Spontaneity, Entropy,

1
Spontaneity, Entropy, Free Energy

• Chapter 16

2
1st Law of Thermodynamics

• The first law of thermodynamics is a statement of
  the law of conservation of energy energy can
  neither be created nor destroyed. The energy of
  the universe is constant, but the various forms
  of energy can be interchanged in physical and
  chemical processes.

3
(No Transcript)
4
Spontaneous Processes and Entropy

• Thermodynamics lets us predict whether a process
  will occur but gives no information about the
  amount of time required for the process.
• A spontaneous process is one that occurs without
  outside intervention.

5
                                                
                           lt TARGET"display"gt
6
Kinetics Thermodynamics

• Chemical kinetics focuses on the pathway between
  reactants and products--the kinetics of a
  reaction depends on activation energy,
  temperature, concentration, and catalysts.
• Thermodynamics only considers the initial and
  final states.
• To describe a reaction fully, both kinetics and
  thermodynamics are necessary.

7
The rate of a reaction depends on the pathway
from reactants to products. Thermodynamics
tells whether the reaction is spontaneous and
depends on initial final states only.
8
What are some examples of spontaneous processes?

• A process that does occur under a given set of
  conditions is considered spontaneous.
• For example, a waterfall flows downhill, but
  never up, spontaneously.
• Heat flows from a warmer object to a cooler one,
  but the reverse never happens spontaneously.
• Iron exposed to water and oxygen forms rust, but
  rust does not spontaneously change back into iron.

9
Entropy

• The driving force for a spontaneous process is an
  increase in the entropy of the universe.
• Entropy, S, can be viewed as a measure of
  randomness, or disorder.
• Nature spontaneously proceeds toward the states
  that have the most spread out energy, or the
  highest probabilities of existing. In other
  words, towards an arrangement where energy can be
  contained in the greatest number of ways.

10
The expansion of an ideal gas into an evacuated
bulb.
11
As the number of molecules increases, the
probability of finding the molecules only in one
bulb becomes VERY small!
12
Positional Entropy

• A gas expands into a vacuum because the expanded
  state has the highest positional probability of
  states available to the system.
• Therefore,
• Ssolid lt Sliquid ltlt Sgas
• Solutions also have high entropy-accounting for
  the solubility of many solids into water to form
  ions in solution.

13
Entropy

• Which of the following has higher entropy?
• a) Solid CO2 or gaseous CO2?
• b) N2 gas at 1 atm or N2 gas at 1.0 x 10-2 atm?

14
Entropy

• What is the sign of the entropy change for the
  following?
• a) Solid sugar is added to water to form a
  solution?
• ?S is positive
• b) Iodine vapor condenses on a cold surface to
  form crystals?
• ?S is negative

15
Predict the sign of the entropy change for each

• Solid sugar is added to water to form a solution
• Iodine vapor condenses on a cold surface to form
  crystals
• Water forms H2O vapor
• Water freezes
• A gas expands
• Student breaks a pyrex beaker

?S
- ?S
?S
- ?S
?S
?S
16
Temperature and Entropy

• Entropy is directly affected by temperature
  changes.
• Recall that kinetic molecular theory tells us
  that matter is made up of particles in motion.
  Temperature is a measurement of the kinetic
  energy of particles in a system.
• Increasing temperature increases particle
  movement, increasing the disorder in a system,
  increasing its entropy.

17
Entropy Values

• We can make generalizations about a reactions
  entropy
• 2KClO3(s) ? 2KCl(s) 3O2(g)
• two mol solids ? two mol solids 3 mol gases
• Entropy appears to increase in this reaction.
• CaO(s) CO2(g) ? CaCO3(s)
• one mol solid one mol gas ? one mol solid
• Entropy appears to decrease in this reaction.

18
More factors that influence entropy
19
Although many factors influence the entropy of a
system, there is usually a DOMINANT source.
20
Entropy for a reaction

• We can assign values to the entropy of formation
  of a substance, called standard entropy, and
  calculate a reactions entropy quantitatively.
  Standard entropy is determined at 25C and 1 atm
  (gas partial pressure) or 1 M (soln
  concentration).
• 2KClO3(s) ? 2KCl(s) 3O2(g)
• 143.7 J/molK ? 82.6 J/molK 205.1 J/molK
• Using ?S Sproducts Sreactants, the reaction
  has a total entropy change of 493.1 J/molK
• NOTE that the units are joules/mol K rather than
  kJ/mol as for enthalpies. Typically smaller
  energies.

