1.1 Equation of state of a perfect gas

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1 THERMODYNAMICS

Lectures on Thermodynamics

1 Thermodynamics

Thermodynamics studies the macroscopic behavior of the many-body systems and describes them in terms of macroscopic parameters characterizing the system which are temperature T , pressure P and volume V . Equation of state is the relation

among these parameters and is written as f (P, T, V ) = 0

The basic assumption is that such an equation always exists. Dierent functions describing thermodynamical systems are divided in two groups:

• functions of state which are complete dierentials and depend only on initial

and nal state. I

df (x) = 0

∫ 2 1

df (x) = f (2)− f(1) Examples are thermodynamical potentials (see later).

• functions of process which are not complete dierentials and depend on the path taken between initial and nal state.

I

d^′f (x)̸= 0 Examples are work and heat.

Let us look at the work done during an isothermal T = const. and an isobaric P = const. process. The work is

A^′_T =

∫ _V₂

V1

n R T

V dV = n R T lnV₂ V₁ A^′_P =

∫ _V₂

V1

P dV = P (V₂− V1) A^′_T ̸= A^′P

Thermodynamical potentials are basic functions of state expressed in terms of conjugate coordinates P, V and T, S. Transformation from one coordinate to

another is know as Laplace transform. They are:

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1 THERMODYNAMICS

• Internal energy U(S, V )

dU = d^′Q− dA^′ = T dS− P dV dU (S, V ) =

(∂U

∂S )

V

dS + (∂U

∂V )

S

dV T =

(∂U

∂S )

V

P =− (∂U

∂V )

S

During an adiabatic ∆Q = 0 process

A|Q =−∆U = Ui− Uf

which means that the work done corresponds to the change of internal energy.

In an isochoric V = const. one nds dU_V ≡ dQV = c_V dT

c_V ≡ (∂U

∂T )

V

= (∂S

∂T )

V

(∂U

∂S )

V

cV = T (∂S

∂T )

V

• Enthalpy H(S, P ) is Laplace transform of U(S, V ) in conjugate coordinates V → P

H = U + P V = F + T S + P V dH = T dS + V dP

T = (∂H

∂S )

P

V = (∂H

∂P )

S

During an isobaric P = const. process

Q|P = ∆H|P = H_f − Hi

which means that the exchange of heat corresponds to the change of enthalpy and

dH_P ≡ dQP = c_PdT cP =

(∂H

∂T )

P

= T (∂S

∂T )

P

Thus, in an isobaric process, enthalpy plays the same role as internal energy in an isochoric process.

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1 THERMODYNAMICS

• Helmoltz free energy F (T, V ) is Laplace transform of U(S, V ) in conjugate coordinates S → T

F = U − T S dF =−S dT − P dV

S =− (∂F

∂T )

V

P =− (∂F

∂V ) ( T

∂S

∂V )

T

=−

( ∂²F

∂T ∂V ) ( T,V

∂P

∂T )

V

=−

( ∂²F

∂T ∂V ) ( T,V

∂P

∂T )

V

= (∂S

∂V )

T

in an reversible isothermal process dF = −P dV so that A_T = F_i − Ff

and the work extracted from the system is equal to the loss of free energy. In case of irreversible process one has

A_T ≤ Fi− Ff

• Gibbs free energy G(T, P ) is Laplace transform of F (T, V ) in conjugate co- ordinates V → P

G = F + P V dG =−S dT + V dP

S = − (∂G

∂T )

P

V = (∂G

∂P )

T

Under constant T and P the Gibbs function satises d G

d t ≤ 0

and is decrising in time reaching its minimum at the thermodynamic equilibrium.

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1.1 Equation of state of a perfect gas 1 THERMODYNAMICS

1.1 Equation of state of a perfect gas

Perfect or ideal gas is a low pressure approximation of a real gas with the following characteristics:

1. molecules are point-like structures 2. no interaction among particles of gas

Equation of state of a real gas can be obtained combining empirical Boyle-Mariott and Guy-Lisac laws

P V = const. T = const.

