MS212 Thermodynamics of Materials (소재열역학의이해)

Lecture Note: Chapter 2

Byungha Shin (신병하)Dept. of MSE, KAIST

1

2017 Spring Semester

Course InformationSyllabus

Chap 1. Introduction and Definition of Terms (1 lecture)Chap 2. The First Law of Thermodynamics (1.5 lectures)Chap 3. The Second Law of Thermodynamics (3 lectures)Chap 4. The Statistical Interpretation of Entropy (2 lectures)Chap 5. Auxiliary Functions (2.5 lectures)Chap 6. Heat Capacity, Enthalpy, Entropy and the Third

Law of Thermodynamics (3 lectures)Chap 7. Phase Equilibrium in a One-Component System (3 lectures)Chap 8. The Behavior of Gases (3 lectures)Chap 9. The Behavior of Solutions (3 lectures)Chap 10. Gibbs Free Energy Composition and Phase Diagram (3 lectures)

• Heat and work are two different forms of energy• Energy: “the ability to do work” or “the ability to exert an action on

surroundings”

Energy

Energy transfer results from displacement under a force

Work

dW = (force)(displacement)• Mechanical work dWmechanical = F ∙ dX• Electrostatic work dWelectrostatic = dq (V / L) ∙ L = V ∙ dq• Magnetic work dWmagnetic = H ∙ dM

a

bm

Mechanical work (in mechanics)

�⃑�𝐹𝑒𝑒𝑒𝑒𝑒𝑒

𝑑𝑑𝑙𝑙

work done by the surrounding on the system

𝑑𝑑𝑤𝑤𝑚𝑚𝑒𝑒𝑚𝑚𝑚 = �⃑�𝐹𝑒𝑒𝑒𝑒𝑒𝑒 � 𝑑𝑑𝑙𝑙

𝑤𝑤𝑚𝑚𝑒𝑒𝑚𝑚𝑚 = �𝑎𝑎

𝑏𝑏𝑑𝑑𝑤𝑤𝑚𝑚𝑒𝑒𝑚𝑚𝑚 = �

𝑎𝑎

𝑏𝑏�⃑�𝐹𝑒𝑒𝑒𝑒𝑒𝑒 � 𝑑𝑑𝑙𝑙

For w, �⃑�𝐹𝑒𝑒𝑒𝑒𝑒𝑒 should be known as a function of positions on the path w is path-dependent, not state function, 𝜹𝜹𝑤𝑤𝑚𝑚𝑒𝑒𝑚𝑚𝑚 = �⃑�𝐹𝑒𝑒𝑒𝑒𝑒𝑒 � 𝑑𝑑𝑙𝑙

WorkSign convention of work in engineering thermodynamics (opposite to the sign convention of mechanics)• Work done by the system on the surrounding >0• Work done by the surrounding on the system <0

Raising work (system = an object of mass, m)

m

m h1

h2 𝑤𝑤𝑚𝑚𝑒𝑒𝑚𝑚𝑚 = ��⃑�𝐹𝑒𝑒𝑒𝑒𝑒𝑒 � 𝑑𝑑𝑙𝑙 = �𝑚1

𝑚2𝑚𝑚𝑚𝑚𝑑𝑑𝑚𝑚 = 𝑚𝑚𝑚𝑚 ℎ2 − ℎ1 > 0

�⃑�𝐹𝑒𝑒𝑒𝑒𝑒𝑒 = 𝑚𝑚𝑚𝑚�̂�𝑚

(work done by the surrounding on the system)

𝒘𝒘 = −𝒘𝒘𝒎𝒎𝒎𝒎𝒎𝒎𝒎𝒎 = −𝒎𝒎𝒎𝒎 𝒎𝒎𝟐𝟐 − 𝒎𝒎𝟏𝟏 < 𝟎𝟎

(work done by the system on the surrounding)

WorkExtension work (system = an elastic rod)

L L+ΔL

𝐹𝐹𝑒𝑒𝑒𝑒𝑒𝑒 = 𝑇𝑇𝑤𝑤𝑚𝑚𝑒𝑒𝑚𝑚𝑚 = 𝑇𝑇∆𝐿𝐿 > 0(work done by the surrounding on the system)

𝒘𝒘 = −𝒘𝒘𝒎𝒎𝒎𝒎𝒎𝒎𝒎𝒎 = −𝑻𝑻∆𝑳𝑳 < 𝟎𝟎(work done by the system on the surrounding)

P

dz

area = A

�⃑�𝐹𝑒𝑒𝑒𝑒𝑒𝑒 = 𝑃𝑃𝑃𝑃 −�̂�𝑚 ,𝑑𝑑𝑙𝑙 = 𝑑𝑑𝑚𝑚

PV work (system = gas inside the cylinder under the piston)