Check out values in Appendix what are values for
elements? Cmpds?
21
The Third Law of Thermodynamics

• . . . the entropy of a perfect crystal at 0 K is
  zero.
• Because S is known ( 0) at 0 K,
• S values at other temps can be calculated.
• See Appendix 4 for values of S0, which is the
  entropy value of substances at 1 atm when heated
  to 298K.
• Review What is the zero for enthalpy values?

22
Practice problems

• Calculate the standard entropy changes for the
  following reactions at 25?C
• (a) CaCO3(s) ?CaO(s) CO2(g)
• ?Sorxn So(CaO) So(CO2) So(CaCO3)
• ?Sorxn 39.8 J/K.mol 213.6 J/K.mol (92.9
  J/K.mol)
• ?Sorxn 160.5 J/K.mol
• (b) N2(g) 3H2(g) ? 2NH3(g)
• ?Sorxn 2So(NH3) So(N2) 3So(H2)
• ?Sorxn (2)(193) J/K.mol 192 J/K.mol
  (3)(131 J/K.mol)
• ?Sorxn -199 J/K.mol

?S favorable
-?S unfavorable
23
More Practice ?Soreaction

• Calculate ?S? at 25 oC for the reaction
• 2NiS(s) 3O2(g) ---gt 2SO2(g) 2NiO(s)
• ?S? ?npS?(products) ? ?nrS?(reactants)
• ?S? (2 mol SO2)(248 J/Kmol) (2 mol NiO)(38
  J/Kmol) - (2 mol NiS)(53 J/Kmol) (3 mol
  O2)(205 J/Kmol)
• ?S? 496 J/K 76 J/K - 106 J/K - 615 J/K
• ?S? -149 J/K gaseous molecules decreases!

24
Comparing So values

• Substances with a greater freedom of motion (or
  number of possible ways to move) have a greater
  absolute entropy.
• example I2(g) (So 261 J/K.mol) and
• I2(s) (So 117 J/K.mol)
• example CH4(g) (So 186 J/K.mol) and
• C2H6(g) (So 230 J/K.mol)

25
(No Transcript)
26
The Second Law of Thermodynamics

• . . . in any spontaneous process there is always
  an increase in the entropy of the universe.
• ?Suniv gt 0
• for a spontaneous process.

27
?SUniverse

• ?Suniverse is positive -- reaction is
  spontaneous.
• ?Suniverse is negative -- reaction is spontaneous
  in the reverse direction.
• ?Suniverse 0 -- reaction is at equilibrium.

28
How can we predict whether a chemical reaction
will be spontaneous?

• Thermodynamics can help us determine the
  direction of a spontaneous reaction, but cannot
  tell us about the speed of the process (for that
  we need kinetics).
• Is this chemist observing
  an an an exothermic or endo-
  thermic reaction?
• What is the sign of ?H?

29
What factors are important in predicting the
spontaneity of a process?

• We know that exothermic reactions (negative ?H)
  are favored since the system moves to a lower
  energy state.
• However, not ALL exothermic reactions will occur
  spontaneously, and endothermic reactions may also
  be spontaneous.
• Example ice will spontaneously melt
  (endothermic) at room temperature.

30
ENTROPY

• The other driving force for a spontaneous process
  is an increase in entropy (S) in the universe.
• A reaction tends to be spontaneous if the change
  in entropy (?S) is positive, and change in
  enthalpy (?H) is negative.
• Note -?H is really increasing entropy in
  disguise! Where is entropy increasing?

31
?G -- Free Energy!!!
How can we determine spontaneity using ?S and ?H?

• Two tendencies exist in nature
• tendency toward higher entropy -- ?S
• tendency toward lower energy -- ?H
• If the two processes oppose each other (e.g.
  melting ice cube), then the direction is decided
  by the Free Energy, ?G, and depends upon the
  temperature.

32
Gibbs Free Energy

• ?G ?H ? T?S (allows us to focus
  from the standpoint
  of the system)
  A process (at constant T, P) is
  spontaneous in the direction in which free energy
  decreases
• ??Gsys means ?Suniverse
• But, how do exothermic processes increase the
  entropy of the universe??

