ChE 131
SOLUTIONS THERMODYNAMICS
MODULE 2C
Solutions Thermodynamics:Theory
Single Phase multicomponent system (simple)
-
No magnetic field, induction , etc.
Fundamental new property
𝜇𝑖 (Chemical potential)- plays role in equilibria
-
Principles of phase and chemical-reaction equilibrium depend upon
this
They lead to partial properties ( the mathematical description allows
them to be interpreted as properties of the individual species as they
exist in solution)
𝜕 𝑛𝐺
𝜇𝑖 = ( 𝜕𝑛 )
(11.1)
𝑖
𝑇,𝑃,𝑛𝑗
Constant composition mixture
(
𝜕 𝜇𝑖
𝜕(𝑛𝑆)
) = −[
]
𝜕𝑇 𝑃,𝑥
𝜕𝑛𝑖 𝑇,𝑃,𝑛
𝑗
(
𝜕 𝜇𝑖
𝜕(𝑛𝑉)
) = −[
]
𝜕𝑃 𝑇,𝑥
𝜕𝑛𝑖 𝑇,𝑃,𝑛
𝑗
̅𝑖 = [𝜕(𝑛𝑀)]
𝑀
𝜕𝑛
(11.7)
𝑖
-
≡ 𝑝𝑎𝑟𝑡𝑖𝑎𝑙 𝑚𝑜𝑙𝑎𝑟 𝑝𝑟𝑜𝑝𝑒𝑟𝑡𝑦
𝑇,𝑃,𝑛𝑗
provides for the calculation of partial properties from mixture properties
Viewed as a response function definition of partial molar property
̅ could be 𝑈, 𝐻, 𝑆, 𝑒𝑡𝑐
𝑀
-
A response function, representing the change of total property 𝑛𝑀 due to
additions at constant 𝑇 and 𝑃 of a differential amount of specie 𝑖 to a finite
amount of solution
Total property of a phase, 𝑛𝑀
𝑛𝑀 = 𝑓(𝑇, 𝑃, 𝑐𝑜𝑚𝑝𝑜𝑠𝑖𝑡𝑖𝑜𝑛)
𝑑 (𝑛𝑀) = [
𝜕(𝑛𝑀)
𝜕(𝑛𝑀)
𝜕(𝑛𝑀)
] 𝑑𝑇 + [
] 𝑑𝑃 + ∑ [
]
𝑑𝑛𝑖
𝜕𝑇 𝑃,𝑛
𝜕𝑇 𝑇,𝑛
𝜕𝑛𝑖 𝑇,𝑃,𝑛
𝑖=
𝑖
̅̅̅
𝑀𝑖
The summation is over all species
present
Page 1 of 6
Where:
subs 𝑛 means that all mole numbers are held constant
𝑛𝑗 , that all mole numbers except 𝑛𝑖 are held constant
𝑑𝑛𝑖 = 𝑑(𝑥𝑖 𝑛) = 𝑥𝑖 𝑑𝑛 + 𝑛𝑑𝑥𝑖
𝑑(𝑛𝑀) = 𝑛𝑑𝑀 + 𝑀𝑑𝑛
𝑛𝑑𝑀 + 𝑀𝑑𝑛 = 𝑛 (
𝜕𝑀
𝜕𝑀
) 𝑑𝑇 + 𝑛 ( ) 𝑑𝑃 + ∑ ̅̅̅
𝑀𝑖 (𝑛𝑑𝑥𝑖 + 𝑥𝑖 𝑑𝑛)
𝜕𝑇 𝑃,𝑥
𝜕𝑃 𝑇,𝑥
subs 𝑥 means differentiation at constant composition
grouping 𝑛 terms and 𝑑𝑛 terms together,
𝜕𝑀
𝜕𝑀
[𝑑𝑀 − ( ) 𝑑𝑇 − ( ) 𝑑𝑃 − ∑ ̅̅̅
𝑀𝑖 (𝑑𝑥𝑖 )] 𝑛 + [𝑀 − ∑ 𝑥𝑖 ̅̅̅
𝑀𝑖 ] 𝑑𝑛 = 0
𝜕𝑇 𝑃,𝑥
𝜕𝑃 𝑇,𝑥
𝑑𝑀 = (
𝜕𝑀
𝜕𝑀
) 𝑑𝑇 + ( ) 𝑑𝑃 + ∑ ̅̅̅
𝑀𝑖 (𝑑𝑥𝑖 )
𝜕𝑇 𝑃,𝑥
𝜕𝑃 𝑇,𝑥
𝑀 = ∑ 𝑥𝑖 ̅̅̅
𝑀𝑖
Summability relations
(11.10)
(11.11)
(A solution property 𝑀 is the sum of the contribution of the components multiplied by 𝑛𝑓
of component.)
