Chapter 3. Second and third law of thermodynamics

Important Concepts Review Entropy; Gibbs Free Energy • Entropy (S) – definitions Law of Corresponding States (ch 1 notes) • Entropy changes in reversible and Reduced pressure, temperatures, volumes irreversible processes • Entropy of mixing of ideal gases • 2nd law of thermodynamics • 3rd law of thermodynamics Math • Free energy Numerical integration by computer • Maxwell relations (Trapezoidal integration • Dependence of free energy on P, V, T https://en.wikipedia.org/wiki/Trapezoidal_rule) • Thermodynamic functions of mixtures Properties of partial differential equations • Partial molar quantities and chemical Rules for inequalities potential Major Concept Review • Adiabats vs. isotherms

p1V1  p2V2 • Sign convention for work and heat w done on c=C /R   vm system, q supplied to system : + p1V1  p2V2  =Cp/CV w done by system, q removed from system : c c V1T1 V2T2 - • Joule-Thomson expansion (DH=0); • State variables depend on final & initial state; not Joule-Thomson coefficient, inversion path. temperature • Reversible change occurs in series of equilibrium V states  T  TT   V       P  p  • Adiabatic q = 0; Isothermal DT = 0  H CP • Equations of state for enthalpy, H and internal • Formation reaction; enthalpies of energy, U reaction, Hess’s Law; other changes D rxn H   iD f Hi i T   D rxn H  Drxn Href   DrxnCpdT Tref • Calorimetry Spontaneous and Nonspontaneous Changes

First Law: when one form of energy is converted to another, the total energy in universe is conserved. • Does not give any other restriction on a process • But many processes have a natural direction

Examples • gas expands into a vacuum; not the reverse • can burn paper; can't unburn paper • heat never flows spontaneously from cold to hot

These changes are called nonspontaneous changes.

It is useful to predict whether a physical/chemical change will go spontaneously in the forward or backward direction.

The second law and entropy, S, a state function of the system, is what tells us this.

Carnot, investigating heat engines, was the first to answer the question of spontaneity. Nicolas Léonard Sadi Carnot 1796-1832 "father of thermodynamics" The Carnot Engine and Carnot Cycle A heat engine is an engine that uses heat to generate mechanical work by carrying a "working substance" through a cyclic process.

A quantity of heat qh is withdrawn from the hot reservoir (the boiler) A quantity of heat qc (negative by convention) is discharged to the cold reservoir (the condenser) the work done by the engine is w (negative by convention). The Carnot Engine and Carnot Cycle

http://galileoandeinstein.physics.virginia.edu/mor e_stuff/flashlets/carnot.htm Carnot Heat Engines

Step 1: Reversible, isothermal expansion from A to B at Th ∆U1 = 푞ℎ + 푤1 = 0 (isothermal, ideal gas) Heat= 푞ℎ 푉퐵 Work 푤1 = −푅푇ℎ푙푛 = −푞ℎ 푉퐴

Step 2: Reversible, adiabatic expansion from B to C; Th→Tc Heat= 푞 = 0(adiabatic) 푇푐 Work 푤2 = ∆푈2 = ׬ 퐶푣푑푇 푇ℎ 푇푐 퐶 푚 푉 න 푣 푑푇 = −Rln 퐶 푇 푉퐵 푇ℎ Step 4: Reversible, adiabatic compression from D to

Step 3: Reversible, isothermal compression from C to D at Tc A; Tc→Th Heat= 푞 = 0(adiabatic) ℎ ∆U1 = 푞푐 + 푤3 = 0 (isothermal, ideal gas) Work 푤4 = ∆푈4 = ׬ 퐶푣푑푇 Heat= 푞 (negative because heat is leaving the system) 푇푐 푐 푇ℎ 푉퐷 퐶 푚 푉 Work 푤3 = −푅푇푐푙푛 = −푞푐 푣 퐴 푉퐶 න 푑푇 = −Rln 푇 푉퐷 푇푐 Efficiency of Heat Engine

