Kombinasi Hukum I Dan II Termodinamika - En

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THE COMBINATION OF FIRST AND SECOND LAW OF

THERMODYNAMIC

A. A Combination of First And Second Law of Thermodynamics

Analytic formulation of the thermodynamics law is:

W d qd dU −= ......................................................................(1)

and analytic formulation of Thermodynamics Law II is:

dS T qd rev = ..........................................................................(2)

If both of them first and second law combined then obtained equations as follows.

W d qd dU −=

W d dS T dU −= ...................................................................(3)

It is known that dV pW d = so:

W d dS T dU −=

dV pdS T dU −= .................................................................(4)

By using equations (4), then the relationships of other thermodynamic can be

determined by taking a pair of coordinates of thermodynamics as a free variable. It

is well known that the U and S is a function of the system state can be expressed by

two coordinates thermodynamic anywhere. For example the U and S are expressed as

a function of T and V then mathematically can be written: ),( V T f U = and

),( V T f S = , then the total differential is:

dV V

U dT

T

U dU

T V

∂∂

+

∂∂

= ................................................(5)

dV V

S dT

T

S dS

T V

∂

∂+

∂

∂= ..................................................(6)

If the equations (5) substitute into equations the combination of law I and II

of thermodynamics (equation (4)) then retrieved:

pdV TdS dU −=

pdV dU TdS +=

pdV dV V

U dT

T

U TdS

T V

+

∂∂

+

∂∂

=

Combination of First And Second Law of Thermodynamic

By Tenth Group



dV V

U pdT

T

U TdS

T V

∂∂

++

∂∂

=

dV V

U

pT dT T

U

T dS T V

∂

∂

++

∂

∂

=

11

So,

V V T

U

T T

S

∂∂

=

∂∂ 1

...............................................................(7)

∂∂

+=

∂∂

T T V

U p

T V

S 1.......................................................(8)

So that equation (7) and (8) can be expressed with quantities measured, can be done

by applying a mathematical concept i.e. If z is a function of x and y. then:

∂∂

∂∂

=

∂∂

∂∂

x

z

y y

Z

x

or x y

z

y x

z

∂∂

∂=

∂∂

∂22

so:

V T T V V

S

T T

S

V

∂∂

∂∂

=

∂∂

∂∂

..............................................(9)

By substituting equation (7) and (8) into equations (9) then retrieved the equation as

follows:

V T T V V

U p

T T T

U

T V

∂∂

+∂∂

=

∂∂

∂∂ 11

∂∂

+−

∂∂

∂+

∂∂

=∂∂

∂

T V V

U p

T V T

U

T

p

T T V

U

T 2

22 111

V T T

p

T V

U p

T

∂∂

=

∂∂

+11

2

V T T

pT V

U p

∂∂=

∂∂+ .........................................................(10)

pT

pT

V

U

V T

−

∂∂

=

∂∂


By Tenth Group



Using the relations:1−=

∂∂

∂∂

∂∂

V pT p

T

T

V

V

p

or pT

V

T

V

V

p p

T

∂∂

∂∂

−=

∂∂ 1

or

pT V T

V

V

p

T

p

∂∂

∂∂

−=

∂∂

then the retrieved:

pT

V

V

pT

V

U

pT T

−

∂∂

∂∂

−=

∂∂

p

p

V

T

V

T V

U

T

p

T

−

∂

∂−

∂∂

=

∂∂

where V T

V

p

β =

∂∂

and V p

V

T

κ =

∂∂

− , then the retrieved:

pT

V

U

T

−=

∂∂

κ

β .................................................................(11)

If this equation applied to the ideal gas system, whereT

1=β

and p

1=κ then the

retrieved:

p

p

T T

V

U

T

−=

∂∂

1

1

0=

∂∂

T V

U

So, the energy in an ideal gas is independent of the system volume. Previously it was

known that the: pT V p T

V

pV

U

C C

∂

∂

+

∂

∂

=− . By substituting it with the equation

(10) in the equation is then derived the equation:

pT

V pT

V p

V

U C C

∂∂

+

∂∂

=−

pV

V pT

V

T

pT C C

∂∂

∂∂

=−


By Tenth Group



Using the relations1−=

∂∂

∂∂

∂∂

V pT p

T

T

V

V

p

or pT

V

T

V

V

p p

T

∂∂

∂∂

−=

∂∂ 1

or

pT V T

V

V

p

T

p

∂∂

∂∂

−=

∂∂

then the retrieved:

p pT

V pT

V

T

V

V

pT C C

∂∂

∂∂

∂∂

−=−

T

p p

V p

V

p

T

V

T

V

T C C

∂

∂−

∂∂

∂∂

=−

Where V T

V

p

β =

∂∂

and V p

V

T

κ =

∂∂

− , then obtained:

