Lecture Notes in Advanced Thermodynamics

Lecture Notes in Advanced Thermodynamics
Part 1.
Ván Péter and Antali Máté
February 13, 2013
Contents
1 Introduction 2
1.1 Subject of thermodynamics . . . . . . . . . . . . . . . . . . . 2
1.2 Mathematical models in thermodynamics . . . . . . . . . . . . 2
1.2.1 Basic concepts . . . . . . . . . . . . . . . . . . . . . . . 2
1.2.2 Classification of models . . . . . . . . . . . . . . . . . . 2
2 Thermostatics of discrete bodies 3
2.1 Simple compressible body . . . . . . . . . . . . . . . . . . . . 3
2.1.1 Basic notations . . . . . . . . . . . . . . . . . . . . . . 4
2.1.2 Definition of the model . . . . . . . . . . . . . . . . . . 4
2.1.3 Important relations from the model . . . . . . . . . . . 6
2.2 New variable sets and functions . . . . . . . . . . . . . . . . . 8
2.2.1 Change of variables . . . . . . . . . . . . . . . . . . . . 8
2.2.2 New quantities with Legendre transformation . . . . . 9
2.3 Densities, specific quantities . . . . . . . . . . . . . . . . . . . 12
2.3.1 Densities . . . . . . . . . . . . . . . . . . . . . . . . . . 12
2.3.2 Specific quantities . . . . . . . . . . . . . . . . . . . . . 13
2.4 Simple model with specific quantities . . . . . . . . . . . . . . 15
2.4.1 Reformulating the model with specific quantities . . . . 15
2.4.2 Examining the constitutive equations . . . . . . . . . . 16
2.4.3 Examples . . . . . . . . . . . . . . . . . . . . . . . . . 18
1

1 Introduction
1.1 Subject of thermodynamics
What is the subject of thermodynamics?
– elementary approach: physics of heat, temperature and energy
→ engineering advantage: performing practical calculations of energy
transfer processes (classical thermodynamics course)
– advanced level: general background and framework of macroscopic
physics
→ engineering advantage: better understanding of other macroscopic
physics courses (solid mechanics, fluid mechanics, classical thermodynamics,
electrodynamics, physical chemistry)
1.2 Mathematical models in thermodynamics
1.2.1 Basic concepts
thermodynamic system : a mathematical abstraction which is used for
modelling a piece of reality, a system is a collection of interacting bodies
thermodynamic body
: the smallest individual piece of a system
state : the set of all pieces of information which determines our whole
knowledge from the system, time and space dependent.
state variables : mathematical objects which are used to represent the
state, they can be numbers, functions, etc. . .
1.2.2 Classification of models
different areas of thermodynamics → different features of state variables
phenomenology - statistics
– phenomenological: state variables are related directly to macroscopic
features of the material
– statistical: state variables are related to the behaviour of the microscopic
particles, and macroscopic features are derived from them by
statistical methods (not covered in this subject)
2

thermostatics - thermodynamics
– thermostatics: state variables do not depend on time
(analogy: mechanical statics)
– thermodynamics: state variables can change in time
(analogy: mechanical dynamics)
discrete thermodynamics - continuum thermodynamics
– discrete thermodynamics: state variables are homogeneous
(analogy: mechanics of point masses and rigid bodies)
– continuum thermodynamics: state variables are space dependent
(analogy: continuum mechanics)
2 Thermostatics of discrete bodies
other names:
– thermostatics of homogeneous bodies
– equilibrium ,,thermodynamics”
we can imagine a homogeneous body as a very small amount of material
or a box of material where all the properties are constant in space and time
2.1 Simple compressible body
simple compressible body: a thermodynamic body, which can be characterized
in equilibrium state by the mass (M), the volume (V ) and the
internal energy (E)
remarks:
– in the following ,,body” means this simple compressible body
– this is an appropriate model for thermal and mechanical behaviour of
single-component gases
– for other material types (solids, fluids, multi-component gases, radiation,
etc.) or other physical properties (chemical, electrical, etc.) different,
usually more complicated models are used
3