̅̅̅𝑖 + ∑𝑖 𝑀𝑖 𝑑𝑋𝑖
Differentiating, 𝑑𝑀 = ∑𝑖 𝑋𝑖 𝑑𝑀
Comparison of this equation with (11.10), another equation for dM, yields the
Gibbs/Duhem equation
(
𝜕𝑀
𝜕𝑀
̅̅̅𝑖 = 0
) 𝑑𝑃 + ( ) 𝑑𝑇 − ∑ 𝑥𝑖 𝑑𝑀
𝜕𝑃 𝑇,𝑥
𝜕𝑇 𝑃,𝑥
(11.13)
Partial properties are denoted by an overbar, with a subscript to identify the species.
Solution properties 𝑀: 𝑈, 𝑆, 𝐻, 𝐺
(11.8)
Partial properties
̅: 𝑈
̅ , 𝑆̅, 𝐻
̅ , 𝐺̅
𝑀
Pure species
𝑀𝑖 : 𝑈𝑖 , 𝑆𝑖 , 𝐻𝑖 , 𝐺𝑖
𝜇𝑖 ≡ 𝐺𝑖
- The chemical potential and the partial molar Gibbs energy are identical.
11.11 & 11.12 - Vital equations known as summability relations
- They allow calculation of mixture properties from partial properties
Multiply both sides by n
Page 2 of 6
𝑛𝑀 = ∑ 𝑛𝑖 𝑀𝑖
(11.12)
Example:
50 𝑛% 𝑛 − 𝑝𝑟𝑜𝑝𝑎𝑛𝑜𝑙, 25 𝑛% 𝑛 − 𝑝𝑒𝑛𝑡𝑎𝑛𝑜𝑙, 25 𝑛% ℎ𝑒𝑝𝑡𝑎𝑛𝑒 } each is pure
Each vat is equipped with stirrer and heating coil
Addition is slow enough, hence isothermal at 294 K
What is the cooling load per vat?
Basis: 1 gmole of product
After the process:
0.5 pentanol
0.5 heptane
𝑄 = ∆ℎ = 𝑛𝐻𝑓 − 𝑛𝐻𝑖
Sum of the n of pentanol and heptane and its partial molar property
𝑛𝐻𝑓 = 0.25 𝐻𝑝𝑒𝑛𝑡𝑎𝑛𝑜𝑙 + 0.25 𝐻ℎ𝑒𝑝𝑡𝑎𝑛𝑒
0.5
𝐻𝑓 = 0.5 𝐻𝑝𝑒𝑛𝑡𝑎𝑛𝑜𝑙 + 0.5 𝐻ℎ𝑒𝑝𝑡𝑎𝑛𝑒
heptane
0.0
0.25
0.25
0.5
1

Mole fraction
propanol
0.0
0.25
0.5
0.0
0
heptane
1155.5
1165.5
846.8
0.0
Partial molar enthalpy (J/g mole)
propanol
pentanol
0.0
167.4
53.5
136.2
74.5
237.1
-
Partial molar H changes with composition
𝐻𝑓 = 0.5(−237.1) + 0.5(−864.8) = −551 𝐽
𝑛𝐻𝑖 = 0 (𝑓𝑜𝑟 𝑝𝑢𝑟𝑒 𝑝𝑒𝑛𝑡𝑎𝑛𝑜𝑙)
In vat 2:
𝑄 = 0.5(𝐻𝑝𝑟𝑜𝑝 ) + 0.25(𝐻𝑝𝑒𝑛𝑡𝑎𝑛𝑜𝑙 ) + 0.25(𝐻ℎ𝑒𝑝𝑡𝑎𝑛𝑒 ) − 0.5(−551)
𝑄 = 103.1 𝐽
𝑄 = 103.1 𝐽
Page 3 of 6
