1. Review of Thermodynamics Chapter 3
Review of Thermodynamics Chapter The Second Law Unit 1 The second law of thermodynamics and Entropy
Fall 2012

2. Spontaneous Change
The purpose of this chapter is to explain the origin of the spontaneity of physical and chemical change.
Spontaneous change does not require work to be done to bring it about.
Examples of spontaneous change:
A gas expands to fill the available volume,
a hot body cools to the temperature of its surroundings
a chemical reaction runs in one direction rather than another.

3. The second law of thermodynamics
Kelvin’s statement
No process is possible in which the sole result is the absorption of heat from a reservoir and its complete conversion into work.

4. The second law of thermodynamics
Engine: heat is drawn from a hot reservoir and converted into work.
All real heat engines have both a hot source and a cold sink; some energy is always discarded into the cold sink as heat and not converted into work.

5. The second law of thermodynamics
In terms of entropy
The entropy of an isolated system increases in the course of a spontaneous change: ∆S total > 0.
(Stotal is the total entropy of the system and its surroundings)

6. Entropy
The definition of entropy instructs us to find the energy supplied as heat for a reversible path between the stated initial and final states regardless of the actual manner in which the process takes place.
when the energy transferred as heat is expressed in joules and the temperature is in kelvins, the units of entropy are joules per kelvin (J K−1).
Entropy is an extensive property.
The molar entropy is an intensive property.
Molar entropy, the entropy divided by the amount of substance, is expressed in joules per kelvin per mole (J K−1 mol−1).

7. Example 3.1  Calculating the entropy change for the isothermal expansion of a perfect gas
Calculate the entropy change of a sample of perfect gas when it expands isothermally from a volume Vi to a volume Vf.
Answer:
For isothermal process DU=0,
According to the first law of thermodynamics DU=q + w,so q = -w
For a reversible change wrev= - nRT ln (Vf / Vi)
qrev= nRT ln (Vf / Vi)

8. Self Test 3.1
Calculate the change in entropy when the pressure of a perfect gas is changed isothermally from pi to pf.
∆S = nR ln(pi/pf)

9. entropy of the surroundings, ∆Ssur

11. Self Test 3.2
Calculate the entropy change in the surroundings when 1.00 mol N2O4(g) is formed from 2.00 mol NO2(g) under standard conditions at 298 K.

12. Entropy changes in expansion
Isothermal Expansion
the change in entropy of a perfect gas that expands isothermally from Vi to Vf is
The logarithmic increase in entropy of a perfect gas as it expands isothermally.

13. Entropy changes in expansion
Isothermal Expansion
∆Stot = 0
∆Stot = 0,which is what we should expect for a reversible process.

14. Entropy changes in expansion
If the isothermal expansion occurs freely (w = 0) and irreversibly, then
For system
q = 0 (because ∆U = 0); ∆Ssys = nRln(Vf/Vi)
For surrounding
∆Ssur = 0,
the total entropy change is
∆Stot > 0, as we expect for an irreversible process

15. Entropy in phase transition
If the phase transition is exothermic (∆trsH < 0,as in freezing or condensing), then the entropy change is negative. This decrease in entropy is consistent with localization of matter and energy that accompanies the formation of a solid from a liquid or a liquid from a gas.
If the transition is endothermic (∆trsH > 0, as in melting and vaporization), then the entropy change is positive, which is consistent with dispersal of energy and matter in the system.

18. Trouton’s rule
a wide range of liquids give approximately the same standard entropy of vaporization (about 85 J K−1 mol−1): this empirical observation is called Trouton’s rule. Restriction: no hydrogen bond or metallic bond Application: predict the standard enthalpy of vaporization of a liquid.

19. Illustration 3.3 Using Trouton’s rule
There is no hydrogen bonding in liquid bromine and Br2 is a heavy molecule that is unlikely to display unusual behaviour in the gas phase, so it is probably safe to use Trouton’s rule. To predict the standard molar enthalpy of vaporization of bromine given that it boils at 59.2°C, we use the rule in the form
Substitution of the data then gives
The experimental value is kJ mol−1.