2.1.1 Basic notations
basic state variables: (E, V, M) ∈ (R + ) 3
(they can be chosen intuitively or considering physical meaning)
state function: an f(E, V, M) : (R + ) 3 → R function which creates a new
state variable from the basic state variables
extensive scaling of a state function (f◦λ)(E, V, M) := f(λE, λV, λM),
where λ is an arbitrary real number
(resizes the system, giving the value of the state function for a λ times
,,larger” body )
intensive state function: f ◦ λ = f for ∀λ ∈ R
(the state function remains constant if the body is resized)
extensive state function: f ◦ λ = λ · f for ∀λ ∈ R
(the state function is increasing linearly with the resizing)
remarks:
– the function itself and the value of the function are not strictly distinguished
in notation
– in the definitions and following calculations equal sign (=) between
state functions means equality for all possible (E, V, M) values
– E, V and M can be considered also as extensive state functions, e.g.
E(E, V, M) = E
– the partial derivation of a function can be denoted in shorter way:
∂f(E, V, M) ∂f
=:
∂E ∂E ∣ =: ∂ E f
V,M
2.1.2 Definition of the model
Definition(discrete thermostatic body):
– a set of ⎧ ⎫
⎨ T (E, V, M) ⎬
p(E, V, M)
⎩ ⎭
µ(E, V, M)
state functions over the (E, V, M) state space
4

– for that there exists a S(E, V, M) state function called entropy, which
has to fulfil the following properties:
(P 1 ) (entropy is a potential)
∂S
∂E ∣ = 1
V,M
T ;
∂S
∂V
∣ = p
E,M
T ;
∂S
∂M ∣ = − µ
E,V
T
(1)
(P 2 ) (entropy is increasing with energy)
∂S
∂E ∣ > 0 (2)
V,M
(P 3 ) (entropy is extensive)
S ◦ λ = λ · S (3)
Consequence: The entropy is extensive if and only if there is a specific
entropy function: s(E/M, V/M) = S(E,V,M)
M
(P 4 ) (specific entropy is concave)
D 2 s(e, v) is a negative definite matrix (4)
Remarks:
– T denotes the temperature, p is for the pressure and µ is called chemical
potential
– T, p and µ functions are called state functions, they correspond to the
thermostatic system, the properties restrict the set of possible constitutive
functions due to physical principles
– the properties are related, but cannot be derived from the laws of thermodynamics,
the dynamical laws have an effect on the structure of the
system even in equilibrium state
5

Example: ideal gas
– the well-known equations of the ideal gas (R is the gas constant and c
is the isochoric specific heat, they are real constants):
E = cMT ;
pV = MRT
– from that the state functions are:
T (E, V, M) = E
cM
p(E, V, M) = MRT
V
= RE
cV
– let us choose the chemical potential function as:
µ(E, V, M) = c ln E M + R ln V M
– for these functions an appropriate entropy function:
S(E, V, M) = Mc ln E M + MR ln V M
+ M(c + R)
– it can be easily checked that for c > 0 this entropy function fulfils
properties (P 1 ), (P 2 ) and (P 3 ), for proving the validity of (P 4 ) a longer
calculation is required
2.1.3 Important relations from the model
Gibbs relation
– the total differential of the entropy function (d can be imagined as a
very small change in the quantity)
dS = ∂S
∂E ∣ dE + ∂S
V,M
∂V ∣ dV + ∂S
E,M
∂M ∣ dM
E,V
– using the (P 1 ) property:
dS = 1 T dE + p T dV − µ dM (5)
T
– after rearranging the Gibbs relation is obtained:
dE = T dS − pdV + µdM (6)
6

