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11.1 Fundamental Property Relation 11.1 Fundamental Property Relation 11.1 Fundamental Property Relation 11.1 Fundamental Property Relation

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(1)

Thermodynamics IIIIIIII

Department of Chemical Engineering

Prof. Kim, Jong Hak

(2)

Solution Thermodynamics : theory Solution Thermodynamics : theory Solution Thermodynamics : theory Solution Thermodynamics : theory

■ Objective

: lay the theoretical foundation for applications of thermodynamics to gas mixture and liquid solution

○ Most of chemical process undergo composition changes by mixing separation.

=> compositions become essential variable along with T and P.

○ Fundamental property relation become more comprehensive than eqn. (6.10) dG = VdP – SdT.

(3)

11.1 Fundamental Property Relation 11.1 Fundamental Property Relation 11.1 Fundamental Property Relation 11.1 Fundamental Property Relation

○ Eqn. (6.6) d(nG) = (nV)dP – (nS)dT for closed system of single phase nV

P ]

∂ ) nG

∂ (

[

T,n

=

and

] - nS

∂ P ) nG

∂ (

[

P,n

=

○ For a single-phase, open system

nG = f ( P, T, n

1

, n

2

, ….)

i

i n , T , P i n

, P n

,

T

] dn

∂ n ) nG

∂ ( [ dT

T ]

∂ ) nG

∂ ( [ dP P ]

∂ ) nG

∂ ( [ ) nG (

d = + +

j

By definition chemical potential of species i in the mixture

i

i n

, T , P i

i

] dn

∂ n ) nG

∂ ( [

µ =

j (11.1)

i

i i

dn µ dT

) nS ( - dP ) nV ( ) nG (

∴ d = +

(11.2)

(4)

For special case of one mole of solution n=1, ni = xi

i

i i

dx µ SdT

- VdP

dG = +

(11.3)

,....) x

,..., x

, x , P , T ( G

G =

1 2 i

From (11.3)

x ,

)

T

∂ P

∂ G

V = ( )

P,x

∂ T

∂ G S = (

x ,

)

P

∂ T

∂ G ( T - G TS

G

H = + =

※ Gibbs energy plays a role of a generating function, providing the means for calculation of all other thermodynamic properties by simple mathematical operations.

(5)

11.2 The Chemical Potential and Phase 11.2 The Chemical Potential and Phase 11.2 The Chemical Potential and Phase

11.2 The Chemical Potential and Phase Equilibria Equilibria Equilibria Equilibria

■ Closed systems consisting of two phase in equilibrium

each individual phase is open to the other => mass transfer occur b/w phases

β i β i β

β β

α i α i α

α α

µ dn dT

) nS ( - dP ) nV ( )

nG ( d

dn µ dT

) nS ( - dP ) nV ( )

nG ( d

=

=

=

=

) ) nM ( ) nM ( nM

( =

α

+

β

∑ µ

αi

dn

iα

µ

βi

dn

βi

(nS)dT

- dP ) nV ( ) nG (

d = + +

at equilibrium

Eqn.(11.2) applies to each phase

Because two phases are in equilibrium -> T and P is uniform

Change of total Gibbs energy = sum of two equation

d ( nG )

α

+ d ( nG )

β

∑ µ

iα

dn

αi

+ ∑ µ

βi

dn

βi

= 0

by mass conservation,

dn

αi

+ dn

βi

= 0

∑ ( µ

αi

- µ

βi

) dn

αi

= 0

(6)

) N , 3 , 2 , 1 i

( µ

µ

∴ µ

αi

=

βi

==

iπ

=

(11-6) At equilibrium μof each phase is same

Thus, multiple phase at same T&P are in equilibrium when the chemical potential of each species is the same in the all phase

µ ) µ

(

αi

=

βi

(7)

11.3 Partial Properties 11.3 Partial Properties 11.3 Partial Properties 11.3 Partial Properties

■ Partial property is defined by

nj

, T , P i

i

]

∂ n ) nM

∂ (

≡ [ M

Chemical potential is partial molar property of Gibbs Energy

※ Equations relating Molar and Partial molar properties

○ From the knowledge of the partial properties, we can calculate solution properties or we can do reversely.