An American physicist and founder of
thermo- dynamics at Yale! (1839-1903. 2005 stamp)
went to Hopkins and graduated from Yale at the
age of 19. He was praised by Albert Einstein as
"the greatest mind in American history".
Entropy changes in the surroundings are primarily
determined by the heat flow. An exothermic
process in the system increases the entropy of
the surroundings.
33
?G, ?H, ?S

• Spontaneous reactions (shift to RIGHT) are
  indicated by the following signs
• ?G negative
• ?H negative
• ?S positive

34
How are you feeling at this point?
The entropy of your brain must be increasing
35
Temperature Dependence

• ?Ho is not temperature dependent. ?So is entropy
  measured at constant 298K.
• ?Go is temperature dependent, and can be
  calculated for any constant temperature using the
  equation
• ?G ? ?H ? ? T?S ?

36
?G ? ?H ? ? T?S ?
Calculations showing that the melting of ice is
temperature dependent. As temperature
increases, the Free Energy becomes lower (more
negative) -the process is spontaneous above 0oC.
37
Predicting spontaneity
?H ?S ?G
- - ALWAYS SPONTANEOUS
- NEVER SPONTANEOUS
- only spontaneous when T?S is greater than ?H (at high temp.)
- - - only spontaneous when T?S is less than ?H (at low temp.)
38
Free Energy ?G

• ?G ? ?H ? ? T?S ?
• ?G negative spontaneous (RIGHT shift)
• ?G positive -- spontaneous in opposite
  direction (SHIFT left)
• ?G 0, system is at equilibrium (with products
  and reactants in std states)

39
Summary of 3 ways to calculate ?G at std state

• 1. ?Grxn ?Ho - T?So
• 2. ?G? ?np?Gf?(products) ? ?nr?Gf?(reactants)
• 3. Hesss Law

40
1. Free Energy Change and Chemical Reactions

• (a) Calculate ?H?, ?S?, ?G? for the following
  reaction at 25C
• 2 SO2(g) O2(g) ----gt 2 SO3(g)
• ?H? ?np?Hf?(products) ? ?nr?Hf?(reactants)
• ?H? (2 mol SO3)(-396 kJ/mol)-(2 mol
  SO2)(-297 kJ/mol) (0 kJ/mol)
• ?H? - 792 kJ 594 kJ
• ?H? -198 kJ

41
?G? CalculationsContinued

• ?S? ?npS?(products) ? ?nrS?(reactants)
• ?S? (2 mol SO3)(257 J/Kmol)-(2 mol SO2)(248
  J/Kmol) (1 mol O2)(205 J/Kmol)
• ?S? 514 J/K - 496 J/K - 205 J/K
• ?S? -187 J/K

42
?G? CalculationsContinued

• ?Go ?Ho ? T?So
• ?Go - 198 kJ - (298 K)(-187 J/K)(1kJ/1000J)
• ?Go - 198 kJ 55.7 kJ
• ?Go - 142 kJ
• The reaction is spontaneous in forward direction
  at 25 oC and 1 atm.
• Now try this using the second method

43

• (b) Find ?G for the following reaction at
  constant pressure (1 atm) and temperature
  (900C)
• CaCO3(s) ? CaO(s) CO2(g)
• ?Grxn ?Ho - T?So
• ?Sorxn So(CaO) So(CO2) So(CaCO3)
• ?Sorxn 39.8 J/K.mol 213.6 J/K.mol (92.9
  J/K.mol)
• ?Sorxn 160.5 J/K.mol
• ?Horxn ?Hfo(CaO) ?Hfo(CO2) ?Hfo(CaCO3)
• ?Horxn (-635.6 kJ/mol) (-393.5 kJ/mol)
  (-1206.9 kJ/mol)
• ?Horxn 177.8 kJ/mol

44
Continued

• ?Grxn ?Ho T?S
• ?Grxn 177.8 kJ/mol 1173K(0.1605 kJ/K.mol)
• ?Grxn -10.5 kJ/mol
• Try this at what Temp (in C) will this become
  nonspontaneous?

45
2. More ?G? Calculations

• ?G? standard free energy change that occurs if
  reactants in their standard state are converted
  to one mole of product in its standard state.
  Values given in textbook appendix. Note that ?Gf?
  for elements in their standard states is zero.
• ?G? ?np?Gf?(products) ? ?nr?Gf?(reactants)
• The more negative the value of ?G?, the further
  a reaction will go to the right to reach
  equilibrium.