( )( )( )V

V V T C C V p

κ

β β =−

κ

β V T C C V p

2

=− ..............................................................(12)

By using the equation (12), the difference between the Cp and CV can be

calculated for each solution are already known and his only β κ. Price T, V, and κ is

positive, but the price could be positive, β is negative or zero For water at

atmospheric pressure and temperatures of 40 c, β = 0 and between 0oC to 40 c β

value is negative (anomalous properties of water). Therefore always positive or zero

β2 and Cp > CV.

B. The Differential Partial Entropy

As known that entropy is a function of the State of the system, so it can be

expressed as a function of two variables to another.

1. S as a function of volume (V) and temperature (T)

In mathematically: S = f (T , V ), the total differential is:

dV V

S dT

T

S dS

T V

∂∂+

∂∂= ..........................................(13)


By Tenth Group



Known that:V V T

U

T T

S

∂∂

=

∂∂ 1

and

V

V

C T

U =

∂∂

so,

κ

β V

T

C

T

C

T

S pV

V

2

−==

∂∂

............................................(14)

By substitution the equation (11) to the equation (8), so can obtained:

∂∂

+=

∂∂

T T V

U p

T V

S 1

κ

β =

∂∂

T V

S

....................................................................(15)

By the return differentiated of partial differential S in the equations (14) and

(15), can get the equation:

T V V T T

S

V V

S

T

∂∂

∂∂

=

∂∂

∂∂

T

V

V T

C

V T

∂∂

=

∂∂

κ

β

V T

V

T T

V

C

∂∂

=

∂∂

κ

β .................................................(16)

If the equations (14) and (15) are substitution to the equation (13), ao we can

get the equation:

dV V

S dT

T

S dS

T V

∂∂

+

∂∂

=

dV dT T

C dS V

κ

β += .......................................................(17)

dV T

dT C TdS V

κ

β += ..................................................(18)

2. S as a function of temperature (T) and pressure (p)

In mathematically: S = f (T , p), the total differential is:

dp p

S dT

T

S dS

T p

∂∂

+

∂∂

= ...........................................(19)

If U and V in the combination of The First and The Second Law of

Thermodynamics expressed as a function p and T so:

dp p

U

dT T

U

dU T p

∂

∂

+

∂

∂

= ........................................(20)


By Tenth Group



dp p

V dT

T

V dV

T p

∂∂

+

∂∂

= ..........................................(21)

If the equation (20) and the equation (21) substituted o the equation (4), so

obtained the equation as follow:

∂∂

+

∂∂

+

∂∂

+

∂∂

= dp p

V dT

p

V pdp

p

U dT

T

U TdS

T pT p

dp p

V p

p

U dT

p

V p

T

U TdS

T T p p

∂∂

+

∂∂

+

∂∂

+

∂∂

=

dp p

V p

p

U

T dT

T

V p

T

U

T dS

T T p p

∂∂

+

∂∂

+

∂∂

+

∂∂

=11

∂∂+

∂∂=

∂∂

p p p T

V pT

U

T T

S 1.....................................(21)

∂∂

+

∂∂

=

∂∂

T T T p

V p

p

U

T p

S 1....................................(22)

By the return differentiated of partial differential in the equation (21) and the

equation (22), we can get:

pT

T

pp

S

T T

S

p

∂∂

∂∂

=

∂∂

∂∂

∂∂

∂+

∂∂

∂+

∂∂

+

∂∂

−=

∂∂

∂+

∂∂

+

∂∂

∂+

pT

V p

pT

U

T

p

V p

p

U

T pT

V p

T

V

T p

U

T T T p

22

2

22

1

110

∂∂

+

∂∂

−=

∂∂

T T p p

V p

p

U

T T

V

T 2

11

pT T T

V

T p

V

p p

U

∂

∂

−

∂

∂

−=

∂

∂

Where V T

V

p

β =

∂∂

and V p

V

T

κ =

∂∂

− , so obtained:

V T V p p

U

T

β κ −=

∂∂

................................................................(23)

Known that:

V pC T

U p

p

β −=

∂∂

...................................................................(24)


By Tenth Group



By substituted the equation (23) to the equation (22), so we get:

∂∂

+

∂∂

=

∂∂

T T T p

V p

p

U

T p

S 1

( )

∂∂

+−=

∂∂

T T p

V pV T V p

T p

S β κ

1

( )[ ] pKV V T V pT p

S

T

−−=

∂∂

β κ 1

V p

S

T

β −=

∂∂

...............................................................(25)

If the equation (24) substituted to the equation (21), we get:

∂∂+

∂∂=

∂∂

p p p T V p

T U

T T S 1

[ ]V pV pC T T

S p

p

β β +−=

∂∂ 1

T

C

T

S p

p

=

∂∂

...................................................................(26)

If the equation (25) and (26) substituted to the equation (19), so we can get:

dp p

S

dT T

S

dS T p

∂

∂

+

∂

∂

=

VdpdT T

C dS

pβ −= ......................................................(27)

VdpT dT C TdS p β −= ...................................................(28)

3. S as a function of pressure (p) and volume (V)

The combination of The First and The Second Law of Thermodynamics

provided that: pdV dU TdS += . In the differential partial from U against p

and V the independent can be written: dV V

U dp

p

U dU

pV

∂∂

+

∂∂

= . It is

obvious that change the energy in by changes in pressure on the process of

isochoric is:

β

κ V

V

C

p

U =

∂∂

The change internal energy by the change volume on the isobaric is


By Tenth Group



pV

C

V

U p

p

−=

∂∂

β

So,dV p

V

C dp

C dU

pV

−+=

β β

κ

If dU subsituted to the combination of The First and The Second Law of

Thermodynamics, so can obtained:

TdS dU pdV = +

pdV dV pV

C dp

C TdS

pV +

−+=

β β

κ

dV

V

C dp

C TdS

pV

β β

κ += ................................................(29)

dV V T

C dp

T

C dS

pV

β β

κ += ................................................(30)

C. Entropy Ideal Gas and Gas Van der Walls

The equations dV dT T

C dS V

κ

β += , VdpdT

T

C dS

pβ −= , and

dV V T

C dpT C dS pV

β β κ += can be used to calculate the entropy change between two

equilibrium. If ideal gas system already known C p and C V only a function of

temperature,T

1=β ,

p

1=κ , then each of the equation above would be:

•

dV dT

T

C dS V

κ

β +=

( ) dV p

T dT T

C dS V

+=1

1

dV T

pdT

T

C dS V +=

For Ideal Gas:V

nRT p =


By Tenth Group



dV T

V nRT

dT T

C dS V

+=

V dV nRdT

T C dS V +=

If the initial state of the system stated the temperature = T 0, Pressure = p0,

volume = V 0, entropy = S 0, and the final state of the expressed by the temperature =

T , pressure = p, volume = V , entropy = S , then by integrating with the equation

above, we can obtained:

V

dV nRdT

T

C dS V +=

∫ ∫ ∫ +=V

V

T

T

V

S

S V

dV nRdT

T

C dS

000

∫ ∫ ∫ +=V

V

T

T

V

S

S V

dV nRdT

T C dS

000

1

00

0 lnlnV

V nR

T

T C S S V ++= ...................................................(31)

• VdpdT T

C dS

pβ −=

VdpT

dT T

C dS

p

+=1

dpT

V dT

T

C dS

p +=

For Ideal Gas: p

nRT V =

dpT

pnRT

dT T

C dS

p

+=

p

dpnRdT

T

C dS

p +=

If the initial state of the system stated the temperature = T 0, pressure = p0,



By Tenth Group





p

dpnRdT

T

C dS

p +=

∫ ∫ ∫ += p

p

T

T

pS

S p

dpnRdT

T

C dS

000

∫ ∫ ∫ += p

p

T

T

p

S

S p

dpnRdT

T C dS

000

1

00

0 lnln p

pnR

T

T C S S p ++=

......................................................(32)