20. Self Test 3.3
Predict the enthalpy of vaporization of ethane from its boiling point, −88.6°C.

21. Entropy when temperature change
Temperature change from Ti →Tf
At constant pressure
If Cp is constant in the temperature range Ti ,Tf
At constant volume

22. Example 3.2 Calculating the entropy change
Calculate the entropy change when argon at 25°C and 1.00 bar in a container of volume 0.500dm3 is allowed to expand to dm3 and is simultaneously heated to 100°C
Step 1 : Vi → Vf at constant T
Step 2 : Ti → Tf at constant V
Step 1 + Step 2

23. Self Test 3.4
Calculate the entropy change when the same initial sample is compressed to dm3 and cooled to −25°C.

24. Calculating Changes in Entropy1. 2. 3. 4.
Reversible adiabatic Process qreversible=0
For any cyclic process

25. 5.4 Calculating Changes in Entropy

26. Chapter 3 The Second Law Unit 2 TheCycle process and Clausius imequality

27. Entropy
Entropy is a state function

28. Carnot Cycle Step 1 A → B Reversible isothermal expansion Step 2 B → C
Reversible adiabatic expansion
Step 3 C → D
Reversible isothermal compression
Step 4 D → A
Reversible adiabatic compression
Nicolas L. Sadi Carnot ( )

29. Carnot Cycle
Step 1. Reversible isothermal expansion from A to B at Th; the entropy change is qh/Th,where qh is the energy supplied to the system as heat from the hot source.
Step 2. Reversible adiabatic expansion from B to C. No energy leaves the system as heat, so the change in entropy is zero. In the course of this expansion, the temperature falls from Th to Tc, the temperature of the cold sink.
Step 3. Reversible isothermal compression from C to D at Tc. Energy is released as heat to the cold sink; the change in entropy of the system is qc/Tc; in this expression qc is negative.
Step 4. Reversible adiabatic compression from D to A. No energy enters the system as heat, so the change in entropy is zero. The temperature rises from Tc to Th.

30. Reversible isothermal expansion qh qh/Th Th
steps
processes q
DS ( = q/T )
temperature
Step 1
Reversible isothermal expansion qh
qh/Th Th
Step 2
Reversible adiabatic expansion
Th→Tc
Step 3
Reversible isothermal compression qc
qc/Tc Tc
Step 4
Reversible adiabatic compression
Tc→Th
Total

31. Prove
Step 1 is an reversible isothermal expansion from VA to VB at Th Step 3 is an reversible isothermal compression from VC to VD at Tc

32. Prove
Step 2 is an reversible adiabatic expansion from Th, VB to Tc, VC
Step 4 is an reversible adiabatic expansion from Tc, VD to Th, VA
Multiplication of the first of these expressions by the second

33. Prove

34. Carnot Cycle efficiency (ε) of a heat engine
the greater the work output for a given supply of heat from the hot reservoir, the greater is the efficiency of the engine.
(Remember that qc < 0. )

35. efficiency (ε) of a heat engine
Suppose an energy qh (for example,20 kJ) is supplied to the engine and qc is lost from the engine (for example, qc = −15 kJ) and discarded into the cold reservoir. The work done by the engine is equal to qh + qc (for example, 20 kJ + (−15 kJ) = 5 kJ). The efficiency is the work done divided by the energy supplied as heat from the hot source.

36. efficiency (ε) of a heat engine
(a) The demonstration of the equivalence of the efficiencies of all reversible engines working between the same thermal reservoirs is based on the flow of energy represented in this diagram.
(b) The net effect of the processes is the conversion of heat into work without there being a need for a cold sink: this is contrary to the Kelvin statement of the Second Law.

37. Thermodynamic Temperature
This expression enabled Kelvin to define the thermodynamic temperature scale in terms of the efficiency of a heat engine.
The zero of the scale occurs for a Carnot efficiency of 1.
The size of the unit is entirely arbitrary, but on the Kelvin scale is defined by setting the temperature of the triple point of water as K exactly.
If the heat engine has a hot source at the triple point of water, the temperature of the cold sink (the object we want to measure) is found by measuring the efficiency of the engine.
This result is independent of the working substance.