P = P₀(1 + α_P∆t) V = const.

V = V₀(1 + α_V ∆t) P = const.

α_P = α_V = 1 273 K P = P₀

T₀ (T₀+ T₀α_P (T − T0)) = P₀ T₀T P

T = P₀ T₀ V T = V0

T₀

Let us take a gas from the state P1, V₁, T₁ to a state P1, V₂^′, T₂. Guy-Lisac gives V₂^′ = T₂

T₁V₁

now we take it from P1, V₂^′, T₂ ot the nal stateP2, V₂, T₂. Boyle-Marott gives P₁V₂ = P₂V₂

combining together the two results one nds P₁V₁

T₁ = P₂V₂ T₂

leading to the perfect gas law

P · V = N k T (1)

where N is the number of molecules and k = 1, 38 · 10⁻²³J/K is the Boltzman constant. It is much easier to calculate the number of moles than a number of

molecules for a gas so it is convenient to rewrite the gas equation as follows P · V = N k T = n NAk T

R = N_Ak P · V = n R T

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1.1 Equation of state of a perfect gas 1 THERMODYNAMICS

where n is number of moles, NA= 6, 02· 10²³ Avogadro's number and R = N_Ak = 8, 3· J/K is gas constant.

• Internal energy of perfect gas does not depend on volume (Joule's law) so

dU = (∂U

∂T )

V

dT + (∂U

∂V )

T

dV = c_V dT U = U₀+ c_V T

• Enthalpy

H = U + P V = U₀+ c_V T + nRT H = c_PT + U₀

• The entropy of ideal gas is dS(T, V ) = ∂S

∂TdT + ∂S

∂V dV dS(T, V ) = c_VdT

T +∂P

∂TdV dS(T, V ) = cV

dT

T +nR V dV S(T, V ) = S₀+ c_V ln T

T0

+ nR ln V V0

S(T, P ) = S0+ (cP − nR) ln T

T₀ + nR ln V

V₀ = S0+ cPln T

T₀ − nR ln P P₀ S(V, P ) = S₀+ c_P ln V

V₀ + c_V ln P P₀

• Helmoltz free energy

F = U − T S = U0 + cV T − T (

S0+ cV ln T

T₀ + nR ln V V₀

)

F = cV T (

1− ln T T₀

)

− nR lnV

V₀ − S0T + U0

• Gibbs free energy

G = U − T S + P V = U0+ c_V T − T (

S₀+ c_Pln T

T₀ + nR ln P P₀

)

+ nRT G = c_PT

(

1− ln T T₀

)

− nR ln P

P₀ − S0T + U₀

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1.2 PV diagrams 1 THERMODYNAMICS

1.2 PV diagrams

These diagrams describe work performed A =

∫ P dV A|T =

∫

P dV = nRT

∫ _V₂

V1

dV /V = nRT lnV₂ V₁ A|P =

∫

P dV = P (V₂− V1) A|V = 0

1.3 TS diagrams

It is usual to dene γ = cP/c_V, c_V = nR/γ− 1, cP = nRγ/γ− 1 and one can write

S− S0 = (cP − cV) ln V

V₀ + cV ln T T₀ S− S0 = ln

(V V₀

)cP ( P P₀

)cV

= ln (V

V₀

)nRγ/γ−1( P P₀

)nR/γ−1

S− S0 = nR γ− 1ln

(V V₀

)γ( P P₀

)

P V^γ = const.e^(γ^−1)S/nR T V^γ⁻¹ = const.e^(γ^−1)S/nR T P^γ^−1/γ = const.e^(γ^−1)S/nRγ

thus one obtains T, S diagrams describing the exchange of heat as

∆Q =

∫ S2

S1

T (S)dS T (S)|V = const.e^(γ−1)S/nR T (S)|P = const.e^(γ^−1)S/nRγ

1.4 Adiabatic processes

A thermodynamical transformation is called adiabatic if no heat is exchanged during the process dQ = 0. A real transformation taking place very quickly is a good approximation of an adiabatic process. The work is done on the expenses of

internal energy

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1.5 Polytropic equation 1 THERMODYNAMICS

0 = c_VdT + P dV dV

V =−c_V nR

dT T

0 = d ln (T

T₀ ) (V

V₀ )γ−1

T V^γ⁻¹ = const.

