Chapter 3 The second law A spontaneous direction of change: the direction of change that does not require work to be done to bring it about. Clausius statement: No cyclic process is possible in which the sole result is the transfer energy from a cooler to a hotter body. All real process is irreversible process. 3.1 The second law

Second kind of perpetual motion machine A machine that converts heat into with 100 percent efficiency. It is impossible to built a second kind of perpetual motion machine. Kelven statement : No cyclic process is possible in which the sole result is the absorption of heat from a reservoir and its complete conversion into work.

3.1.1 Carnot principle 1. Efficiency of heat engine 

2. Carnon cycle Process AB ： (i) CD ： (ii) BC ： (iii) DA ： (iv) (iii), （ iv) therefore (v)

3. Carnot principle No heat engine can be more efficient than a reversible heat engine when both engines work between the same pair of temperature T 1 and T 2. Reversible engine (=) Irreversible engine(<) Reversible engine (=) Irreversible engine(<)

3.1.2 Entropy 1. Definition of entropy 0≤ T Q 1 1 T Q Reversible engine (=) Irreversible engine(<)  ≤ 0 δ ex T Q Reversible engine (=) Irreversible engine(<)  ≤ 0 δ ex T Q Reversible engine (=) Irreversible engine(<) Clausius inequality Reversible process Irreversible process

Entropy S (1) S is a state function (2) S is an extensive property (3) unit is J·K － 1 * Reversible process *

2. Clausius inequality (Irreversible cycle)therefore Irreversible Reversible Irreversible Reversible

3. The principle of the increase of entropy For an adiabatic process ， δQ ＝０， The entropy of a closed system must increase in an irreversible adiabatic process. Irreversible Reversible Irreversible Reversible

4. The entropy criterion of equilibrium For an isolated system ， δQ ＝０， Since all real process is irreversible, when processes are occurring in an isolated system, its entropy is increasing. Thermodynamic equilibrium in an isolated system is reached when the system’s entropy is maximized.

5. Calculation of entropy change of surrounding Note: δQ su ＝ - δQ sy T ex =constant

3.1.3 Calculation of entropy change of system  S=0 Reversible adiabatic process. Reversible process Reversible isothermal process.

(perfect gas) Isochoric Isobaric Isothermal Adiabatic  reversible  irreversible liquid and solid 1. p, V, T change (1). p, V, T change

(2) mixing of different perfect gas at constant temperature and constant pressure mixing at constant T, p at constant T, p  mix S = － (n 1 Rlny 1 ＋ n 2 Rlny 2 ) y 1 ＜ 1 ， y 2 ＜ 1 ， so  mix S ＞０ n 1 T, p, V 1 n 2 T, p,V 2 n 1 +n 2 T, p, V 1 +V 2

2. The phase transition (1) The phase transition at transition temperature At constant T, p, the two phase in the system are in equilibrium, the process is reversible, and W′ ＝０， so Q p ＝  H ，  fus H m ＞０，  vap H m ＞０， S m (s) ＜ S m (l) ＜ S m (g)

(2) irreversible phase transition Irreversible B( ,T 1,p 1 )B( ,T 2,p 2 )  S=? B( ,T eq,p eq )B( , T eq,p eq ) reversible S2S2 S1S1 S3S3 S＝S１＋S２＋S３S＝S１＋S２＋S３

S＝S１＋S２＋S３S＝S１＋S２＋S３ For example Irreversible  S =? H 2 O （ l, 90 ℃,100kPa) H 2 O （ g, 90 ℃, 100kPa) S1S1 S3S3 Reversible S2S2 H 2 O （ l, 100 ℃, 100kPa)H 2 O （ g, 100 ℃, 100kPa)

3.2 The third law The third law Nernst heat theorem: The entropy change accompanying any physical or chemical transformation approaches zero as the temperature approaches zero:  S  0,  T  0. The entropy of all perfect crystalline substance is zero at 0 K. S * (perfect crystalline ， 0 K) ＝０Ｊ · Ｋ - １

Conventional molar entropy and standard molar entropy Second law Third law S * ( ０Ｋ )= ０ S m * （ B ， T)— conventional molar entropy of substance at temperature T 。 S  m ( B, ,T) —standard molar entropy (p ＝ 100kPa)

3.2.3 The entropy change of chemical reaction For reaction aA ＋ b B →yY ＋ zZ  r S  m ( T) ＝ Σν Ｂ S  m （ B, ,T )  r S  m (298.15K) ＝ Σν Ｂ S  m （ B, ,298.15K )  r S  m (T) = yS  m (Y, ,T ) ＋ z S  m (Z, ,T ) － a S  m (A, ,T ) － b S  m (B, ,T )

aAaA + bBbB yYyY + zZzZ aAaA + bBbB yYyY + zZzZ  r S  m (T 1 ) =  S  1 ＋  S  2 ＋  r S  m (T 2 ) ＋  S  3 ＋  S  4

3.3 Helmholtz and Gibbs energies Helmholtz energy A T ex （ S 2 － S 1 ） ≥Q at constant T, T 2 S 2 － T 1 S 1 ＝ Δ(TS) Q＝ΔU－WQ＝ΔU－W Δ(TS ） ≥ΔU － W － Δ(U － TS ） ≥ － W

The maximum work done by a closed system in an isothermal process is obtained when the process is carried out reversible. Helmholtz energy A is an extensive state function reversible irreversible   A T ≥  W reversible irreversible  A T ≤ W reversible irreversible dA T ≤  W