(formation)
46
Practice Problem

• Calculate the standard free energy changes for
  the following reactions at 1 atm and 25C.
• (a) CH4 (g) 2O2(g) ? CO2(g) 2H2O(l)
• ?Gorxn ?Gproducts ?G reactants
  
• ?Gorxn Gof(CO2) 2Gof (H2O) Gof(CH4)
  2Gof(O2)
• ?Gorxn (-394 kJ/mol) (2)(-237 kJ/mol)
• 
  (-51 kJ/mol) (2)(0 kJ/mol)
• ?Gorxn -818 kJ/mol

?G is -, SPONTANEOUS
47

• (b) 2MgO(s) ? 2Mg(s) O2(g)
• ?Gorxn ?Gproducts ?G reactants
  
• ?Gorxn 2Gof(Mg) Gof (O2) 2Gof(MgO)
• ?Gorxn (2)(0 kJ/mol) (0 kJ/mol)
  (2)(-570 kJ/mol)
• ?Gorxn 1139 kJ/mol

?G is , NOT SPONTANEOUS
48
3. Hesss Law ?Go

• Cdiamond(s) O2(g) ---gt CO2(g) ?Go -397 kJ
• Cgraphite(s) O2(g) ---gt CO2(g) ?Go -394 kJ
• Calculate ?Go for the reaction
• Cdiamond(s) ---gt Cgraphite(s)
• Cdiamond(s) O2(g) ---gt CO2(g) ?Go -397 kJ
• CO2(g) ---gt Cgraphite(s) O2(g) ?Go 394 kJ
• Cdiamond(s) ---gt Cgraphite(s) ?Go -3 kJ
• Diamond is kinetically stable, but
  thermodynamically unstable.

49
?G? 0, Boiling Point Calculations

• (a) What is the normal boiling point for liquid
  Br2?
• Br2(l) ---gt Br2(g)
• ?Ho 31.0 kJ/mol ?So 93.0 J/Kmol
• At equilibrium (a phase change), ?Go 0
• ?Go ?Ho ? T?S0 0
• ?Ho T?S0
• T ?Ho/?S0
• T 3.10 x 104 J/mol/(93.0J/Kmol)
• T 333K

50

• (b) Find the increase in entropy of the phase
  transition H2O(s) ? H2O(l)
• During a phase transition, ?G0 since the system
  is at equilibrium.
• This transition occurs when T273K
• ?Grxn ?H T?S
• 0 ?H T?S
• ?Srxn ?Hfusion
• T
• ?Srxn 6010 J/mol
• 273 K
• ?Srxn 22 J/K.mol

51
Calculating ?Ssurroundings

• ?Ssurr is positive -- heat flows into the
  surroundings out of the system.
• ?Ssurr is negative -- heat flows out of the
  surroundings and into the system.

?Ssurr - ?Hsystem T
52
?Ssurroundings Calculations

• Sb2S3(s) 3Fe(s) ---gt 2Sb(s) 3FeS(s) ?H
  -125 kJ
• Sb4O6(s) 6C(s) ---gt 4Sb(s) 6CO(g) ?H 778
  kJ
• What is ?Ssurr for these reactions at 250C 1
  atm.
• ?Ssurr - ?Hsystem
• T
• ?Ssurr -(-125kJ/298K)
• ?Ssurr 419 J/K

?Ssurr - ?Hsystem T ?Ssurr
-(778kJ/298K) ?Ssurr -2.61 x 103 J/K
53
Free Energy and partial pressure

• For reactions that occur under partial pressures
  other than standard (1 atm), the ?G at those
  pressures is calculated as follows
• ?G ?G? RT ln(Q)
• Q reaction quotient from the law of mass
  action, using non-standard pressures.
• R 8.314 J/mol K

54
(No Transcript)
55
?G and Equilibrium

• Although a reaction with a negative ?G will move
  forward spontaneously, it does NOT mean that a
  reaction will go 100 to completion.
• Remember, equilibrium occurs when the forward and
  reverse reaction rates are equal (kinetics).
• From a thermodynamic point of view, this occurs
  when the reaction system is also at the lowest
  value of ?G.
• The relationship between K and ?Go
• ?Go -RTln(K)