∆U퐶푌퐶퐿 = 0 = 푞ℎ+푞퐶 + 푤1 + 푤2 + 푤3 + 푤4 = 푞푐푦푐푙+푤퐶푦푐푙 −푤퐶푦푐푙= 푞푐푦푐푙 = 푞ℎ + 푞퐶 푤표푟푘 푝푒푟푓표푟푚푒푑 −푤 푞 푞 푞 Efficiency: = 퐶푦푐푙= ℎ+ 퐶 = 1 − 퐶 ≡ 휂. ℎ푒푎푡 푠푢푝푝푙𝑖푒푑 푞ℎ 푞ℎ 푞ℎ 푇 퐶 푚 푉 푉 Comparing adiabatic steps: ׬ ℎ 푣 푑푇 = −Rln 퐵 = −Rln 퐴 푇푐 푇 푉퐶 푉퐷 푉 푉 푉 푉 푉 푉 Ratios of volumes are equal: 퐵 = 퐴, so 퐶 = 퐵, so −ln 퐷 = ln 퐵 푉퐶 푉퐷 푉퐷 푉퐴 푉퐶 푉퐴 푉퐵 푉퐷 푉퐵 Can not rewrite heats: 푅푇ℎ푙푛 = 푞ℎ ; 푅푇푐푙푛 = 푞푐=−푅푇푐푙푛 푉퐴 푉퐶 푉퐴 푞 푞 푞 푞 Therefore, 푐 = − ℎ 푎푛푑 푐 + ℎ = 0 푇푐 푇ℎ 푇푐 푇ℎ Means two things:

1) Even though q is not a state function, q/T is! 푑푞 We define this as entropy 푑푆 ≡ 푟푒푣 푇 Although a change in S must be calculated along a reversible path, entropy can be defined for any process

푇 푇 −푇 2) Can use T in efficiency calculation. 휂 = 1 − 푐 = ℎ 푐 푇ℎ 푇ℎ Heat Engines and Refrigerators

Operate Heat Engine in Reverse Supply work to system Remove heat from cold reservoir, and add it to hot reservoir

Work required to furnish heat qh with a heat pump

푇ℎ − 푇푐 푤 = −푞ℎ 푇ℎ

Ratio of work required to remove heat qc with a/c or refrigerator

푤 푇ℎ − 푇푐 = 푞푐 푇푐 푞푐 푇푐 Coefficient of performance 푐 = = 푤 푇ℎ − 푇푐 Definition of entropy (S)

Thermodynamic definition – entropy is related to the heat flow

f dS= dq /T , DS  1 T dq rev  rev i

e.g. DS for isothermal reversible expansion of ideal gas:

DU = q+w = 0, so, qrev = -wrev = nRT ln(Vf/Vi)

DS = qrev/T = nR ln(Vf/Vi) Statistical definition of entropy (S) Statistical definition – entropy is related to the probability Entropy is a measure of the number of possible microscopic states of a system. S=k ln W (k: Boltzmann constant, W: number of microstates W)

W is a measure of disorder of the system in a sense. e.g. There is only one way to have 100 coins with all heads up: W=1 There are 100891344545564193334812497256 ways to have half of the coins are heads up: W=100891344545564193334812497256 – a lot more probable.

http://www.bbc.com/earth/story/20150309-why-does-time-only-run-forwards Inequality of Clausius The second law discusses the properties of entropy. • Start with First Law dU  dq  PexdV • For a reversible process, P = Pex & dq = TdS • Gives dU  TdS PdV (true for any process, reversible or not since S and U are state variables) • Now use dU = dq + dw • equate 2 equations for dU TdS PdV  dq  PexdV TdS  dq  P  Pex dV

If P > Pex , sys. will expand, dV > 0, 2nd term is + If P < Pex , sys. will contract, dV < 0, 2nd term is + If P = Pex , 2nd term = 0 (equilibrium) Therefore, TdS  dq Second Law

• If the process is reversible, the equality applies Rudolf Julius • If any part of the process is irreversible, the inequality applies Emanuel Clausius 2 It is impossible for dS  dq T January 1822 – 24 August 1888) In an isolated system, dq = 0 DS > 0 spontaneous, irreversible process DS = 0 reversible process DS < 0 impossible process! Second law of thermodynamics - direction of spontaneous change

Spontaneous changes occurs in the direction where the energy of isolated system is dispersed in a random, chaotic way.