intensive property of T ,p and µ:
– let us calculate the resizing of (5):
dS ◦ λ = 1
T ◦ λ dE ◦ λ + p ◦ λ
T ◦ λ dV ◦ λ − µ ◦ λ
T ◦ λ dM ◦ λ
– E,V ,M are extensive (trivial) and S is extensive too due to property
(P 3 ):
λdS = 1
T ◦ λ λdE + p ◦ λ
T ◦ λ λdV − µ ◦ λ
T ◦ λ λdM
– from that:
∂S
∂E ∣ = 1
V,M
T ◦ λ ;
∂S
∂V
∣ = p ◦ λ
E,M
T ◦ λ ;
– comparing them with property (P 1 ):
∂S
∂M ∣ = − µ ◦ λ
E,V
T ◦ λ
T ◦ λ = T ; p ◦ λ = p; µ ◦ λ = µ (7)
– thus T , p and µ are intensive state functions
Potential relation
– differentiating property (P 1 ) respect to λ:
(
) (
) (
)
∂S
∂S
∂E ∣ ◦ λ E +
∂S
V,M
∂V ∣ ◦ λ V +
E,M
∂M ∣ ◦ λ M = S
E,V
– using property (P 1 ):
– using (7):
1
T ◦ λ E + p ◦ λ
T ◦ λ V − µ ◦ λ
T ◦ λ M = S
1
T E + p T V − µ T M = S
– after rearranging we get the potential relation:
E = T S − pV + µM (8)
7

2.2 New variable sets and functions
2.2.1 Change of variables
introduction of a new variable
– given a function f(x, y)
– the variable x is changed to z with the substitution x = x(z, y)
– the new function is
ˆf(y, z) := f ( x(y, z), z ) (9)
derivatives respect to the new variable
– (partial) differentiating (9) respect to z
– rearranging:
∂ z ˆf|y = ∂ x f| y · ∂ z x| y
∂ x f| y = ∂ z ˆf| y
∂ z x| y
(10)
– special case z = f
∂ x f| y = 1
∂ f x| y
derivatives respect to the remaining variable
– (partial) differentiating (9) respect to y:
– using (10):
– rearranging:
– special case z = f
∂ y ˆf|z = ∂ x f| y · ∂ y x| z + ∂ y f| x
∂ y ˆf|z = ∂ z ˆf| y
∂ z x| y
· ∂ y x| z + ∂ y f| x
∂ y f| x = ∂ y ˆf|z − ∂ yx| z
∂ z x| y
· ∂ z ˆf|y (11)
∂ y f| x = −∂ y x| f · ∂ x f| y
8

Example: internal energy
– let us change the variable space from the standard (E, V, M) to (S, V, M)
– the Gibbs relation remains valid if we change the T (E, V, M) function
to T (S, V, M) and so on, so the internal energy is a potential of the
(T, −p, µ) functions:
∂E
∂S
∣ = T ;
V,M
∂E
∂V
∣ = −p;
S,M
∂E
∂M ∣ = µ (12)
S,V
– in the previous notations x = E, z = S and the ,,y-s” are V and M
– let f be p(E, V, M), therefore (10) becomes:
∂ E p| V,M = ∂ Sp| V,M
∂ S E| V,M
using (12):
– now applying (11):
∂ E p| V,M = ∂ Sp| V,M
T
∂ V p| E,M = ∂ V p| S,M − ∂ V E| S,M
∂ S E| V,M
· ∂ S p| V,M
using (12):
∂ V p| E,M = ∂ V p| S,M + p T · ∂ Sp| V,M
2.2.2 New quantities with Legendre transformation
introduction of the Legendre transform
– given a function f(x, y)
– the variable x is changed to z with the substitution x = x(z, y)
– the Legendre transform of f(x, y) is g(z, y) which is determined by the
following two properties:
g(z, y) + x(z, y) · z = f ( x(z, y), y ) (13)
∂ z g| y = −x(z, y) (14)
9