• dV V T

C dp

T

C dS

pV

β β

κ +=

dV

V T

T

C dp

T T

pC

dS p

V

11

1

+=

dV V

C

dp p

C

dS

pV

+=

V

dV C

p

dpC dS pV +=

If the initial state of the system stated the temperature = T 0, pressure = p0,




V

dV C

p

dpC dS pV +=

∫ ∫ ∫ +=V

V

p

p

p

V

S

S V

dpC dp

pC dS

000

1

∫ ∫ ∫ +=V

V

p

p

p

V

S

S V

dpC dp

pC dS

000

1

00

0 lnlnV

V C

p

pC S S pV ++=

......................................................(33)


By Tenth Group



The equations (31), (32), and (33) are equivalent. If the system is to be

reviewed gas Van der Waals, known23

3

)(2

)(

bva RTv

bv Rv

−−

−=β

and

23

22

)(2

)(

bva RTv

bvv

−−−

=κ , by using the equation (18) expressed in specific quantity is

dV T

dT cTds vκ

β += , so we can get the equation:

dV T

dT cTds vκ

β +=

T

dV T

dT cds

vκ

β +

=

dv

bva RTv

bvv

bva RTv

bvv R

T

dT cvds

23

22

23

2

)(2

)(

)(2

)(

−−−

−−

−

+= , so:

∫ ∫ ∫ −+=

−+=

T

T

v

v

v

s

s

v

bv

dv R

T

dT cds

bv

dv R

T

dT cds

0 00

If cv constant as long as the temperature interval T 0 to T , so the results of the

above integral is:

bv

bv R

T

T c s s v −

−+=−

00

0lnln ....................................................(34)

So it is clear that the constant “a” on the equation of State gas Van der Waals

does not affect the change of entropy.

D. Potential Thermodynamics

It has been known a hydrostatic system, such as a gas has a number of

coordinates that is quantity that can give illustrate of a macroscopic state. Already

introduced p, V, T, and S , and is known also that the system state can be written with


By Tenth Group



two coordinates that are free, in other words each coordinate can be expressed as a

function of two others coordinates

Through the first law of thermodynamics have introduced a quantity of

energy that is in the system (U), where U is a state function of system so that can be

expressed as a function of two coordinates. However, U has what is called natural

coordinates, namely S and V. It can be seen from the first law of Thermodynamics

pdV qd dU −= . If the process is reversible, then this natural law can be written:

pdV TdS dU −= , so ),( V S f U = . Apparently the properties of pure substances

besides can be described by the function U, can also be described by three other

energy functions, namely enthalpy (H), the Helmholtz free energy (F) and Gibbs free

energy (G). The third of this energy is a function of the system state and with U is

called Potential Thermodynamic of a system, each highlighting particular properties

of a process. As with U, other potential Thermodynamics also has its natural

coordinates are discussed as follows.

1. Helmholtz Function (F)

Helmholtz function (often also referred to as the Helmholtz free energy)

is defined as energy in the system reduced the time and temperature with

entropy in mathematics can be formulated:

TS U F −= .....................................................................(35)

In the process of infinite equation (35) can be written

dT S dS T dU dF −−= . From the combination of first and second law

of Thermodynamic obtained equations pdV dU dS T =− . By substitute

this equations to equation (35), then obtained equations below:

pdV dT S dF −−= ........................................................(36)

Then, it can be said to be T and V natural coordinates of Helmholtz

function ( F ). If the isothermal process, dT = 0 then equations (36) can be

written such as below:

pdV dF −= ....................................................................(37)

Thus, the total effort on isothermal process is similar to the Helmholtz

energy change. Equation (37) States if W positive (the system does

work), then the Helmholtz energy decreases and vice versa. A major part


By Tenth Group



of the Helmholtz function is in statistical mechanics that relates closely to

the partition function Z is defined as:

∑ −= kT

iie g Z /ε

with εi and g i stated in a row value of energy and degeneration different

levels of energy of particle system. Because;

pdV dT S dF −−=

Entropy and pressure can be calculated using simple differential:

V T

F S

∂∂

−=dan

T V

F p

∂∂

−=

2. Gibbs Function (G)

Gibbs function is defined as the difference in enthalpy to the product

temperature and entropy, mathematically represented by the equation:

TS H G −= ....................................................................(38)

Already known pV U H += , by substitute this equation to equation (38)

obtained equations below:

TS pV U G −+=

In the equation of infinite, then the above equation becomes:

dT S dS T dpV dV pdU dG −−++= ........................(39)

From the combination of first and second law of Thermodynamic

obtained equations pdV dU dS T += .