2nd Law: Total entropy of an isolated system increases in the course of a spontaneous change.

DStot >0 You can't break even. dU  w  q Can't convert all energy to work. q  TdS Even in a perfectly efficient system, lose some to entropy. For a reversible system, dU  TdS w

http://www.bbc.com/earth/story/20150309-why-does-time-only-run-forwards Entropy of System and Surroundings

The second law does not say that the entropy of a system can never spontaneously decrease.

It does say that the entropy of the universe can never decrease.

So, if in a spontaneous process, the entropy of the system goes down, then the entropy of the surroundings must go up. dSsur= dqsur, rev/T , dSsur= dqsur/T

To predict spontaneity in a system that is not isolated, must calculate both entropy of system and surroundings.

• Surroundings typically very large, and V or P is constant. • Heat transferred to the surrounding is dU or dH, state function – same regardless of reversible path or not. • Entropy change in surroundings can be simply calculated from the heat transferred to the surroundings and its temperature. dSsys = dqsys, rev/T

• For reversible process, heat can be used to calculate the entropy change. • For irreversible process, find a reversible path having the same initial and final state. Example of entropy change system & surroundings (reversible case)

Reversible isothermal compression of ideal gas 22.4 L of gas compressed to 11.2 L at 273K • For system,

DU = 0, qrev = -w = nRTln(Vf/Vi) = 8.314 x 273 ln(1/2) = -1573J

DS = qrev /T, -1573/273 = -5.76 J/K (entropy of system decreased)

• For surroundings,

Heat is added to the surrounding from the system, qsur = - qrev

DSsur = qsurv /T, 1573/273 = 5.76 J/K (entropy of surroundings increased)

• Total entropy change

DStotal = DSsys + DSsur = -5.76 J/K + 5.76 J/K = 0 J/K

Total entropy change is zero for a reversible process ! Example of entropy change system & surroundings (irreversible case)

Isothermal compression of ideal gas (P1V1=P2V2) initially at 1 atm 22.4 L of gas compressed to 11.2 L at 273K with external pressure same as final pressure

• For system, Calculate heat for the reversible path with same initial and final states. Entropy change in the system is the same as in the previous page. -5.76 J/K

• For surroundings, must know how much heat is transferred

DSsur = qsur /T = - qsys/T, qsys can be calculated by looking at system. DU = 0, qsys = -w = Pext DV = 2 atm x101.325 J/L-atm· (-11.2 L)= -2270J DSsur = 2270/273 = 8.31 J/K

• Total entropy change

DStotal = DSsys + DSsur = -5.76 J/K + 8.31 J/K = 2.55 J/K

Total entropy change is positive for an irreversible process ! Calculating Entropy Changes dU  PdV 푑푈 = 푑푞 + 푑푤 dU  TdS PdV to get dS  T I. Changes with Temperature

A. At constant V C dS  V dT dV = 0 dU = CVdT T

T2 C ST ,V ST ,V   V dT 2 1 T T1 B. At constant P U = H – PV dU  dH  PdV VdP

dH  VdP CPdT dS  dP = 0 dH = CPdT dS  T T

T2 C ST , P ST ,P   P dT 2 1 T T1 Example of entropy change using Cp(T)

Molar entropy change for heating of oxygen gas from 300K to 500K under constant pressure of 1 bar. T2 C T2   T T 2 • From DS  p dT  dT  T  T T1 T1 T  1  2  2 2   ln   (T2 T1)   (T2 T1 )  T1  2 • For oxygen,   25.503,  13.612103,   42.553107 DS 15.41J / K mol

Can also use numerical integration if you don’t have a formula for Cp(T) but have data

-1 -1 Cp=(83.7+101.4)/2=93.9JK mole

lnT2 DS  C d ln T  C D ln T  p  p   lnT1 -1 -1 Cp(200K)=83.7JK mole -1 -1 ln(x / y) = ln(x) - ln(y) Cp(240K)=104.1JK mole Changes with P or V at constant T