P V^γ = const.

T P⁽¹^−γ)/γ = const.

The work in an adiabatic process is

A = const.

∫ _V₂

V1

dV /V^γ = const.

1− γV₁¹^−γ [

1− (V₁

V₂

)γ−1]

(2) const. V₁^γ⁻¹ = P₁V₁ = nRT₁ (3)

A = nRT₁ γ− 1

( 1− T₂

T1

)

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1.5 Polytropic equation

This is a process during which the heat capacity, dened in general as dQ = c dT , remains constant. From the rst law and ideal gas law one has

c dT = c_V dT + P dV P dV + V dP = RdT

P dV + V dP = R P dV c− cV

(c− cP)dV

V = (c− cV)dP P P Vⁿ= const.

n = c− cP

c− cV

c = nc_V − cP

n− 1

n is known as polytropic exponent. One can have the following situations

• n = 0 isobaric process c = cP

P = const.

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2 VAN DER WAALS EQUATION FOR REAL GAS

• n = ∞ isochoric process c = cV

V = const.

• n = 1 isothermal process c = ∞

T = const.

n = c_P/c_V adiabatic process c = 0

dQ = 0

Figure 1: PV graph of a polytropic process while T, S graph is obtained from T = const.e⁽ⁿ^−1)S/nc^V^−c^P = const.e^S/c

2 Van der Waals equation for real gas

Real gas has interacting molecules of nite size and these facts have to be taken into account. First of all the real volume available to the molecules is less than V

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due to the volume of interaction of each molecule b. This volume corresponds to the half (facing incoming molecule) of cross section of two molecules given by

b = 1 2N4π

3 (2R)² = 16 3 N π R³ so one nds

P (V − b) = nR T

this is known as Clausius equation. Now take into account intermolecular interactions of order 1/r⁶ (Van der Waals forces) which reduce pressure of gas due to the density of molecules in the last two layers of gas. This net force decreases the

force exerted onto the wall by the particles in the surface layer. The net force on a surface particle, pulling it into the container, is proportional to the number density

C = N/V

The number of particles in the surface layers is, again by assuming homogeneity, also proportional to the density. In total, the force on the walls is decreased by a factor proportional to the square of the density, and the pressure (force per unit

surface) is decreased by

a C² = a (N

V )2

Thus, modied equation is [

P + a (N

V )2]

(V − b) = nR T

if we introduce abbreviations as a = a NA², b = b N_A/N the equation can be written as

R T = (

P + a V²

)

(V − b) P = R T

V − b − a V²

which is Van der Waals equation for one mole. For large volume V → ∞ it becomes the equation of perfect gas. VDW equation is cubic equation having an inexion

point characterized by critical values of Tc, Pc, Vc. Critical values correspond to

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coinciding roots of the cubic equation V1 = V₂ = V₃ that can be found as follows.

0 = (V − Vc)³

0 = V³− 3V²V_c+ 3V V_c²− V_c³ 0 = V³−

(R T P + b

)

V²+ V a P − ab

P 3V_c= R T_c

P_c + b 3V_c² = a

P_c V_c³ = ab Pc

T_c= 8 a 27 b R P_c= a

27 b² Vc= 3 b

VDW equation can be written in reduced form which is independent of type of gas involved. Introduce notation t = T/Tc, p = P/P_c, v = V /V_c to nd

p = 8 t

3 v− 1 − 3 v²

Figure 2: PV graph of Van der Waals gas

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3 STATICS AND DYNAMICS OF A RIGID BODY

3 Statics and dynamics of a rigid body

Rigid body is dened as a collection of particles (atoms and molecules) that maintain xed position with respect to each other. Therefore, a rigid body moves as

a whole, rigid object. To every rigid body we associate a point called centre of mass, where we can consider concentrated the total mass of the object. In other words, the extended object can be replaced by its centre of mass. The motion of a

rigid body can thus be split into a motion of its centre of mass and a motion around an axis passing through its centre of mass. The centre of mass is dened as

⃗r_CM =

∑

nm_n⃗r_n M_tot

⃗r_CM = 1 M

∫

⃗r ρ dV

Example: The position of centre of mass of a circular section of angle φ0.