○ Total thermo properties of homogeneous phase are functions of T, P and the numbers of moles of the individual species which comprise the phase

(11-7)

M

i

i n

, T , P i

i

) G

∂ n ) nG

∂ ( (

µ =

j

=

) n ,..., n

, n , P , T ( µ

nM =

1 2 i

(8)

The total differential of nM is

i i

n , T , P i n

, P n

,

T

] dn

∂ n ) nM

∂ ( [ dT

T ]

∂ ) nM

∂ ( [ dP P ]

∂ ) nM

∂ ( [ ) nM (

d ∑

+

j

+

=

i

i i x

, P x

,

T

) dT M dn

∂ T

∂ M ( n dP P )

∂ ) nM

∂ ( ( n ) nM (

d = + +

∑ M ( x dn ndx ) dT

T )

∂ M ( n dP P )

∂ M ( n Mdn

∴ ndM + =

T,x

+

P,x

+

i i

+

i

(11-9)

Mdn

≡ ndM )

nM ( d

ndx dn

x

→ dn n x

n

i i i i i

+

+

=

=

(9)

The terms containing “n” are collected separated from those containing dn to yield

i

i i i

i x

, P x

,

T

) dT - M dx ] n [ M - x M ] dn 0

∂ T

∂ M ( - dP P )

∂ M ( - dM

[ + =

i

i i x

, P x

,

T

) dT M dx

∂ T

∂ M ( dP P )

∂ M (

∴ dM = + +

i

i

i

M

M = x

i

i i

M nM = n

Multiply Egn.(11.11) by n yield

(11-12) (11-11) (11-10)

(11.10) -> special case of Eqn (11.9) by setting n=1 (11.11), (11.12) -> summability relations

=> allow calculation of mixture from partial property

(10)

From Egn (11.11)

∑ ∑

i i

i i i

i

d M M dx

x

dM = +

i

i i x

, P x

,

T

) dT - x dM 0

∂ T

∂ M ( dP P )

∂ M

( + =

i

i

i

d M 0

x =

(11-14)

(11-13)

∑ x

i

M

i

M =

Compare this with (11.10)

=> Gibbs/Duhem equation

at const T, P

(11)

■ Rationale for partial property

=> Solution property is sum of its partial properties

∑ x

i

d M

i

M =

i 1 i

x

1

x

M lim M M

lim

i i

=

=

(in the limit as a solution become pure in species i)

By definition i i

0

x

lim M M

i

=

■ Summary of partial property

1. Definition

nj

, T , P i

i

]

∂ n ) nM

∂ (

M = [

=> yield partial properties from total property

2. summability

i

i

i

M

M = x

=> yield total properties from partial properties

3. Gibbs/Duhem

) dT

∂ T

∂ M ( dP P )

∂ M ( dM

x

T,x P,x

i

i

i

= +

=> partial properties of species in solution => dependent one another

(12)

■ Partial properties in binary solutions

2 2 1

1

M x M

x

M = + dM = x

1

d M

1

+ M

1

dx

1

+ x

2

d M

2

+ M

2

dx

2

0 M

d x M d

x

1 1

+

2 2

=

1 2 1

1

dx - M dx

dM = M

) M - M ( x - M M

x M ) x - 1 (

M =

2 1

+

2 2

=

1 2 1 2

1 2

1

dx

x dM M

M = +

at constant T, P by Gibbs/Duhem Egn.

Because x1 + x2 = 1, dx1 = -dx2 , Eliminating dx2 in Eq

1 1

2

dx

x dM - M = M

⒜ =>=>=>=>

For binary solution (system), From summability relations M =

xidMi

1 2 2

2 1

1 1

1

d M M dx x d M - M dx

x

dM = + +

2 1

1

M - dx M

∴ dM =

by Egn

From Egn

) M - M ( x M M

) x - 1 ( M x

M =

1 1

+

1 2

=

2

+

1 1 2

insert Egn

So, from the solution properties as a function of composition (at const T, P)

=> Partial properties can be calculated

(13)

G/D can be written in derivative form

dx 0 M x d

dx M x d

1 2 2

1 1

1

+ =

1 2 1

2 1

1

dx M d x - x dx

M d =

2 1 &M M

=>

=>=>

=>

=> When are plotted vs x1 => sign of slope is opposite

dx 0 M lim d

1 1

1

x1

= dx 0

M lim d

1 2

1

x2

=

2

1 and M

M

Moreover from this Equation

,,,,

=> Plots of vs x1 => horizontal as each species approach purity

(14)

■ Relations among partial properties

) µ G

( i = i

∑ µ

i

dn

i

dT

) nS ( - dP ) nV ( ) nG (

d = +

i

i i

dn G (nS)dT

- dP ) nV ( ) nG (

⇒ d = +

n , T n

,

P

)

∂ P

∂ S -(

T )