A. Ideal Gas dU = 0 at constant T Using dU  TdS PdV P dV dS  dV  nR T V

V2 1. Change in V: SV2 SV1 nRln V1

P2 2. Change in P: P1V1  P2V2 SP2SP1 nR ln P1 Two variable change e.g. (P1,V1,T1) to (P2,V2,T2)

• Break calculation into steps

P T2 Change V at dT = 0 then change T at dV = 0 or change P at dT = 0 then change T at dP = 0 T1 V • Ok because S is state function -- independent of path. • Easy for ideal gases: T, P, and V all independent.

Example: Calculate DS for 1 mole of Ar if T1 = 300 K, P1 =1 atm and T2 = 900 K, P2 =20 atm (ideal gas).

 900  20.93 J ln   DS from DT and DS from DP almost cancel.  T  P   Kmole 300  DS  C ln  2  Rln  2  Pm  20 If they did cancel exactly, this would be an  T1  P1  8.31451 J ln   Kmole  1  adiabatic change. (DS = 0) J  1.91 K Phase Changes

• Can add heat which changes phase, but not T. e.g. heat required to melt solid is enthalpy of fusion. • Enthalpy Change: D fusH  Hliquid  Hsolid

• Entropy change for any equilibrium phase change: D H D S  T

• Phase changes associated with discontinuous changes in heat capacity (C), enthalpy (H), entropy (S). Entropy of Mixing • Allow two gases in separate containers at same T & P to mix spontaneously and irreversibly:

V1 V2 V1+V2 • Entropy change of each due to increased volume   V1  V2   1  DSA  nARln  nAR ln   nAR ln XA  V1   XA     1  V1  V2  DSB  nBRln  nBR ln   nBR ln XB  V2   XB 

nA moles of gas A nB moles of gas B V1 volume of gas A V2 volume of gas B nA V1   XA • Number of moles of each proportional to original V. nA  nB V1 V2

• Total Entropy Change: DmixS  RnA ln XA  nB ln XB  Third law of thermodynamics Can calculate S, not just DS. Unlike U and H, S has a reference point. The reference point is entropy at zero Kelvin • An ↑ in S is associated with an ↑ disorder in matter. • Entropy goes to a minimum as T  0. • Third Law of Thermodynamics: The entropy of all pure perfect crystalline substances is zero at T = 0. T CP • Entropy at a given T: S(T)  S0   dT S0 = 0 0 T

• CP → 0 as T → 0 (related to minimum quantum size)

Third law: • Entropy of every element in its stable state at 0 K is zero. • It is impossible to reach 0 K via finite number of processes. •You can’t quit Calculating Absolute Entropies (S, not DS)

푇푓 푇푏 푇 퐶푝,푚 푠, 푇 ∆푓푢푠퐻 퐶푝,푚 ℓ, 푇 ∆푣푎푝퐻 퐶푝,푚 푔, 푇 푆푚 푇 = 푆푚 0 + න 푑푇 + + න 푑푇 + + න 푑푇 푇 푇푓 푇 푇푏 푇 0 푇푓 푇푏 Cp not measured down to 0. Debye Law 3 Gap between T and 0 K filled with theory. 퐶푝,푚 = 푎푇 퐶 푇∗ T/K Method of Calculation So 푆 푇∗ = 푝.,푚 3 0-15 Debye function 1.26 15-197.64 Graphical (trapezoidal 84.18 integration)* 197.64 Fusion 37.45 197.64-263.08 Graphical (trapezoidal 24.94 integration) 263.08 Vaporization 94.79 263.08-298.15 From Cp(T) formula 5.23 247.85

lnT2 DS  C d ln T  C D ln T *  p  p   lnT1 Standard Entropies & Reaction entropies

Can use standard molar entropies to calculate reaction entropies

⊖ ⊝ ⊝ ∆r푆 = ෍ 휈푆m − ෍ 휈푆m Products Reactants ⊖ ⊝ ∆r푆 = ෍ 휈J푆m (J) J Entropy Recap