56
Making connections

• At any point in the reaction, the following
  equation is used
• ?G ?G? RT ln(Q)
• It follows that when all concentrations are
  standard (1M),
• Q1 ln(Q)0 and ?G ?G?
• At equilibrium,
• ?G 0 and ?G? -RT ln(Q)

Qproda reactb
57
A system can achieve the lowest possible free
energy by going to equilibrium, not by going to
completion.
58
Relationship between ?G and K

• ?Go -RTln(K)
• Given this equation, it follows that when
  products and reactants are in their std states
  and
• ?Go 0, K 1
• ?Go lt 0, K gt 1
• ?Go gt 0, K lt 1
• We can use ?Go to calculate K, since K e
  -?Go/RT

The relationship between K and ?G at equilibrium
?Go -RTln(K) 0 ?Ho -T?So
59
Two cases approaching equilibrium
Q products reactants

• ?G ?G? RT ln(Q)
• Case 1 A large (-) value for ?G? will make ?G
  negative also. Reaction will proceed to the
  right, creating MORE products (so lnQ goes from
  (-) with Qlt1 to () with Qgt1 as Q increases).
  Eventually, the RT ln(Q) term becomes () enough
  that ?G? is canceled out and ?G is zero
  (equilib.)
• Case 2 A large () value for ?G? will make ?G
  positive also. Reaction will proceed right to
  left, creating MORE reactants (so lnQ goes from
  () with Qgt1 to (-) with Qlt1 as Q decreases).
  Eventually, the RT ln(Q) term becomes (-) enough
  that ?G? is canceled out and ?G is zero
  (equilib.)

60

• Determine the sign of ?G? for each reaction.
  Which reaction favors reactants? Products? Recall
  ?G? ?Gf?(prod) ? ?Gf?(rcts)

61
But.
If it looks like a free energy diagram then you
are in AP Chemistry!
62
Practice Problem

• Calculate ?G at 25C for the following reaction
  where PCO 5.0 atm and PH2
  3.0 atm
• CO(g) 2H2(g) ?CH3OH(l)
• ?G ?Go RTln(Q) Q 1
  1 2.2x10-2
• (PCO)(PH2)2
  (5.00)(3.00)2
• ?Go ?Go (CH3OH) ?Go (CO) ?Go (H2)
• ?Go -166 kJ/mol -137 kJ/mol 2(0 kJ/mol)
  -29 kJ/mol
• ?G -2.9x104 8.31 J/K.mol)(298K)ln(2.2x10-2)
• -2.9x104 J/mol - 9.4 x103 J/mol
• -3.8 x104 J/mol -38 kJ/mol
• Compare this ?G to ?Go a more negative ?G means
  this reaction is more spontaneous (shift to
  right) than at one atm.

63
Equilibrium Calculations

• 4Fe(s) 3O2(g) lt---gt 2Fe2O3(s)
• Calculate K for this reaction at 25 oC.
• ?Go - 1.490 x 106 J or find this using table,
  is 2(-740 KJ/mol)
• ?G? ?RT ln(K)
• K e - ?G?/RT
• K e 601 or 10261 too big to handle in
  calculator
• try (e200)3
• K is very large because ?G? is very negative.

64
Practice Problem 1

• AP 1971
• Given the following data for graphite and diamond
  at 298K S(diamond) 2 J/mol K
• S(graphite) 6 J/mol K ?Hf CO2(from
  graphite) -395.3 kJ/mol
• ?Hf CO2(from diamond) -393.4
  kJ/mol
• Consider the change C (graphite) ? C(diamond) at
  298K and 1 atm.
• (a) What are the values of ?S and ?H for the
  conversion of graphite to diamond?
• ?S S(dia.) - S(graph.) (2 - 6) J/mol K
  - 4 J/mol K
• CO2 ? C(dia.) O2 ?H 393.4
  kJ/mol
• C(graph.) O2 ? CO2 ?H - 395.3 kJ/mol
• C(graph.) ? C(dia.) ?H -1.9
  kJ/mol

65
1971 AP (Continued)

• (b) Perform a calculation to show whether it is
  thermodynamically feasible to produce diamond
  from graphite at 298K and 1 atmosphere.
• ?G ?H - T?S
• -1.9x103 J/mol - (298K)(-4 J/mol K)
• -708 J/mol
• a ?G is negative, indicates feasible
  conditions
• (c) For the reaction, calculate the equilibrium
  constant
• Keq at 298K. check this one
• Keq e-?G/RT
• e-(-708/(8.314)(298))
  e0.285 1.3