⃗rCM = 1 M

∫

⃗r ρ dV

⃗r_CM = ρ M

∫ R 0

r²dr

∫ φ0

0

(

cos φ⃗i + sin φ⃗j )

dφ

⃗r_CM = ρ R³ 3 M

(

sin φ₀⃗i + (1− cos φ0) ⃗j )

M = ρ

∫ _R

0

r dr

∫ _φ₀

0

dφ = ρR² 2 φ₀

⃗r_CM = 2 R 3φ₀

(

sin φ₀⃗i + (1− cos φ0) ⃗j )

We can consider translation and rotation of a rigid body. Physical quantities characterizing translation are linear momentum ⃗p = m⃗v and force ⃗F = d ⃗p/d t.

Rotation is characterized by angular momentum ⃗L = ⃗r × ⃗p and by the momentum of force or twisting force known as torque (Latin torquere=to twist) M⃗ ≡ ⃗r × ⃗F = d⃗L/d t. The table below put in relation various quantities describing

translation and rotation

Translation Rotation

m-mass I-moment of inertia

⃗

p = m⃗v-linear momentum L = I⃗⃗ ω-angular momentum F = d⃗⃗ p/dt-force M = d⃗⃗ L/dt-torque EK = m v²/2-kinetic energy EK = Iω²/2-kinetic energy

One can see that the quantities describing rotation do not depend only on mass but also on the distance from the axis of rotation.

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The moment of inertia of a rigid body is dened as

I = lim

n→∞

∑n i=1

∆ mi⃗r²_i =

∫

⃗r²ρ d V

where ρ is the volume density of the substance. The angular momentum is Examples:

1. The moment of inertia of a circular section of angle φ0. I =

∫

⃗ r²ρ dV I = ρ

∫ R 0

r³dr

∫ φ0

0

dφ = ρφ₀R⁴ 4 M = ρπ R²

M_φ = ρ R²φ₀ 2 = φ₀

2π M Iφ0 = φ₀

4πM R²

2. The moment of inertia of a circular ring of width R2− R1. I = I_R₂ − IR1 = 1

2

(M₂R²₂− M1R²₁)

M₁ = ρπ R²₁ M₂ = ρπ R²₂

M = M₂− M1 = M1

R²₁

(R²₂− R1²

)

I = 1 2ρπ(

R⁴₂− R⁴1

)

I = 1 2M(

R²₂+ R²₁)

3. Moment of inertia of a cylinder of radius R and hight h around an axis z passing through its centre

I_z = ρ

∫

r²dV = ρ

∫ _R

0

r³dr

∫ _2π

0

dφ

∫ _h/2

−h/2

dz I_z = ρπ

2R⁴h M = ρπ R²h

I_z = 1 2M R²

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4. Moment of inertia of a cylinder of radius R and hight h around an axis x passing through the centre of its height h

I_x = ρ∫ (

z² + y²)

dV = ρ

∫ _R

0

r dr

∫ _2π

0

dφ

∫ _h/2

−h/2

(r²sin²φ + z²) dz

I_x = ρ

∫ _R

0

r dr

∫ _h/2

−h/2

d z(

r²π + 2π z²)

I_x = ρ

∫ h/2

−h/2

d z (π

4R⁴+ π z²R² )

= ρπ 4R²h

(

R²+ h² 3

)

M = ρπ R²h I_x = 1

4M (

R²+ h² 3

)

To show the advantage of choosing a proper basis for calculation let us calculate the area of a circle in Cartesian and polar coordinates.