𝛿𝛿𝑤𝑤𝑚𝑚𝑒𝑒𝑚𝑚𝑚 = −𝑃𝑃𝑃𝑃𝑑𝑑𝑚𝑚 = −𝑃𝑃𝑑𝑑𝑃𝑃(work done by the surrounding on the system)

(< 0, if dV > 0)

𝒘𝒘 = −𝒘𝒘𝒎𝒎𝒎𝒎𝒎𝒎𝒎𝒎 = 𝑷𝑷∆𝑽𝑽 (> 0, if dV > 0)

(work done by the system on on surrounding)

Equivalence of Work and Heat: James Joule’s experiment (1840)

Heat

wmech = mg(h2 – h1)(Work done by the weight on the system)

Immoveable,adiabatic wall

Same change of the state (T1 T2) can be also achieved by with no work.

• Observation: change of the state T1 T2

• Conclusion: ∆T = T2 – T1 = α wmech = α mg(h2 – h1)

Equivalence of Work and Heat: James Joule’s experiment (1840)

Heat

• Observation: change of the state T1 T2

• Conclusion: ∆T = T2 – T1 = α wmech = α mg(h2 – h1) = q / C• We cannot distinguish the case of the same state change

1 J = 0.2389 cal, or 1 cal = 4.186 J(1 cal is the energy required to raise 1 g of water by 1 oC)

(V1, T2)

(V1, T1)↑ ΔT

q > 0

rigid, adiabatic wall

rigid (w=0), diathermal (q≠0) wall

wmech = mg(h2 – h1)(Work done by the weight on the system)

rigid (w=0),adiabatic (q=0) wall

The change in the internal energy of a closed system is equal to the amount of heat supplied to the system, minusthe amount of work done by the system on its surroundings.

The 1st Law of Thermodynamics

∆𝑈𝑈 = 𝑈𝑈2 − 𝑈𝑈1 = 𝑞𝑞 − 𝑤𝑤

if heat is transferred, to the system, >0out of the system, <0

if work is doneby the system, >0on the system, <0

In a differential form, 𝑑𝑑𝑈𝑈 = 𝛿𝛿𝑞𝑞 − 𝛿𝛿𝑤𝑤state function, path-indenpendent path-denpendent

�𝛿𝛿𝑞𝑞 ≠ 0; �𝛿𝛿𝑤𝑤 ≠ 0; � 𝛿𝛿𝑞𝑞 − 𝛿𝛿𝑤𝑤 = �𝛿𝛿 𝑞𝑞 − 𝑤𝑤 = 0

Internal Energy of a Substance, UKinetic energy + potential energy of all the particles constituting the system

Intermolecular PE Molecular KE

Translation of center of mass (c.m.) of molecule with respect to c.m. of small but macroscopic volume element

Intramolecular PE and KE

Note: all the other contribution except the I.M.P.E. are independent of the intermolecular spacing and therefore independent of the specific volume.

electronicnuclear

Total Energy of a Substance, EE = U + KE(macro) + PE(external)

KE(macro)

Macroscopic KE: Total KE is sub-classified into bulk (or macroscopic) & molecular; molecular is counted in U, macroscopic is counted here.

PE(external)

“Potential Energy”: Here we count only the PE associated with the interaction of the system with the surroundings, e.g.- Gravitational PE, - Electrostatic PE (charges or dipoles in external field) - Magnetic PE (magnetic dipoles in external field)

Note: all PE associated with internal interactions goes into U

Differential Forms of U

Choose whichever form appropriate to a given problem.

𝑑𝑑𝑈𝑈 =𝜕𝜕𝑈𝑈𝜕𝜕𝑇𝑇 𝑉𝑉

𝑑𝑑𝑇𝑇 +𝜕𝜕𝑈𝑈𝜕𝜕𝑃𝑃 𝑇𝑇

𝑑𝑑𝑃𝑃

𝑈𝑈 = 𝑈𝑈 𝑇𝑇,𝑃𝑃 = 𝑈𝑈 𝑃𝑃,𝑃𝑃 = 𝑈𝑈 𝑃𝑃,𝑇𝑇 = ⋯⋯

=𝜕𝜕𝑈𝑈𝜕𝜕𝑃𝑃 𝑃𝑃

𝑑𝑑𝑃𝑃 +𝜕𝜕𝑈𝑈𝜕𝜕𝑃𝑃 𝑉𝑉

𝑑𝑑𝑃𝑃

= ⋯⋯⋯⋯

=𝜕𝜕𝑈𝑈𝜕𝜕𝑃𝑃 𝑇𝑇

𝑑𝑑𝑃𝑃 +𝜕𝜕𝑈𝑈𝜕𝜕𝑇𝑇 𝑃𝑃

𝑑𝑑𝑇𝑇

Application of the 1st Law: Isochoric Process

Work

Heat

Heat Capacity (at constant V)