If equations above substitute to equation (39) then obtained equation

below:

dT S dS T dpV dS T dG −−+=

dT S dpV dG −=.........................................................

(40)

So, a natural coordinate G is p and T .

3. Enthalpy (H)

Enthalpy is a term used in thermodynamics which stated the amount of

internal energy of a thermodynamic system plus the energy used to do

work. Already known H is a function of the State system, and is

expressed by the equation pV U H += . In the process of infinite can be


By Tenth Group



written dpV dV pdU dH ++= , where đq = dU + pdV. So dH = đQ +

Vdp …………………………………..(41)

By dividing both sides of equation it with dT , we obtained:

dT

dpV

dT

Qd

dT

dH +=

on p constant,

p

p p

cdT

Qd

dT

dH =

=

....................................................(42)

Because dH = đQ + Vdp, enthalpy change during the process of moving

along the same heat were moved. By applying the combination of first

and second law of Thermodynamic, the equation above can be written as:

dpV dS T dH += .........................................................(44)

E. Maxwell's Equations

It is well known that the nature of the pure substances can easily be expressed

in four functions:

Internal Energy U,

Enthalpy H = U + pV

Helmholtz function F = U – TS

Gibbs function G = H - TS

Each function that can be thought of as a function of p, V and T . Suppose U

and S both can be expressed as functions of V and T , so:

U = function (V , T ) and S = function (V , T ).

The second equation solved for T is expressed in S and V; with enter a value

of T in the first equation, then obtained:U = function (S , V ).

In a similar way can be continued and said that one of the eight quantities p,

V , T , S , U , H , F , and G can be express as function from other spouse.

Imagine a hydraulic system underwent a process of invertible infinitesimals

from one equilibrium to another State.

1. The internal energy change by:

pdV Qd dU −=


By Tenth Group



pdV TdS −=

with U , T , and p viewed as a function of S and V .

2. Enthalpy change by:

Vdp pdV dU dH +−=

VdpTdS +=

with H , T , and V viewed as a function of S and p.

3. Helmholtz function change by:

SdT TdS dU dF −−=

pdV SdT −−=

with F , S , and p viewed as a function of T and V .

4. Gibbs function change by:

dG = dH – T dS – S dT

= -S dT + V dP

with G, S, and V everything is viewed as a function of T and P.

Because U, H, F, and G It is the actual function, differential thoroughly is

dz = M dx + N dy,

with z, M, and N is a function of x and y. So

y xx

N yM

∂∂=

∂∂

By applying this result in differential thoroughly dU, dH, dF and dG, we get

1. ;S V

T pdU TdS pdV So

V S

∂ ∂ = − = − ∂ ∂

2. ;

pS

T V dH TdS Vdp So

p S

∂ ∂ = + = ∂ ∂

3. ;T V

S pdF SdT pdV So

V T

∂ ∂ = − − = ∂ ∂

4. ; pT

S V dG SdT Vdp So

p T

∂ ∂ = − + = − ∂ ∂

So the result of fourth function equation above is known as Maxwell's

equations. The equation does not refer to a process but expressing a relationship in

force at each equilibrium of hydrostatic system.


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Maxwell relations are very useful because it presents the relationship between

the quantity that can be measured and the quantity cannot be measured or is difficult

to measure. For example the four Maxwell relations,

, pT

T

V

p

S

∂∂

−=

∂∂

Maxwell relations are very useful because it presents the relationship between the

quantity that can be measured and the quantity cannot be measured or is difficult to

measure. For example the four Maxwell relationships β from pure substances

according to the following way. If pure substances be used also for isotherm and if

you can't reset a molecular extraordinary (such as the Association or dissociation),

then the molecule would occupy a volume that is smaller so that it is in a more

regular basis. In information theory, our knowledge about this molecule increases. So

entropy decreases and derivativesT

p

S

∂∂

negative. So , pT

V

∂∂

positive and it must

have a coefficient from the positive.


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REFERENCE

Hadi, Dimsiki. 1993. Termodinamika. Jakarta: Departemen Pendidikan Dan

Kebudayaan Derektorat Jendral Pendidikan Tinggi Proyek Pendidikan

Tenaga Guru.

Rapi, K. 2009. Buku Ajar Termodinamika. Singaraja.

Zemansky, Mark W. & Dittman, Richard H. 1986. Kalor dan Termodinamika.

Bandung: ITB.


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Kombinasi Hukum I Dan II Termodinamika - En

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