• Study of heat engines led to concept of entropy, a measure of disorder; related to spontaneity of a process • Second Law TdS  dq

• Third Law S0 = 0 • Can Calculate Changes in Entropy

C Change in T, constant V dS  V dT T C Change in T, constant P dS  P dT T V Change in V, constant T: DS  nRln 2 V1

P2 Change in P, constant T: DS  nR ln P1

Mixing: DS  RnA ln XA  nB ln XB 

Phase changes: DS  DH T Free energy – criterion for spontaneity

From Clausius inequality, dS - dq/T  0 for spontaneous change

• Note that dq may be either dU or dH depending on the conditions (constant volume or constant pressure)

• In those cases, equivalent ways of stating the Clausius inequality: dU - TdS  0 or dH - TdS  0 depending on the conditions

• Let’s define new thermodynamic variables Helmholtz free energy A = U - TS Gibbs free energy G = H - TS

dA = dU-TdS  0 (under constant T, V) Free energy is a criterion dG = dH-TdS  0 (under constant T, P) for spontaneous change! Properties of Free Energy dA=dU-TdS-SdT. Since dU = dq+dw, dA=dq+dw-TdS-SdT For a reversible process dq=TdS and dA=TdS+dw-TdS-SdT

Constant T: dAT=dw Reversible process is maxiumum work, -(dAT)=-(dw)max (sign convention) A=arbeit; A is maximum work done on surroundings dU = dq-PextdV+dwnonPV dG=dH-TdS-SdT=dU+PdV+VdP-TdS-SdT=dq-PextdV+dwnonPV+PdV+VdP-TdS-SdT Constant T, reversible (dq=TdS); P=Pext=constant; Josiah Willard Gibbs dG=TdS-PdV+dwnonPV+PdV+VdP-TdS-SdT 1839-1903 -(dGT, P ) =-(dwnonpv)max

Reversible, constant T &P, Gibbs energy is max non PV work done important in electrochemistry For a closed system in the absence of non-expansion work: dG = VdP -SdT • G is a function of P and T- usually the variables under our control • G carries the combined consequences of 1st and 2nd law in a way suitable for chemistry • The equilibrium composition of a system depends on G Pressure Dependence of Gibbs Free Energy

dG   SdT VdP (basic equation)

G  V P T

If V ind. of P (when?)

If ideal gas

Can relate Gibbs energy of a perfect gas at pressure p to standard ⊖ value by 퐺m = 퐺m + 푅푇 ln(푝/푝⊖) Gibbs Free Energy of Reaction

⊖ ⊖ ∆r퐺 = ෍ 휈J∆f퐺m (J) J Basic Equations of Thermodynamics dU  TdS  PdV dH  TdS VdP dA   SdT  PdV dG   SdT VdP

H

8 state functions: P, V, T, U, H, S, A, G 336 slopes e.g. TP

over 1011 equations! best to learn how to derive them Working equations Properties of Matter

  P   dU  Cv dT  T   PdV   T v  Maxwell relations

 T   P  Four thermodynamic functions (U, H, A, G), which are state      functions that can be manipulated to give useful relations among  V   S  S V thermodynamic variables.  S   P        V T  T V These relations provide convenient way to measure certain  S   V  thermodynamic quantity by changing the variables involved.       P T  T P

 T   V  Some Derived Relationships:       P S  S P Deriving Maxwell Relations

Maxwell relations can be derived simply from the fact that dU, dH, dA and dG are exact differentials. Compare basic equation with slope equation with the same independent variables.  U   U  dU  TdS  PdV dU    dS    dV  S V  V S

 U   U  Reveals T   ,  P     S V  V S Take derivatives of the first wrt V and the second wrt S  T   2U   P   2U             V S  VS   S V  SV 

 T   P  Order of derivatives doesn't matter for state function, so the two are equal       V S  S V Deriving Maxwell relation Recall dU  TdS  PdV Derive dH  TdS VdP  S   P       dA   SdT  PdV  V T  T V dG   SdT VdP  S   V  Which basic equation can be used to derive the following      Maxwell Relationship?  P T  T P