∂ V

( =

i n

, T , P i n

, P

i

) ⇒ - S

∂ n ) nS

∂ ( -(

T )

∂ G

( =

j

i n

, T , P i n

, T

i

) ⇒ V

∂ n ) nV

∂ ( -(

P )

∂ G

( =

j

Apply criterion of exactness for differential expression

※ Property relations used in const. composition solution has their counterpart equations for partial properties

For example, H = U + PV, for n mole nH = nU + P(nV)

Differentiation with respect to ni at const T, P, nj

j j

j P,T,n

i n

, T , P i n

, T , P i

n ]

∂ ) nV

∂ ( [ P n ]

∂ ) nU

∂ ( [ n ]

∂ ) nH

∂ (

[ = +

=>=>=>=>

H

i

= U

i

+ P V

i

dT S - dP V G

d

i

=

i

(15)

11.4 Ideal Gas Mixtures 11.4 Ideal Gas Mixtures 11.4 Ideal Gas Mixtures 11.4 Ideal Gas Mixtures

※ Ideal gas mixture model : Basis to build the structure of solution thermodynamics

P V = RT

=> All ideal gas, whether pure or mixture have same molar volume at the same T, P

P ) RT

∂ n

∂ n P (

] RT

∂ n

) P / nRT

∂ ( [ n ]

∂ ) nV [ (

V

j nj

i i

n , P , T i

ig ig

i

= = =

∂ ∂ ∂

=

since

=> partial molar volume = pure species molar volume = mixture molar volume Molar volume of Ideal Gas :

※ Partial molar volume of species i in ideal gas mixture

j

i

n

n n = +

P V RT

V

V

igi

=

iig

=

ig

=

(16)

Partial pressure of ideal gas for n mol of ideal gas

i i

i

y

n n P P

=

=

for species i , t i

i V

RT P = n

V

t

P = nRT

yP

∴ P

i

=

t i

t

V

(since in ideal

V =

In ideal gas, thermodynamic properties independent of one another

※ Gibbs theorem

A partial molar property (except volume) of a constituent in an ideal gas mixture is equal to the corresponding molar property of the species as a pure ideal gas but at a pressure equal to its partial pressure

) P , T ( M )

P , T (

M

igi

=

igi i (11.21)

(17)

※ H of Ideal Gas => independent of P

) P , T ( H ) P , T ( H )

P , T (

∴ H

igi

=

igi i

=

igi

ig i ig

i

H

∴ H =

-> ① (11.22)

dT C H =

P

※ S of an ideal gas -> dependent on P and T P

R dP T -

C dT

dS

ig

=

igP (6.24)

at const T.

P -R dP -RdlnP

dS

igi

= =

Integration from Pi to P i

i i

i ig

i ig

i

R ln y

P y -Rln P p

-Rln P )

p , T ( S - ) P , T (

S = = =

i ig

i i

ig

i

( T , p ) S ( T , P ) - Rlny

∴ S =

Since,

S

iig

( T , P ) = S

igi

( T , P

i

)

Gibbs theorem

i ig

i ig

i

S - Rlny

S =

-> ② (11.23)

(18)

For the Gibbs Energy of ideal gas mixture

ig i ig

ig ig ig

ig

H - TS ⇒ G H - T S

G = =

i ig

i ig

i ig

i

H - TS RT ln y

G = +

i ig

i ig

i ig

i

≡ G G RT ln y

µ = +

or

(by using egn. ① ② ② ) -> ③ (11.24)

∑ ∑

∑ ∑

= x M ) M

( i i

By summability eqn

i i

ig i i ig

i i

ig i i ig

ig i i ig

y ln y RT

G y G

y ln y R

- S y S

H y H

+

=

=

=

H y -

H

ig

i igi

-> ④ -> ⑤ -> ⑥

i i

ig i i ig

y ln 1 y R

S y -

S =

(enthalpy charge of mixing) = 0

=> no heat transfer for ideal gas mixing

(entropy charge of mixing) > 0

=> agree with second law, mixing is irreversible

(19)

ig

µ

i

⇒ VdP dT

S - dP V

dG

igi

=

iig igi

Alternative expression for

P ln RTd P dP

dP RT V

∴ dG

igi

=

iig

= =

(at const T)

By integration

P ln RT )

T Γ (

G

igi

=

i

+

) P y ln(

RT )

T ( Γ

∴ µ

igi

=

i

+

i

) P y ln(

y RT

) T ( Γ y

∴ G

ig

=

i i

+

i i

(11.28) (11.29)

(11.30) Applying summability relation

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