66
Practice Problem 2

• AP 1999
• Answer the following question in terms of
  thermodynamic principles and concepts of kinetic
  molecular theory.
• (a) Consider the reaction represented below,
  which is spontaneous at 298 K.
• CO2(g) 2 NH3(g) ? CO(NH2)2(s) H2O(l)
  ?Hº 134 kJ
• (i) For the reaction, indicate whether the
  standard entropy change, ?Sº, is positive,
  negative, or zero. Justify your answer.
• (i) ?Sº is negative because (1) two different
  gases make a solid and a liquid (both with
  smaller entropies) and (2) three molecules of
  reactant make two molecules of product (a
  decrease in entropy).

67
AP 1999 (Continued)

• (a) Consider the reaction represented below,
  which is spontaneous at 298 K.
• CO2(g) 2 NH3(g) ? CO(NH2)2(s)
  H2O(l) ?Hº 134 kJ
• (ii) Which factor, the change in enthalpy, ?Hº,
  or the change in entropy, ?Sº, provides the
  principle driving force for the reaction at 298
  K? Explain.
• (ii) natural tendency is to maximize entropy
  and since this reaction decreases entropy and is
  spontaneous (-?Gº), then ?Hº must be negative to
  overcome the entropy change and drive this
  reaction.
• (iii) For the reaction, how is the value of the
  standard free energy change, ?Gº, affected by an
  increase in temperature? Explain.
• (iii) ?Gº ?Hº T?Sº as T increases, the
  value of T?Sº increases (it becomes more
  positive since ?S is negative) and the value of
  ?Gº becomes a smaller negative number (i.e.,
  moves toward zero).

68
AP 1999 (Continued)

• b) Some reactions that are predicted by their
  sign of ?Gº to be spontaneous at room temperature
  do not proceed at a measurable rate at room
  temperature.
• (i) Account for this apparent contradiction.
• (i) the sign of ?Gº (thermodynamics) does not
  account for activation energy (kinetics) a large
  activation energy would effectively prevent a
  reaction even though there is a favorable free
  energy change.
• (ii) A suitable catalyst increases the rate of
  such a reaction. What effect does the catalyst
  have on ?Gº for the reaction? Explain.
• (ii) a catalyst changes neither the ?Hº nor
  the ?Sº for a reaction, therefore, it will have
  no effect on the ?Gº.

69
EXTRA Temperature Dependence of K

• y mx b
• (?H? and S? ? independent of temperature over a
  small temperature range)
• If the temperature increases, K decreases for
  exothermic reactions, but increases for
  endothermic reactions.

70
Summary

• Entropy is usually described as a measure of the
  disorder of a system. Any spontaneous process
  must lead to an increase in entropy of the
  universe.
• The standard entropy of a chemical reaction can
  be calculated from the absolute entropies of
  reactants and products.
• Under conditions of constant temperature and
  pressure, the free-energy change ?G is less than
  zero for a spontaneous process and greater than
  zero for a non-spontaneous process. For an
  equilibrium process, ?G 0.

71
Summary

• 4. For a chemical and physical change at constant
  temperature and pressure, ?G ?H T?S. This
  equation can be used to predict the spontaneity
  of a process.
• 5. The standard free energy change for a
  reaction, ?Go, can be calculated from the
  standard free energies of formation of reactants
  and products.
• 6. The relationship between ?G and equilibrium
  position is given by ?Go -RTln(K).
• 7. The value of ?G for conditions other than
  standard is calculated from ?G ?Go RTln(Q).

72
Tutorials

• http//www.wwnorton.com/college/chemistry/gilbert2
  /contents/ch13/studyplan.asp
• THE END

73
Reversible vs. Irreversible Processes

• Reversible The universe is exactly the same as
  it was before the cyclic process.
• Irreversible The universe is different after
  the cyclic process.
• All real processes are irreversible -- (some work
  is changed to heat). ? w lt ?G
• Work is changed to heat in the surroundings and
  the entropy of the universe increases.

74
Laws of Thermodynamics

• First Law You cant win, you can only break
  even.
• Second Law You cant break even.

75
Review. So

• So increases with
• solid ---gt liquid ---gt gas
• greater complexity of molecules (have a greater
  number of rotations and vibrations)
• greater temperature (if volume increases)
• lower pressure (if volume increases)