derivatives respect to the new variable
– (partial) differentiating (13) respect to z
– using (14):
∂ z g| y + x(z, y) + z · ∂ z x| y = ∂ x f| y · ∂ z x| y
so z can be also considered as a z(x, y) function
z = ∂ x f| y (15)
– from (14) and (15) comes that the derivatives of f and g are inverse
functions
– x and z are called conjugated variables
derivatives respect to the variable, that is no transformed
– (partial) differentiating (13) respect to y:
– using (15):
∂ y g| z + ∂ y x| z · z = ∂ x f| y · ∂ y x| z + ∂ y f| x
∂ y g| z = ∂ y f| x (16)
the exact equality is valid if the x(z, y) substitution is done on the right
side after the derivation
– thus the Legendre transformation keeps constant the derivatives respect
to the remaining variables
example: free energy
– let us start with the f = E(S, V, M) function and x = E, then according
to (15):
z = ∂ S E| V,M
due to (12):
z = T
– let the new function be F (T, V, M) and called Helmholtz free energy,
then the formula of the Legendre transformation in (13) becomes:
F + ST = E
so the definition of the free energy is:
F = E − T S (17)
10

– substituting to the potential relation (8) :
(potential relation for free energy)
– differentiating (17)
– using the Gibbs-relation (6):
F = −pV + µM
dF = dE − d(T S) = dE − SdT − T dS
dF = T dS − pdV + µdM − SdT − T dS
(Gibbs relation for free energy)
dF = −SdT − pdV + µdM
– thus after the S ↔ T interchange the free energy is an F (T, V, M) function,
which is a potential of the S(T, V, M), p(T, V, M) and µ(T, V, M)
functions:
∂F
∂T
∣ = −S;
V,M
∂F
∂V
(we can check the validity of (16))
∣ = −p;
T,M
∂F
∂M ∣ = µ
T,V
– it can be checked easily that F is not an extensive function over its
(T, V, M) variable space, but it is extensive if we transform the variables
to the standard (E, V, M) variables
– the whole procedure can be applied similarly to get the enthalpy (H)
and the Gibbs free energy (G), the results can be found in the tables,
E, F, H and G are called thermodynamic potentials
quantity definition potential relation
energy - E = T S − pV + µM
Helmholtz free energy F := E − T S F = −pV + µM
enthalpy H := E + pV H = ST + µM
Gibbs free energy G := E − T S + pV G = µM
Table 1: Important thermodynamic potentials - definitions
11

quantity variables Gibbs relation
energy E(S, V, M) dE = T dS − pdV + µdM
Helmholtz free energy F (T, V, M) dF = −SdT − pdV + µdM
enthalpy H(S, p, M) dH = T dS + V dp + µdM
Gibbs free energy G(T, p, M) dG = −SdT + V dp + µdM
Table 2: Important thermodynamic potentials - variables
2.3 Densities, specific quantities
– the extensive and intensive properties are constraints
– one of the extensive variables can be eliminated
– it can be useful to choose a reference variable
– usually the volume (densities) and the mass (specific quantities) are
used for reference
2.3.1 Densities
transformation of the basic state variables
– energy density:
ε := E V
– mass density:
ρ := M V
transformation of the state functions
– the (E, V, M) state space can be simplified to (ε, ρ)
– considering an extensive state function:
( E
ˆf(ε, ρ) := f (ε, 1, ρ) = f
V , 1, M )
= V f ( E
, 1, )
M
V V
=
V V
f(E, V, M)
V
specially:
Ê = ε;
ˆM = ρ
12