Case II:
n-prop
n-pent
heptane

Choose one that involves the least energy
- Interpolate/ exterpolate
 safer
Problem to ponder on:
Determine the work (W) needed to separate air (assumed binary) into O2 and N2
𝑊 = −𝑃𝑑𝑉
V is for mixture
Intensive variables:
[
𝜕(𝑛𝑀)
𝜕𝑛
𝜕𝑀
]
= 𝑀( )
+𝑛( )
𝜕𝑛𝑖 𝑇,𝑃,𝑛
𝜕𝑛𝑖 𝑇,𝑃,𝑛
𝜕𝑛𝑖 𝑇,𝑃,𝑛
𝑗
𝑗
𝑗
(only 𝑛𝑖 is varying, the rest is constant)
𝜕𝑛
subscript 𝑛𝑗 indicates that all mole
( )
=1
𝜕𝑛𝑖 𝑇,𝑃,𝑛
numbers expect the ith are held constant
𝑗
̅ = 𝑀 +𝑛(
𝑀
𝜕𝑀
)
𝜕𝑛𝑖 𝑇,𝑃,𝑛
𝑗
intensive 𝑀 = 𝑓(𝑇, 𝑃, (𝑚 − 1)𝑚𝑜𝑙𝑒 𝑓𝑟𝑎𝑐𝑡𝑖𝑜𝑛𝑠)
at constant T, P:
𝜕𝑀
𝜕𝑀 = ∑ (
)
𝑑𝑥𝐴
𝜕𝑥𝑘 𝑇,𝑃,𝑥
𝑘≠𝑖
𝐿
Where:
𝑥𝐿 means all mole fractions other than 𝑥𝑘 are held constant
𝜕𝑀
𝜕𝑀
𝜕𝑥𝑘
( )
= ∑(
)
(
)
𝜕𝑛𝑖 𝑇,𝑃,𝑛
𝜕𝑥𝑘 𝑇,𝑃,𝑥 𝜕𝑛𝑖 𝑛
𝑗
𝑘≠𝑖
𝐿
𝑗
Page 4 of 6
𝑥𝑘 𝑛 = 𝑛𝑘
𝑥𝑘 =
𝑛𝑘
𝑛
𝜕𝑛
𝜕𝑛
𝑛 ( 𝑘 ) − 𝑛𝑘 ( )
𝜕𝑋𝑘
𝜕𝑛𝑖
𝜕𝑛𝑖
(
) =
2
𝜕𝑛𝑖 𝑛𝑗
𝑛
=−
𝑛𝑘
𝑥𝑘
=
−
𝑛2
𝑛
𝜕𝑀
̅𝑖 = 𝑀 − ∑ (
𝑀
)
𝜕𝑥𝑘 𝑇,𝑃,𝑋
𝑤ℎ𝑒𝑟𝑒: 𝑙 ≠ 𝑘𝑖
2
𝑘≠1
For Binary mixture at constant T&P
̅1 = 𝑀 − 𝑋2
𝑀
𝑑𝑀
𝑑𝑋2
̅1 = 𝑀 + (1 − 𝑋1 )
𝑀
𝑑𝑀
𝑑𝑋1
(for partial molar property of component 1 in a binary mixture)
̅2 = 𝑀 − 𝑋1
𝑀
𝑑𝑀
𝑑𝑋1
(𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑎 𝑙𝑖𝑛𝑒)
𝑑𝑀 𝑀 − 𝐼2
=
𝑑𝑥1
𝑋1
𝑑𝑀
= 𝐼1 − 𝐼2
𝑑𝑥1
* the equation to use depends on the given values.
Page 5 of 6
𝑑𝑀
𝐼1 = 𝑀 + (1 − 𝑋1 ) (
)
𝑑𝑥1
𝐼2 = 𝑀 − 𝑋1
𝐼1 = ̅̅̅̅
M1
∞
∞
𝐼2 = ̅̅̅̅
M2
𝑑𝑀
𝑑𝑥1
(𝑥1 = 0)
(𝑥2 = 0)
Problem: For a given T and P (binary System)
𝑉 = 100𝑋1 + 80𝑋2 + 2.5𝑋1 𝑋2
𝑐𝑚3
(
)
𝑚𝑜𝑙
Solve for 𝑉̅1 as a function of X1; f(X1) and 𝑉̅2 as a function of X2; f(X2) using
2 different methods:
𝑉̅1 = 100 + 2.5(1 − 𝑋1 )2
𝑉̅2 = 80 + 2.5(1 − 𝑋2 )2
Page 6 of 6