Internal Energy, U

Isochoric (or isometric): constant V or dV = 0; how? rigid wall

𝛿𝛿𝑤𝑤 = 𝑃𝑃𝑑𝑑𝑃𝑃 = 0, or 𝑤𝑤 = �𝑃𝑃𝑑𝑑𝑃𝑃 = 0

𝑑𝑑𝑈𝑈 =𝜕𝜕𝑈𝑈𝜕𝜕𝑇𝑇 𝑉𝑉

𝑑𝑑𝑇𝑇 +𝜕𝜕𝑈𝑈𝜕𝜕𝑃𝑃 𝑇𝑇

𝑑𝑑𝑃𝑃 =𝜕𝜕𝑈𝑈𝜕𝜕𝑇𝑇 𝑉𝑉

𝑑𝑑𝑇𝑇 = 𝛿𝛿𝑞𝑞)𝑉𝑉

Q: path-independent dU = path-dependent δq ??

𝜕𝜕𝑈𝑈𝜕𝜕𝑇𝑇 𝑉𝑉

=𝛿𝛿𝑞𝑞)𝑉𝑉𝑑𝑑𝑇𝑇

≡ 𝐶𝐶𝑉𝑉

𝑑𝑑𝑈𝑈 = 𝐶𝐶𝑉𝑉𝑑𝑑𝑇𝑇; ∆𝑈𝑈 = 𝑈𝑈2 − 𝑈𝑈1 = �𝑇𝑇1

𝑇𝑇2𝐶𝐶𝑉𝑉 𝑇𝑇 𝑑𝑑𝑇𝑇 = 𝑞𝑞

U, CV: extensive propertiesCV / unit mass: specific heat capacity; CV / # of mole: molar heat capacity

Application of the 1st Law: Isobaric Process

Work

Heat

Enthalpy

Heat Capacity (at constant P)

Isobaric: constant P or dP = 0; how? via wall movement (non-rigid boundary) + pressure reservoir

𝑤𝑤 = �1

2𝑃𝑃𝑑𝑑𝑃𝑃 = 𝑃𝑃 𝑃𝑃2 − 𝑃𝑃1 = 𝑃𝑃∆𝑃𝑃

𝑞𝑞)𝑃𝑃 = ∆𝑈𝑈 + 𝑃𝑃∆𝑃𝑃 = 𝑈𝑈2 − 𝑈𝑈1 + 𝑃𝑃 𝑃𝑃2 − 𝑃𝑃1 = ∆𝐻𝐻

𝐻𝐻 ≡ 𝑈𝑈 + 𝑃𝑃𝑃𝑃• Q: H is a state function. Why?• Meaning of H: energy needed to create a rabbit by a magician =

rabbit (U) + space for the rabbit (PV)

𝛿𝛿𝑞𝑞)𝑃𝑃 = 𝑑𝑑𝐻𝐻 =𝜕𝜕𝐻𝐻𝜕𝜕𝑇𝑇 𝑃𝑃

𝑑𝑑𝑇𝑇 +𝜕𝜕𝐻𝐻𝜕𝜕𝑃𝑃 𝑇𝑇

𝑑𝑑𝑃𝑃 =𝜕𝜕𝐻𝐻𝜕𝜕𝑇𝑇 𝑃𝑃

𝑑𝑑𝑇𝑇

𝜕𝜕𝐻𝐻𝜕𝜕𝑇𝑇 𝑃𝑃

=𝛿𝛿𝑞𝑞)𝑃𝑃𝑑𝑑𝑇𝑇

≡ 𝐶𝐶𝑝𝑝; 𝑞𝑞)𝑃𝑃 = �𝑇𝑇1

𝑇𝑇2𝐶𝐶𝑝𝑝 𝑇𝑇 𝑑𝑑𝑇𝑇

Heat Capacity

𝛿𝛿𝑞𝑞)𝑉𝑉 =𝜕𝜕𝑈𝑈𝜕𝜕𝑇𝑇 𝑉𝑉

𝑑𝑑𝑇𝑇 = 𝐶𝐶𝑉𝑉𝑑𝑑𝑇𝑇

Constant V

Heat used only to increase T (or U)