– considering an intensive state function:
( E
ĝ(ε, ρ) := g (ε, 1, ρ) = g
V , 1, M V
)
= g(E, V, M)
– the functions over the new state space (ε, ρ) are intensive functions
potential relation for densities:
– let us divide (8) by V :
E
V = T S V − p + µM V
– using the definition of the densities:
Gibbs relation for densities:
– let us substitute the densities to (6):
ε = T s − p + µρ (18)
d(εV ) = T d(sV ) − pdV + µd(ρV )
– using the derivation rule d(xy) = xdy + ydx:
V dε + εdV = T V ds + T sdV − pdV + µV dρ + µρdV
dV (ε − T s + p − µρ) + V (dε − T ds − µdρ) = 0
– the expression in the first bracket is zero due to (18), therefore the
Gibbs relation for densities:
2.3.2 Specific quantities
transformation of the basic state variables
– specific energy:
– specific volume:
dε = T ds + µdρ (19)
e := E M
v := V M
13

transformation of the state functions
– the (E, V, M) state space is simplified to (e, v)
– considering an extensive state function:
˜f(e, v) := f (e, v, 1) = f
specially:
( E
M , V M , 1 )
= Mf ( E
M , V M , 1)
M
Ẽ = e;
Ṽ = v
=
f(E, V, M)
M
– considering an intensive state function:
( E
˜g(e, v) := g (e, v, 1) = g
M , V )
M , 1 = g(E, V, M)
– the functions over the state space (e, v) are neither extensive not intensives
potential relation for specific quantities:
– let us divide (8) by M:
– using the definitions:
E
M = T S M − p V M + µ
e = T s − pv + µ (20)
Gibbs relation for specific quantities:
– let us substitute the specific quantities to (6):
d(eM) = T d(sM) − pd(vM) + µdM
– using the derivation rule:
Mde + edM = T vds + T sdM − pvdM − pMdv + µdM
dM(ε − T s + pv − µ) + M(de − T ds + pdv) = 0
– from (20) the first bracket equals zero, thus the Gibbs relation:
de = T ds − pdv (21)
14

extensive quantity (A) V M E S F G H
density (A/V ) 1 ρ ε s
specific quantity (A/M) v 1 e s f g h
Table 3: Densities and specific quantities
2.4 Simple model with specific quantities
2.4.1 Reformulating the model with specific quantities
Definition(discrete thermostatic body for specific quantities):
– a set of { } T (e, v)
p(e, v)
state functions over the (e, v) state space
– for that there exists a s(e, v) state function called specific entropy, which
has to fulfil the following properties:
(B 1 ) (specific entropy is a potential)
∂s
∂e∣ = 1 v
T ; ∂s
∂v ∣ = p
e
T
(22)
(B 2 ) (specific entropy is increasing with energy)
∂s
∂e∣ > 0 (23)
v
remarks:
(B 3 ) (specific entropy is concave)
D 2 s(e, v) is a negative definite matrix (24)
– if the mass M is constant, this and the previous models are equivalent
– the property corresponding to (P 3 ) is missing, this property was used
to reduce the three variables to two
– we do not require the complicated extensive and intensive properties
15

2.4.2 Examining the constitutive equations
motivation
– T (e, v) and p(e, v) functions are often given as empirical approximations
– it is useful to check if they fulfil the propertys of thermostatics
– sometimes the propertys can be used to make restrictions to the parameters
of the model
Condition of the potential property of entropy
– if the entropy function is twice differentiable then the mixed second
partial derivatives must be equal
∂ 2 s(e, v)
∂v∂e
= ∂2 s(e, v)
∂e∂v
– using property (B 1 ):
( )
∂
1
∂v ∣ = ∂ ( )
p(e, v)
e
T (e, v) ∂e∣ v
T (e, v)
(25)
– this condition is useful to check the T and p functions against the (B 1 )
property, without calculating the entropy function
– if (25) is not valid, then there exists no s(e, v) function, if the condition
fulfils, then s exists and the gas described by T and p functions is called
entropic
Condition of the increasing property of entropy
– property (B 2 ) is simple, it requires only
∂s
∂e∣ > 0
v
– if the property (B 1 ) is substituted, a very simple condition is obtained:
T (e, v) > 0 (26)
therefore the range of the temperature function must be positive
16