In a closed system doing volume work only at constant T and V, the Helmholtz energy A decrease in a spontaneous change. At constant T and VW ＝ 0 δW′＝0δW′＝0 reversible irreversible dA T, V ≤ 0 reversible irreversible  A T, V ≤ 0 In a closed system capable doing only volume work, the constant temperature and volume equilibrium condition is the minimization of the Helmholtz energy A. dA T, V = 0 Helmholtz energy criterion

3.3.2 Gibbs energy G at constant T Δ(TS)≥ΔU － W Δ(H － TS ） ≤W′ at constant p p 1 ＝ p 2 ＝ p ex W ＝－ p ex （ V 2 － V 1 ）＋ W ′ ＝－ p 2 V 2 ＋ p 1 V 1 ＋ W′ ＝－ Δ(pV) ＋ W′ Δ(TS)≥ΔU ＋ Δ(pV) － W′ －［ ΔU ＋ Δ(pV)-Δ(TS) ］ ≥ － W′ － Δ(U+pV-TS)≥ － W′

G H － TS ＝ U ＋ pV － TS ＝ A ＋ pV Gibbs energy G is an extensive state function reversible irreversible  G T, p ≤ W reversible irreversible dG T, p ≤  W The maximum possible non-volume work done by a closed system in a constant temperature and pressure process is equal to the  G

reversible irreversible  G T, p ≤ W reversible irreversible dG T, p ≤  W W′ ＝ 0, δW′ ＝ 0 reversible irreversible  G T, p ≤ 0 reversible irreversible dG T, p ≤ 0 In a closed system doing volume work only at constant T and p, the Gibbs energy G decrease in a spontaneous change. In a closed system capable doing only volume work, the constant temperature and pressure equilibrium condition is the minimization of the Gibbs energy G. dG T, p = 0 Gibbs energy criterion

3.3.3 Calculation of  A and  G (1).change p,V at constant temperature at constant T, reversible, dA T ＝ δW r ＝－ pdV ＋ δW r ' If δW r ' ＝ 0 ， then dA T ＝－ pdV closed system, change p, V at constant T, W ′ ＝ 0 reversible process reversible irreversible dA T ≤  W For perfect gas ΔA＝ΔU－TΔSΔA＝ΔU－TΔS ΔG＝ΔH－TΔSΔG＝ΔH－TΔS

G ＝ A ＋ pV, dG ＝ dA ＋ pdV ＋ Vdp at constant T, reversible, δW′ r ＝ 0, dA ＝－ pdV dGT＝VdpdGT＝Vdp closed system, change p, V at constant T, W ′ ＝ 0 reversible process For an perfect gas

(2) phase transition (i) reversible phase transition (ii) irreversible phase transition reversible phase transition at constant T and p ΔU = ΔH-Δ(pV) = ΔH- pΔV = ΔH- nRT ΔA = ΔU -Δ(TS) = ΔH - nRT - TΔS = - nRT G =H - (TS) A = U - (TS) ΔG =ΔH-Δ(TS) = ΔH- TΔS ΔA = ΔU -Δ(TS) = ΔU - TΔS ΔH =TΔS reversible vaporization or sublimation at constant T and p, and vapor is an perfect gas ΔG= ΔH- TΔS = 0

3.3.4 Funtdantal relations of thermodynamic functions U, H, S, A, G, p, V and T H = U + pV H UpV ATS G A = U - TS G = H - TS = U + pV -TS = A +pV

1. Master equation of thermodynamic If δW r ′ ＝ 0 ， then δW r ＝－ pdV ， dU=δQ r +δW r ， For a reversible change δQr＝TdSδQr＝TdS dU = TdS  pdV dH = TdS + Vdp dA =  SdT  pdV dG =  SdT + Vdp H =U+pV A =U-TS G =H-TS closed system of constant composition ， reversible process ， volume work only. Master equation of thermodynamic

dU = TdS － pdV dH = TdS + Vdp dA = － SdT － pdV dG = － SdT + Vdp dZ = M dx + N dy

dU = TdS － pdV dH = TdS + Vdp dA = － SdT － pdV dG = － SdT + Vdp 2. Maxwell’s relation

3. Gibbs - helmholtz equation

3.4 Chemical potential Chemical potential Total differential G ＝ f (T ， p ， n A ， n B ……) Consider a system that consists of a single homogeneous phase whose composition can be varied

Chemical potential is an intensive state function.  B of pure substance G * (T ， p ， n B ) = n B G * m,B (T ， p ， )

Heterogeneous system

(1) Condition of phase equilibrium T，pT，p B(  )B(  ) dn B Condition of phase equilibrium   spontaneous Equilibrium criterion of substance

(2) Condition of chemical reaction equilibrium Chemical reaction Σ B μ B ＜ 0 ， dξ ＞ 0 spontaneous Homogeneous system dξ For example aA ＋ bB ＝ yY ＋ zZ aμ A +bμ B ＝ yμ Y +zμ Z Σ B μ B ＝ 0 ， equilibrium

A—potential of chemical reaction A ＜ 0  left spontaneous 。 A ＝ 0  at equilibrium ； A ＞ 0  right spontaneous ； A= (aμ A +bμ B ) -( yμ Y +zμ Z ) aA ＋ bB ＝ yY ＋ zZ