𝛿𝛿𝑞𝑞)𝑃𝑃 =𝜕𝜕𝐻𝐻𝜕𝜕𝑇𝑇 𝑃𝑃

𝑑𝑑𝑇𝑇 = 𝐶𝐶𝑃𝑃𝑑𝑑𝑇𝑇

Constant P

Heat used to increase T (or U) + to do PV work<

𝐶𝐶𝑉𝑉 < 𝐶𝐶𝑃𝑃

𝑑𝑑𝑈𝑈 = 𝐶𝐶𝑉𝑉𝑑𝑑𝑇𝑇 𝑑𝑑𝐻𝐻 = 𝐶𝐶𝑃𝑃𝑑𝑑𝑇𝑇

𝑇𝑇1 → 𝑇𝑇2𝑃𝑃1 = 𝑃𝑃2

𝐶𝐶𝑉𝑉=𝛿𝛿𝑞𝑞)𝑉𝑉𝑑𝑑𝑇𝑇

𝑑𝑑𝑃𝑃 = 0

𝛿𝛿𝑊𝑊)𝑃𝑃𝑇𝑇1 → 𝑇𝑇2𝑃𝑃1 → 𝑃𝑃2

𝐶𝐶𝑃𝑃=𝛿𝛿𝑞𝑞)𝑃𝑃𝑑𝑑𝑇𝑇𝛿𝛿𝑞𝑞)𝑃𝑃

𝑑𝑑𝑃𝑃 = 0

𝛿𝛿𝑞𝑞)𝑉𝑉

Heat Capacity

work done by the system per degree rise in T in expanding against constant external P

𝐶𝐶𝑃𝑃 − 𝐶𝐶𝑉𝑉 = �𝜕𝜕𝑈𝑈𝜕𝜕𝑇𝑇 𝑃𝑃

+ 𝑃𝑃 �𝜕𝜕𝑃𝑃𝜕𝜕𝑇𝑇 𝑃𝑃

− �𝜕𝜕𝑈𝑈𝜕𝜕𝑇𝑇 𝑉𝑉

= 𝑃𝑃 �𝜕𝜕𝑃𝑃𝜕𝜕𝑇𝑇 𝑃𝑃

+ �𝜕𝜕𝑃𝑃𝜕𝜕𝑇𝑇 𝑃𝑃

�𝜕𝜕𝑈𝑈𝜕𝜕𝑃𝑃 𝑇𝑇

𝑑𝑑𝑈𝑈 = �𝜕𝜕𝑈𝑈𝜕𝜕𝑃𝑃 𝑇𝑇

𝑑𝑑𝑃𝑃 + �𝜕𝜕𝑈𝑈𝜕𝜕𝑇𝑇 𝑉𝑉

𝑑𝑑𝑇𝑇 �𝜕𝜕𝑈𝑈𝜕𝜕𝑇𝑇 𝑃𝑃

= �𝜕𝜕𝑈𝑈𝜕𝜕𝑃𝑃 𝑇𝑇

�𝜕𝜕𝑃𝑃𝜕𝜕𝑇𝑇 𝑃𝑃

+ �𝜕𝜕𝑈𝑈𝜕𝜕𝑇𝑇 𝑉𝑉

work done per degree rise in T in expanding against internal cohesive forces acting between particlesQ: value of this term for an ideal gas?

1 mole of monoatomic ideal gas (PV = 1*RT):CP = (5/2) R; CV = (3/2) R; CP – CV = R

(𝝏𝝏𝑼𝑼/ 𝝏𝝏𝑽𝑽)T =?Ideal gas: (𝜕𝜕𝑈𝑈/ 𝜕𝜕𝑃𝑃)T = 0, Real gas: (𝜕𝜕𝑈𝑈/ 𝜕𝜕𝑃𝑃)T ≈ 0, Condensed matter: (𝜕𝜕𝑈𝑈/ 𝜕𝜕𝑃𝑃)T ≫ 0

adiabatic wall

Joule’s experimentUpon opening a stopcock, Joule found no change in T and concluded that U is function of only T

system = gasw = 0 (why? because it is ______ expansion, P = 0)q = 0 (why? because of the adiabatic wall)

∆U = 0

𝑑𝑑𝑈𝑈 = �𝜕𝜕𝑈𝑈𝜕𝜕𝑃𝑃 𝑇𝑇

𝑑𝑑𝑃𝑃 ≠ 0 + �𝜕𝜕𝑈𝑈𝜕𝜕𝑇𝑇 𝑉𝑉

𝑑𝑑𝑇𝑇 = 0 = 0 �𝜕𝜕𝑈𝑈𝜕𝜕𝑃𝑃 𝑇𝑇

= 0

• True for only an ideal gas.• There was very small δT, which Joule wasn’t able to detect; later,

Thompson showed that (𝜕𝜕𝑈𝑈 / 𝜕𝜕𝑃𝑃)T ≠ 0 for a real gas.

