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Thermodynamics Potentials
Equation of State
Consider an isolated system in thermal equilibrium. It has volume V and
contains N molecules whose total energy is E.
The 2nd law tells that every equilibrium state has well defined entropy
S = S(N, V , E ) – this is equation of state and specific to the material or
gas under consideration. For example, the equation of state of ideal gas
is given by Sackur-Tetrode equation (won’t be derived here),
S(N, V , E ) = Nk
[
ln
V
N
4 πmE 3 Nh^2
+^5
]
We can also express E = E (N, V , S) as equation of state.
S, N, V , E are all extensive state variables and any three of them
uniquely specifies a state. Because these are state variables, their
differentials are exact.
dS =
( ∂S
∂N
V ,E
dN +
( ∂S
∂V
N,E
dV +
( ∂S
∂E
N,V
dE
or, dE =
∂E
∂N
V ,S
dN +
∂E
∂V
N,S
dV +
∂E
∂S
N,V
dS
Now consider passing δQ heat to the gas container keeping N, V fixed
i.e. dE = δQ, dN = dV = 0. Knowing that TdS = δQ
T ≡
( ∂E
∂S
N,V
Suppose we do mechanical work δW by changing volume adiabatically
i.e. dS = dN = and dE = δW and knowing δW = −p dV ,
p ≡ −
( ∂E
∂V
N,S
We can also do chemical work by injecting more gas molecules keeping
the volume and entropy (i.e. adiabatically) fixed. In this case,
dS = dV = 0, dE = δW = μ N, where μ is defined as chemical potential.
μ ≡
∂E
∂N
V ,S
Therefore, we write
dE = T dS − p dV + μ dN
where temperature E , pressure p and chemical potential μ are all
intensive variables.
From Virial theorem, therefore, it follows that
E (N, V , S) =
∂E
∂N
V ,S
N +
∂E
∂V
N,S
V +
∂E
∂S
N,V
S
= μ · N − p · V + T · S ⇒ dE = N dμ + μ dN − V dp − p dV + S dT + T dS
But since dE = TdS − pdV + μdN, we arrive at Gibbs-Duhem relation,
S dT − V dp + N dμ = 0
i.e. it is impossible to vary all intensive variables freely, this is in contrary
to the ability to vary all extensive variables freely.
Two more mathematical identities needed for Maxwell relations.
f (x, y ) : df = M dx + N dy , where M ≡
∂f ∂x
y
, N ≡
∂f ∂y
x ⇒
∂M
∂y
x
∂N
∂x
y
z = z(x, y ) :
( (^) ∂x
∂y
z
( (^) ∂y
∂z
x
( (^) ∂z
∂x
y
Note the minus sign, it is NOT a chain rule!
Employ these identities to dE = TdS − pdV + μdN. The first one yields,
∂T
∂V
N,S
∂p ∂S
N,V
∂T
∂N
V ,S
∂μ ∂S
N,V
∂p ∂N
V ,S
∂μ ∂V
N,S
Fix N i.e. E = E (V , S), then the second identity yields,
∂E
∂V
N,S
∂E
∂S
N,V
∂S
∂V
N,E
⇒ p = T ·
∂S
∂V
N,E
These relations, although true, are not always useful and their
implications are also not always obvious.
The equation of state E (N, V , S) is not very convenient to work with.
Mostly because it is difficult to control S in lab and we usually do not do
experiment at adiabatic condition δQ = 0.
Very often, experiments are performed at constant temperature dT = 0
or pressure dp = 0.
Hence, we would like to have a different equation of state, say, A(N, V , T ).
Legendre transform helps us to achieve that.
Helmholtz Free Energy F (T , V ) : perhaps more important than U.
F (V , T ) = U(V , S) − S ·
( ∂U
∂S
V
= U − TS
dF = −S dT − p dV
⇒ S = −
∂F
∂T
V
, p = −
∂F
∂V
T
Helmholtz free energy is commonly used for thermodynamics system at
constant volume and measures the maximum amount of work obtainable
at constant volume. Helmholtz free energy is minimized at equilibrium.
N.B. To prove any of the above statements, we have to include the N dependence of F i.e. dF (N, V , T ) = −SdT − pdV + μdN.
Gibbs Free Energy G (T , p)
G = H − TS = F + pV ⇒ dG = −S dT + V dp + μ dN
⇒ S = −
∂G
∂T
p
, V =
∂G
∂p
T
Gibbs free energy is commonly used for thermodynamics system at
constant pressure and measures the maximum amount of work obtainable
at constant pressure. Gibbs free energy is minimized at equilibrium.
