ChE 311 final exam review
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Intermolecular force
Gravitational force: Is not only existing between objects and earth, any matters that have a mass will
have gravitational forces.
Formular:
Force: Fij = −
Gmi mj
r2
Rf
R f Gmi mj
Gmi mj
dr = −
Potential energy: τij = − i Fij dr = − i
r2
r
Electrostatic force:Generated from charged materials, for example anion and cations
Qi Qj
Force: Fij =
r2
Rf
R f Qi Qj
Qi Qj
Potential energy: τij = − i Fij dr = − i
dr =
2
r
r
Attention!For Q has different signs in positive charged and negative charged matter, two different charge
will result in attractive force but same charge(no matter ”-” or ”+”) will result in repulsive force. And
that is the reason why force equation doesn’t have a sign with it.
Electric dipole:Mainly generated from different electron-negativity from atoms, different electron-negativity
will result in different ability for attracting electrons. During this tug of war competition, molecules perform dielectric dipole property. The ratio of the number of dipoles in any two states, N1 , N2 can be described as:
N1
= e[(τ1 −τ2 )/(kT )]
N2
Potential energy: τ̄ij = −
2 µ2i µ2j
3 r6 kT
Induction Forces: Result for dipole moment(Dipole caused unevenly distributed electron)
Potential energy: τ̄ij = −
αi µ2j
r6
Dispersion Forces: This is result from the instantaneous nonsymmetry of the electron cloud surrounding a nucleus.
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Potential energy: τij ≈ −
3 αi αj Ii Ij
(
)
2 r6 Ii + Ij
I, α and µ are represented for Ionization potential, polarizability, and dipole moment. These can be find
in table 4.1 in text book.
σ
σ
Lennard-jones Potential energy: τ = 4[( )12 − ( )6 ]
r
r
, σ stand for energy and distance parameter. C12 = 4σ 12 , C6 = 4σ 6
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Equation of State
The difference between ideal gas and real gas is the forces and volume from molecules cannot be neglected.
And that is the reason why we will have ”a”, and ”b” in all equation of states. ”a” represents for intermolecular forces, while ”b” stands for volume correction.
a
RT
− 2
Van der Waals: P =
v̂ − b v̂
”a” and ”b” from critical point data are equal to:
27 (RTc )2
64 Pc
a =
RTc
8Pc
b =
√
RT
a/ T
Redlich-Kwong: P =
−
v̂ − b v̂(v̂ + b)
”a” and ”b” from critical point data are equal to:
Peng-Robinson: P =
a =
0.42748R2 Tc2.5
Pc
b =
0.08664RTc
Pc
aα(T )
RT
−
v̂ − b v̂(v̂ + b) + b(v̂ − b)
”a” and ”b” from critical point data are equal to:
a =
0.45724R2 Tc2
Pc
0.07780RTc
Pc
√
α(T ) = [1 + k(1 − Tr )]2
b =
k = 0.37464 + 1.54226ω − 0.26992ω 2
2
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Compressibility
Compressibility: z =
P v̂
= z (0) + ωz (1)
RT
Tr =
T
Tc
Pr =
P
Pc
vr =
v
vc
ω = −1 − log10 [P sat (T = 0.7 ∗ Tc )/Pc ]
All parameters can be find in Appendix A.1.
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Thermodynamic Web I-Iv
We need to understand that Internal energy, Entropy, Enthalpy, Gibbs free energy and Helmholz energy
are all state function. And this indicate that these terms are only a function of volume(v̂), pressure(p),
and temperature(T). If we know any two factors from above and the EOS, we can get the third factor. If
initial state and final state can be determined, those state functions can definitely be found.
u ≡ q + w
s ≡ qrev /T
h ≡ u + pv
a ≡ u − Ts
g ≡ h − Ts
3
Thus,
ds
= dqrev /T
du
= dq + dw
= T ds − pdv
dh = du + d(pv)
= du + vdp + pdv
= T ds − pdv + vdp + pdv
= T ds + vdp
da
= du − d(T s)
= T ds − pdv − ( T ds + sdT )
= −sdT − pdv
dg
= dh − d(T s)
= T ds + vdp − ( T ds + sdT )
= vdp − sdT
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Equation setup:
du
= (
∂u
∂u
)v dT + ( )T dv
∂T
∂v
= Cv dT + [T (
dh = (
∂h
∂h
)p dT + ( )T dp
∂T
∂p
= Cp dT + [−T (
ds
= (
=
ds
∂v
)p + v]dp
∂T
∂s
∂s
)v dT + ( )T dv
∂T
∂v
∂P
Cv
dT + ( )v dv
T
∂T
= (
=
∂p
)v − p]dv
∂T
∂s
∂s
)p dT + ( )T dp
∂T
∂p
Cp
∂v
dT − ( )p dp
T
∂T
Normally, we should have two paths to calculate ∆h and ∆u, the first path is to use the equation directly, but we need to remember that if we use the equation above directly, we need to get the real heat
capacity:”CPreal ” and ”Cvreal ” from ”CPideal ” and ”Cvideal ”Thus we will get:
du
= Cvreal dT
dh = Cpreal dT
ds
ds
= (
∂s
∂s
)v dT + ( )T dv
∂T
∂v
=
Cv
∂P
dT + ( )v dv
T
∂T
= (
∂s
∂s
)p dT + ( )T dp
∂T
∂p
=
Cp
∂v
dT + ( )p dp
T
∂T
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Where ”Cpreal ” and ”Cvreal ” are equal to:
Cpreal
R real
∂2v
= Cpieal − ideal [T ( 2 )p ]dp
∂T
Cvreal
R real
∂2p
= Cvieal + ideal [T ( 2 )v ]dv
∂T
If you are wondering how to find the Temperature, I will suggest you to read Example 5.3.
Figure 1: Path to find heat capacity
∂2p
∂2v
Do not worry about T in the equation of [T ( 2 )v ]dv and [T ( 2 )p ]dp, we will keep the T term in the
∂T
∂T
expression, since we will integrate it with respect of temperature to get ∆u1 or ∆h1 .
Alternative path to find state function: Since we are talking about ∆h, by fabricating this hypo-
Figure 2: Path to find enthalpy change
thetical path, we will be able to use Cpideal at step 2, for step 1 and 3 we can solve them by setting up
∂v
∂u
∂p
∂h
( )T dP = [−Ti ( )p + v]dp, Ti = (T1 , T2 )(for internal we need to find ( )T dv = [Ti ( )v − P ]dv,Ti
∂p
∂T
∂v
∂T
= (T1 , T2 ))
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Helpful information
Cylic relation:
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−1 = (
∂z
∂x
∂y
)x · ( )y · ( )z
∂y
∂z
∂x
Inversion:
(
∂z
1
)x =
∂y
∂y
( )x
∂z
Figure 3: Road map
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