0TdS dE PdV


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Variable T and V

See page 305

Dependence of Entropy on Variable of a System

V T

S SdS dT dV

T V

T T

E ST P

V V

v

V

C PdS dT dV

T T


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Variable T and P

See page 306


P T

S SdS dP dP

T P

1

T T

S HV

P T P

p

P

C VdS dT dP

T T


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Also on rearrangements we have

and


T V

E PT P

V T

T P

H VV T

P T


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Entropy Change in Ideal Gases

In the case of T and V, we have

Or if we deal with n moles of gas, it is more convenient to write Cv

as nCv

.

And for ideal gas, PV = nRT, finally

v

V

C PdS dT dV

T T

2 2

1 1

ln lnT V

S nCv nRT V


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In the case of T and P, we have

Or if we deal with n moles of gas, it is more convenient to write Cv

as nCv

.

And for ideal gas, PV = nRT, finally

p

P

C VdS dT dP

T T

2 2

1 1

ln lnpT P

S nC nRT P

i i i l


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Example

1. Calculate the entropy change involved in the isothermal reversibleexpansion of 5 moles of an ideal gas from a volume of 10 liters to avolume of 100 liters at 3000 K.

2. For a certain ideal gas Cp = 5/2 R cal mole-1 degree-1. Calculate thechange in entropy suffered by 3 moles of the gas on being heated from300 to 6000 K at

(a) Constant pressure(b) Constant volume

i it i l


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Entropy of Mixing for Ideal Gases

Suppose : n1, P10 + n2, P2

0 P1 and P2

For any constant temperature the change in entropy accompanying a changein pressure is given by:

1

1 1 0

1ln

P

S n R P 0

11

1

lnP

n RP

0

2

2 22

ln

P

S n R P

S f ii it i l


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Entropy of Mixing for Ideal Gases

And the total entropy of mixing is

But according to Daltons law : P1 = N1Pt and P2 = N2Pt

And for special instance where P10 = P2

0 = Pt,

1 2mS S S

0 0

1 21 2

1 2

ln lnP P

n R n R

P P

0 0

1 21 2

1 2ln lnm

t t

P P

S n R n RN P N P

1 1 2 2( ln ln )mS n R N n R N

S d L f Th d ini ersitas airlangga


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Entropy Change in Physical Transformations

The transition enumerated take place reversibly at constant temperature Tand pressure P and are accompanied by an absorption or evolution of DH calof heat for a given quantity of substance.

Therefore, for all such processes

Of necessity this equation is valid only when reversible conditions obtainduring the transformation, i.e., when equilibrium exists between the two

forms.

rq HST T

S d L f Th d iuniversitas airlangga


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Example

Find the entropy change for the transition :

H2O(l, 1 atm) = H2O(g, 1 atm) H 373.2 K = 9717 cal mole-1

S = H/T = 9717/373.2 eu = 26.04 eu



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Example

Find the entropy change for the transition :

H2O(l, 1 atm) = H2O(g, 0.1 atm)

S = S vaporization + S expansion

1

2

lnH P

RT P

9717 14.58log 30.62373.2 0.1

eu



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Kimia Fisika I



Entropy Change in Chemical Reactions

The entropy change accompanying a chemical reaction is defined as thedifference between the sum of the entropies of all the products and the sumof the entropies of all reactants.

aA + bB + . . . = cC + dD + . . .

The entropy change is given by

S = (cSc + dSD + . . . ) (aSA + bSB + . . . )

Or in another form :2

12 1

T

pT

dTS S C

T

2

1

lnp

TC

T

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05_01_The Second Law of Thermodynamics