38. Clausius inequality
internal energy is a state function, its change is the same for irreversible and reversible paths between the same two states,
dU= dw + dq = dqrev+ dwrev
−dwrev ≥ −dw
dqrev − dq = dw − dwrev ≥ 0
dqrev ≥ dq
When energy leaves a hot reservoir as heat, the entropy of the reservoir decreases. When the same quantity of energy enters a cooler reservoir, the entropy increases by a larger amount. Hence, overall there is an increase in entropy and the process is spontaneous.
dqrev/T ≥ dq/T

39. Spontaneous cooling
When |dq| leaves the hot source (dqh < 0), the Clausius inequality implies that
dS ≥ dqh/Th
When |dq| enters the cold sink (dqc > 0), the Clausius inequality implies that
dS ≥ dqc/Tc
Overall
Since dqh = −dqc, dqc > 0 and Th ≥ Tc
When energy leaves a hot reservoir as heat, the entropy of the reservoir decreases. When the same quantity of energy enters a cooler reservoir, the entropy increases by a larger amount. Hence, overall there is an increase in entropy and the process is spontaneous.
Hence, cooling (the transfer of heat from hot to cold) is spontaneous

40. Chapter 3 The Third Law Unit 3 The Third Law of Thermodynamics

41. The measurement of entropy

42. When temperature approaches zero Cp = a T3
Debye extrapolation
When temperature approaches zero Cp = a T3

43. The standard molar entropy of nitrogen gas at 25°C has been calculated from the following data:

44. Calculating the entropy at low temperatures
The molar constant-pressure heat capacity of a certain solid at 4.2 K is 0.43 J K−1 mol−1. What is its molar entropy at that temperature?

45. At T = 0,
all energy of thermal motion has been quenched,
in a perfect crystal all the atoms or ions are in a regular, uniform array. The localization of matter and the absence of thermal motion suggest that such materials also have zero entropy.
This conclusion is consistent with the molecular interpretation of entropy, because S = 0 if there is only one way of arranging the molecules and only one microstate is accessible (the ground state).

46. The Nernst heat theorem
The entropy change accompanying any physical or chemical transformation approaches zero as the temperature approaches zero:
∆S → 0 as T → 0 provided all the substances involved are perfectly crystalline.

47. The Nernst heat theorem
orthorhombic sulfur, S(a) → monoclinic sulfur, S(b), transition enthalpy (DH=−402 J mol−1) at the transition temperature (T=369 K)
Sm(a) = Sm(a,0) + 37 J K−1 mol−1
Sm(b) = Sm(b,0) + 38 J K−1 mol−1
Sm(a)- Sm(b) = Sm(a,0)- Sm(b,0)-1 J K−1 mol−1= J K−1 mol−1
Sm(a,0)- Sm(b,0) ～ 0 J K−1 mol−1

48. The Third Law of thermodynamics
The entropy of all perfect crystalline substances is zero at T = 0.

49. Third-Law entropies
When the substance is in its standard state at the temperature T, the standard (Third-Law) entropy is denoted S °(T).
The standard reaction entropy, ∆rS °, is defined, like the standard reaction enthalpy, as the difference between the molar entropies of the pure, separated products and the pure, separated reactants, all substances being in their standard states at the specified temperature

50. Calculating a standard reaction entropy
Calculate the standard reaction entropy of H2(g) + ½ O2(g) → H2O(l) at 25°C,

51. Calculating a standard reaction entropy
Calculate the standard reaction entropy for the combustion of methane to carbon dioxide and liquid water at 25°C.

52. Solution of ions
solutions of cations cannot be prepared in the absence of anions, the standard molar entropies of ions in solution are reported on a scale in which the standard entropy of the H+ ions in water is taken as zero at all temperatures
A positive entropy means that an ion has a higher molar entropy than H+ in water.
A negative entropy means that the ion has a lower molar entropy than H+ in water.