# Ocw Chapter 11 Essay

Submitted By druck2720@gmail.com
Words: 4755
Pages: 20

Chemical Engineering Thermodynamics

Solu%on Thermodynamics:

Theory and Deriva%on

Single component system !

Example: Steam Line

We know that for a single component system, the most common thermodynamic variables are……

T and P
By now, we know how to determine Tsat, Psat,
H (liquid and ideal/non-ideal gas),
V (liquid and ideal/non-ideal gas) and ∆Hvap.

Ok….we also learned about mixture before!!! H = ∑ yi H i

etc.

………we will see that this equa%on is only valid for ideal mixture!!!

Consider the following systems,

A

B

Both systems at the sameT, P and n but with different compositions. Assuming an ideal mixture,

(H )A = ∑ yiA H i yiA ≠ yiB

so
(H )A ≠ (H )B

(H )B = ∑ yiB H i

So we have new thermodynamic variables

for mul%component system………….

Composition!! !
(x1, x2, x3) or (y1, y2, y3)!

BUT typical chemical processes involve a mul%component system whereby we can’t assume ideal mixture or ideal solu%on behavior

Consider mixing 1000 cm3 methanol and 1000 cm3 water at 25oC. +

=

1000 cm3 MeOH + 1000 cm3 H2O !
! !
!
In! fact,!

!
!

!≠ 2000 cm3 solution !!!!!!
!!

1000 cm3 MeOH + 1000 cm3 H2O !
! !
!
!= 1970 cm3 solution!

Discuss with a person next to you why the volume of the solu%on is less than 2000 cm3?

Let’s determine the density

of the solu%on Assume ideal solution 1 ρ= =
V

1
∑ x iVi

From above eqn, the calculated solu%on molar density is 0.0396 mol/cm3 The actual molar density is 0.0410 mol/cm3 that

could be calculated from 1 ρ= =
V

1
∑ x iVi

Partial Properties

We just introduced a new thermodynamic property for a mul%component system……

Partial properties!! !

11.1 Fundamental Property Rela%on

(for variable composi%on mixture)

Deﬁni%on of Chemical Poten%al Consider the following system,
- Open system and multicomponent
- Single phase at T and P
- Variable composition (n1, n2, n3 etc)
So Gibbs energy (G) could be written as a function of T, P, n1, n2, n3 etc.

(nG) = g ( P,T , n1 , n2 ....ni )

Apply partial differentiation and FPR (Chapter 6),

( ) ⎤⎥

⎡ ∂ nG d nG = ⎢
⎢⎣ ∂ P

( )

( ) ⎤⎥

⎡ ∂ nG dP + ⎢
⎥⎦T ,n
⎢⎣ ∂T

( ) ⎤⎥

⎡ ∂ nG dT + ∑ ⎢ i ⎢ ∂ ni
⎥⎦ P,n

( ) ⎤⎥

⎡ ∂ nG d nG = nVdP − nSdT + ∑ ⎢ i ⎢
⎣ ∂ ni

( )

Let's define

( ) ⎤⎥

⎡ ∂ nG µi = ⎢
⎢⎣ ∂ ni

⎥⎦ P,T ,n j ⎥⎦ P,T ,n j ⎥⎦ P,T ,n j dni

dni

≡ Chemical potential for species i in the mixture

This equation forms the basis for the definition of partial properties.
So,

( )

d nG = nVdP − nSdT + ∑ µi dni i (11.2)

How do we determine the phase of a mul%component system?

If it is two phases, how do we determine the composition of each phase?

Before answering these, we need to establish the concept of phase equilibria for multicomponent system.

11.2 The Chemical Poten%al and Phase Equilibria nα

Consider the following system,

- Closed system and multicomponent
- Two phases in equilibrium.
- Mass transfer occurs if the equilibrium is disturbed. Each phase (α and β) is actually an open system, so eqn 11.2 becomes, d ( nG) = ( nV ) dP − ( nS ) dT + ∑ µαi dnαi α α

α

i

d ( nG) = ( nV ) dP − ( nS ) dT + ∑ µ iβ dn iβ β β

β

i

Add the two equations for a VL system
(α is replaced by v and β is replaced by l). d( nG ) + d( nG ) = ( nV ) dP + ( nV ) dP − ( nS ) dT − ( nS ) dT + ∑ µiv dniv + ∑ µil dnil v l

v

l

v

l

i

For system in equilibrium,

d( nG ) = ( nV )dP −