Isothermal ProcessIsobaric: constant T; assuming 1 mole of ideal gas (constant T, ∆U = 0)

Work 𝑤𝑤 = �1

2𝑃𝑃(𝑃𝑃)𝑑𝑑𝑃𝑃

P

V

P1

P2

1

2

V1 V2

wn=1 = P2 (V2 – V1)

P1

P2

n = 1

P1

P2

n = 2

P

V

P1

P2

2(i= 2)

V1 V2

i= 1P1-∆𝑃𝑃

Vi=1

P1 – ∆P

wn=2 = (P21 – ∆P) (Vi=1 – V1) + P2 (V2 – Vi=1)

Upon removing the weight, the gas expands • naturally, • spontaneously,• irreversibly.

Heat 𝑞𝑞 = 𝑤𝑤

System (the gas)• does work (w > 0), • absorbs heat (q > 0)

<

Isothermal Process

Work 𝑤𝑤 = �1

2𝑃𝑃(𝑃𝑃)𝑑𝑑𝑃𝑃 Heat 𝑞𝑞 = 𝑤𝑤

P

V

P1

P2

i= 1

V1 V2

i= 2i= 3

ii= n

⋮

⋮

P

V

P1

P2

V1 V2

1

2

P

V

P1

P2

V1 V2

n ∞n = n reversible process (quasi-static

process)

𝑤𝑤𝑟𝑟𝑒𝑒𝑟𝑟 = �𝑉𝑉1

𝑉𝑉2𝑃𝑃𝑑𝑑𝑃𝑃 = �

𝑉𝑉1

𝑉𝑉2 𝑅𝑅𝑇𝑇𝑃𝑃𝑑𝑑𝑃𝑃 = 𝑅𝑅𝑇𝑇 ln

𝑃𝑃2𝑃𝑃1

Isobaric: constant T; assuming 1 mole of ideal gas (constant T, ∆U = 0)

• wn ≤ wrev• q ≤ qrev• Maximum work than can be done by the system during

isothermal expansion = wrev only when the process is reversible

Adiabatic ProcessIsobaric: no heat transfer; assuming 1 mole of ideal gas

δq = 0Heat Work δw = P dV = (R T(V) / V) dV

Internal Energy dU = – δw

dU = CV dT = – δw = – P dV

Q: Can we use this equation even if the process is adiabatic (not isochoric)?

𝐶𝐶𝑉𝑉𝑑𝑑𝑇𝑇 = −𝑃𝑃𝑑𝑑𝑃𝑃 = −𝑅𝑅𝑇𝑇𝑃𝑃 𝑑𝑑𝑃𝑃

𝐶𝐶𝑉𝑉𝑑𝑑 ln𝑇𝑇 = −𝑅𝑅𝑑𝑑 ln𝑃𝑃

𝐶𝐶𝑉𝑉 ln𝑇𝑇2𝑇𝑇1

= −𝑅𝑅 ln𝑃𝑃2𝑃𝑃1

= 𝑅𝑅 ln𝑃𝑃1𝑃𝑃2

𝑇𝑇2𝑇𝑇1

=𝑃𝑃1𝑃𝑃2

𝑅𝑅𝐶𝐶𝑉𝑉

=𝑃𝑃1𝑃𝑃2

𝛾𝛾−1

=𝑃𝑃2𝑃𝑃2𝑃𝑃1𝑃𝑃1

𝐶𝐶𝑃𝑃 − 𝐶𝐶𝑉𝑉 = 𝑅𝑅,𝐶𝐶𝑃𝑃𝐶𝐶𝑉𝑉

≡ 𝛾𝛾 = 1 +𝑅𝑅𝐶𝐶𝑉𝑉

( )

𝑃𝑃1𝑃𝑃1𝛾𝛾 = 𝑃𝑃2𝑃𝑃2

𝛾𝛾 = 𝑃𝑃𝑃𝑃𝛾𝛾 = constant(only for reversible adiabatic process of ideal gas!)

Isothermal vs Adiabaticassuming 1 mole of ideal gas

P1

P

V

P2

V1 V2V3

1

3 2

Isothermal reversible process, PV = P1V1 = RT

Adiabatic reversible process, PVγ = P1V1

γ

isotherm at Tisotherm at T’ T > T’

• wisothermal reversible > wadiabatic reversible

• After adiabatic expansion, T of the system drops. Why?