A dU  TdS  PdV

B dH  TdS VdP

C dA   SdT  PdV

D dG   SdT VdP Application of Maxwell relations

We used this relation without derivation before.  U   p  Internal pressure     T   P  V T  T V

 U   U   U   S  Change in constant             V   V   S   V   z   z   z   x  T S V T            y v  y x  x  y  y v From Maxwell relations  U   U   P   S  T   ,  P          S V  V S  T V  V T

 U   U   S   U   P              T   P  V T  S V  V T  V S  T V Useful Relationships Among Partial Derivatives

• Consider four variables: w, x, y, z • Any 2 may be independent, other 2 are fns of these 2w(x,y) z(x,y) or x(y,z) w(y,z) etc. • 24 possible slopes y w x ,  ,    etc. x z y x yw • Relationship between slopes given by Chain Rule: z  z x  (similar to chain rule for total derivatives)  wy xywy z x • Reciprocal Rule: 1 xyzy x x z       • Cyclic Rule:   yz z y yx z z z y • Change of constant:           x w x y yx x w Variation of G with T and P

Fundamental equation of chemical thermodynamics: dG = VdP  SdT

(G/T)P = S (G/P)T = V Temperature Dependence of Gibbs Free Energy

퐺 = 퐻 − 푇푆 Divide through by T 퐻 퐺 = − 푆 푇 푇 휕 퐺 휕 퐻 휕 Take deriv wrt T = − 푆 푃 휕푇 푇 푃 휕푇 푇 푃 휕푇 Use product rule 휕 퐺 1 휕퐻 퐻 휕푆 = − 2 − 휕푇 푇 푃 푇 휕푇 푃 푇 휕푇 푃 휕퐻 휕푆 Be definition and derivation 퐶푝 = = 푇 휕푇 푃 휕푇 푃 휕 퐺 퐻 Give Gibbs-Helmholtz equation =− 2 휕푇 푇 푃 푇 휕 퐺 Can be rewritten 1 =퐻 휕 Τ푇 푇 푃 Pressure Dependence of Gibbs Free Energy

dG   SdT VdP (basic equation)

G  V P T

If V ind. of P (when?)

If ideal gas

Can relate Gibbs energy of a perfect gas at pressure p ⊖ ⊖ 퐺m = 퐺m + 푅푇 ln(푝/푝 ) to standard value by Fugacity Fugacity, f – effective pressure ⊖ ⊖ Replaces pressure, same units 퐺m = 퐺m + 푅푇 ln(푓/푝 ) Used to discuss real systems Allows us to discuss DG (and K) in the same way for a real and ideal gas.

For an ideal gas fi = Pi. lim fi  Pi Since all gases become ideal at low pressure: P0

Need to connect back to actual pressure, so define fugacity coefficient,  f = p Ideal gas: =1; Real gas: →1 as P → 0

Fugacity coefficient related to compressibility, Z, of a gas

푝 ln 휙 = න 푍 − 1 Τ푝 d푝 0 Fugacity coefficients Major Concept Review C • Study of heat engines led to concept of entropy, a Change in T, constant V dS  V dT measure of disorder; related to spontaneity of a process T CP • Second Law TdS  dq Change in T, constant P dS  dT T V Change in V, constant T: 2 • Third Law S0 = 0 DS  nRln V1 P • Can Calculate Changes in Entropy Change in P, constant T: DS  nR ln 2 P1 Mixing: DS  RnA ln XA  nB ln XB  Basic Equations Phase changes: DS  DH T dU  TdS  PdV dH  TdS VdP dG   SdT VdP dA   SdT  PdV  T   P   S   P   S   V   T   V  Maxwell Relations                     V S P T  S  V  V T  T V  T  P  P S  S P Deriving thermodynamic equation; using working equations Free energy and criteria for spontaneity; Helmholtz free energy A = U - TS relationship between work and A & G Gibbs free energy G = H - TS G G Temperature and pressure dependence of Gibbs free energy  S    V T P P T Fugacity coefficient f  P