R² = x²+ y²circle in Cartesian coordinates A =

∫

dx dy =

∫ _R/2

−R/2

dx

∫ ^√_R²_−x²_/2

−√

R²−x²/2

dy = 2

∫ _R/2

−R/2

√R²− x²dx

A =|x = R sin α| = 2 R

∫ π/2

−π/2

sin²α dα = R² (

α + sin 2α 2

)

|^π/2_−π/2 A = π R²

R = const. circle in polar coordinates A =

∫ R 0

r dr

∫ 2π 0

dφ = π R²

General form of moment of inertia following from the denition of angular momentum is

L = m ⃗⃗ r× ⃗vt= m ⃗r× (⃗ω × ⃗r) = m

∑3 i=1

(x²_i ⃗ω− (⃗ω · ⃗r) ⃗xi

)

L = m⃗ [

⃗ ω(

x²+ y²+ z²)

− (ωx· x + ωyy + ω_zz) ⃗x_i] L = m⃗

[ ωx

(y²+ z²)

⃗i + ωy

(x²+ z²)

⃗j + ωx

(y²+ x²)

⃗k

− (ωyy + ω_zz) x⃗i− (ωxx + ω_zz) y ⃗j− (ωyy + ω_xx) z ⃗k ]

L_x = m(

y² + z²)

ω_x− m x y ωy− m x z ωz = I_xxω_x− Ixyω_y− Ixzω_z L_y = m(

x²+ z²)

ω_y − m y x ωx− m y z ωz = I_yyω_y− Iyxω_x− Iyzω_z L_z = m(

y² + x²)

ω_z − m z y ωy− m z x ωx = I_zzω_z− Izyω_y− Izxω_x

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One can see that the moments of inertia are components of a second-rank tensor (matrix)



 I_xx I_xy I_xz I_yx I_yy I_yz I_zx I_zy I_zz





Fortunately, it is always possible to choose three axis in a way that the mixed momenta are zero. These axis are known as principal axis and coincide with axis

of symmetry of the rigid body. The torque is

M =⃗ d⃗L

d t =∑

i

I_ii (dω_i

d t⃗e_i+d⃗e_i d tω_i

)

d⃗e_i

d t = ⃗ω× ⃗ei

M = I⃗ · ⃗α + ⃗ω × ⃗L → Euler equations

in the principal axis system one has M_x = I_xω˙_x+ (I_z− Iy) ω_yω_z M_y = I_yω˙_y+ (I_x− Iz) ω_zω_x M_z = I_zω˙_z+ (I_y − Ix) ω_xω_y

To dene equilibrium conditions for translation and rotation one requires F⃗_tot ≡ ⃗F1+ ⃗F₂+· · · + ⃗Fn = 0

M⃗_tot ≡ ⃗M₁+ ⃗M₂+· · · + ⃗M_n = 0

The above condition lead also to the conservation of linear and angular momentum, which, together with conservation of energy represent fundamental

conservation laws of physics F⃗_tot = 0 ⇒ ⃗ptot = const..

M⃗tot = 0 ⇒ ⃗Ltot = const..

Torque can be written in terms of of the components of force ⃗F and its arm ⃗r in the Cartesian coordinate system as

M = ⃗⃗ r× ⃗F M_x = y F_z− z Fy

M_y = z F_x− x Fz

M_z = x F_y− y Fx

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Assume that we have two forces ⃗F1, ⃗F₂, in the y, z plane, producing two torques M⃗₁, ⃗M₂ around a centre of rotation O having an arm ⃗r. The resulting (total) torque

is M⃗_R = ⃗M₁+ ⃗M₂

M_x,R = y F_z,1− z Fy,1+ y F_z,2− z Fy,2

Mx,R = y (Fz,1+ Fz,2)− z (Fy,1+ Fy,2) M_x,R = y F_R,z− z FR,y = M_x,1+ M_x,2

this simple property of vectors goes under the name of Varignon theorem in civil engineering. The proof is based on equivalence of areas which simply means that the geometrical meaning of a cross-product is an area formed by the two vectors in

the product.