Condition of the concave property of specific entropy
– the condition of property (B 2 ) is called thermodynamic stability, which
requires the concavity of the specific entropy function
– the concavity requires that the Hesse-matrix of the specific entropy
function, this D 2 s(e, v) has to be negative definite (or −D 2 s(e, v) positive
definite) whenever the function is two times continuously differentiable
– the positive definiteness of a symmetric matrix can be investigated by
the Sylvester criteria, a matrix
[
a11 a 12
a 12 a 22
]
is positive definite if and only if a 11 > 0 and a 11 a 22 − a 2 12 > 0
– applying to the entropy function:
⎡
= ⎣
]
p
T
Ds = [ ] [
∂ e s| v ∂ v s| e = 1
T
⎡
−D 2 s = ⎣ −∂ ( 1
) ∣ (
∣v
e −∂ 1
) ∣ ⎤
∣e
T
v T
(
−∂ p
) ∣ (
∣v
e −∂ p
) ∣ ⎦
∣e
=
T v T
1
1
∂
T 2 e T | v ∂
T 2 v T | e
− 1 T ∂ ep| v + p
T 2 ∂ e T | v
⎤
− 1 ∂ ⎦
T vp| e + p ∂
T 2 v T | e
– using the a 11 > 0 condition:
1
T 2 ∂ eT | v > 0
– using the a 11 a 22 − a 2 12 > 0 condition:
∂ e T | v > 0 (27)
1
T 4 (
∂e T | v (p ∂ v T | e − T ∂ v p| e ) − ∂ v T | e (p ∂ e T | v − T ∂ e p| v ) ) > 0
1 ( )
∂v T |
T 3 e ∂ e p| v − ∂ v p| e ∂ e T | v > 0
− ∂ (
eT | v
∂
T 3 v p| e − ∂ )
vT | e
∂ e p| v > 0
∂ e T | v
17

– the bracket equals to ∂ v p| T because of (11), therefore:
− ∂ eT | v
T 3 ∂ v p| T > 0
– ∂ e T | v is positive due to (27) and T 3 is positive due to (26), thus
∂ v p| T < 0 (28)
– (27) and (28) are a particular example of the le Chatelier-Brown principle.
2.4.3 Examples
ideal gas
– the thermic equation of state:
– the caloric equation of state:
pv = RT
e = cT
– from that the standard state functions:
– the condition of (25):
T (e, v) = e c
p(e, v) = Re
cv
( )
∂
1
∂v ∣ = 0
e
T (e, v)
( )
∂
p(e, v)
∂e∣ = 0
v
T (e, v)
they are equal, hence the ideal gas model is entropic
– the condition (26): because the energy (e) is positive by definition, the
condition for the positive temperature is:
18
c > 0

– the condition (27):
– the condition (28):
from (26) we get:
∂ e T | v = 1 c > 0
c > 0
p(T, v) = RT
v
∂ v p| T = − RT
v 2
R > 0
– therefore if c > 0 and R > 0 then an ideal gas fulfils the properties of
thermostatics
van der Waals gas
– the termic equation of state is:
– the caloric equation of state is:
p = RT
v − b − a v 2
T = e c + a cv
– therefore the standard state functions:
– the condition (25):
p(e, v) =
p
T =
T (e, v) = e c + a cv
Re
c(v − b) + Ra
cv(v − b) − a v 2
1
T =
c
e + a v
R
v − b − a v · 2
e + a v
it can be checked that
( )
∂
1
∂v ∣ = ∂ ( )
p(e, v)
e
T (e, v) ∂e∣ =
v
T (e, v)
therefore the van der Waals gas model is entropic
19
c
ac
(ev + a) 2

– the condition (26):
e
c + a cv > 0
if c > 0 then ev > −a, if c < 0 then ev < a
– the condition (27):
∂ e T | v = 1 c > 0
c > 0
– the condition (28):
∂ v p| T = −
RT
(v − b) + 2a
2 v < 0 3
– Therefore the van der Waals gas body fulfils (B 2 ) and (B 3 ) properties
for a given domain of (e, v) restricted by the parameters.
20