Maxwell Relations
Here we re-collect all the differential form of thermodynamic potentials,
dU(S, V ) = T dS − p dV + μ dN, dH(S, p) = T dS + V dp + μ dN, dF (T , V ) = −S dT − p dV + μ dN, dG (T , p) = −S dT + V dp + μ dN.
The exactness of differentials of U, H, F , G leads to the following most
commonly stated Maxwell relations respectively,
∂T
∂V
S
∂p ∂S
V
∂T
∂p
S
∂V
∂S
( p ∂S ∂V
T
∂p ∂T
V
∂S
∂p
T
∂V
∂T
p
Mnemonic for thermodynamic potentials:
G ood Physicists Have Studied Under V ery F ine T eachers
P
H
U
G
S
V
T F
+ − (^) +
−
Expansion of ideal gas
An isolated container of volume Vf is so partitioned that N ideal gas
molecules are confined within a volume Vi , while the remaining volume
Vf − Vi is in vacuum. The corresponding state variables of this state are
Ei , Si , Ti , pi , Hi , Fi , Gi. What happens when (a) the partition is suddenly
removed? (b) the partition is removed infinitely slowly allowing the gas
to expand gradually? Use Sackur-Tetrode equation of state of ideal gas.
S(N, V , E ) = Nk
[
ln
V
N
( (^4) πmE
3 Nh^2
]
First we consider sudden expansion of gas. Since there is no heat flow or
work done, the change of total energy is zero i.e.
Ei = Ef
The expressions for entropy in the initial and final state are
Si = Nk
[
ln
Vi N
4 πmEi 3 Nh^2
]
, Sf = Nk
[
ln
Vf N
4 πmEf 3 Nh^2
]
Sf = Si + Nk ln
Vf Vi
The temperature is
T
∂S
∂E
N,V
3 Nk 2 E
⇒ Tf = Ti =
2 Ei 3 Nk
Because pV = NkT , we have
pi Vi = NkTi , pf Vf = NkTf ⇒ pf Vf = pi Vi
Enthalpy is H = E + pV ,
Hf = Ef + pf Vf =
NkTf + NkTf =
NkTf =
NkTi ⇒ Hf = Hi
Helmholtz free energy F = E − TS
Ff = Ef − Tf Sf = Ei − Ti
Si + Nk ln
Vf Vi
⇒ Ff = Fi − NkTi ln
Vf Vi
Finally Gibbs free energy G = H − TS
Gf = Hf − Tf Sf = Hi − Ti
Si + Nk ln
Vf Vi
⇒ Gf = Gi − NkTi ln
Vf Vi
Enthalpy H = E + pV ,
Hf = Ef + pf Vf = Ei
Vi Vf
Vi Vf
= (Ei + pi Vi )
Vi Vf
⇒ Hf = Hi
( V
i Vf
Helmholtz free energy F = E − TS,
Ff = Ef − Tf Sf = Ei
Vi Vf
− Si Ti
Vi Vf
= (Ei − Ti Si )
Vi Vf
⇒ Ff = Fi
( V
i Vf
And similarly, Gibbs free energy G = F + pV
Gf = Gi
Vi Vf
Heat capacities
For ideal gas CP − CV = nR, but what about other systems? We use
thermodynamic i.e. relations to arrive at it. We know,
CV =
δQ dT
V
= T
∂S
∂T
V
∂U
∂T
V
and CP =
δQ dT
p
= T
∂S
∂T
p
∂H
∂T
p
If we consider S = S(T , V ), then
dS =
∂S
∂T
V
dT +
∂S
∂V
T
dV ( (^) ∂S
∂T
p
( ∂S
∂T
V
( ∂S
∂V
T
( ∂V
∂T
p
(× T )
T
( ∂S
∂T
p
= T
( ∂S
∂T
V
+ T
( ∂S
∂V
T
( ∂V
∂T
p
CP = CV + T
∂S
∂V
T
∂V
∂T
p
= CV + T
∂p ∂T
V
∂V
∂T
p
CP − CV = −T
∂V
∂T
p
∂p ∂V
T
where we used Maxwell relation
∂S
∂V
T
∂p ∂T
V
and
∂p ∂T
V
∂T
∂V
p
∂V
∂p
T
