1. Pid = RT/Vm Vm,id = RT/P Pvdw = RT/(Vm – b) – a/Vm2
Do Problem 40 in Chapter 1
Pid = RT/Vm
Vm,id = RT/P
Pvdw = RT/(Vm – b) – a/Vm2
Add Vid into VdW equation.
Adjust Vm to change P value until target P reached.
Have class do handout
Assign problem 20 and 35 for Friday

2. What is the composition of the Universe?
Thermodynamics – The study of the interactions
between matter and energy.
The Laws of Thermodynamics
1. Conservation of Energy
or … “You can’t Win”
2. The Entropy of the Universe is Increasing
or … “You can’t break even”
3. A Pure Substance has 0 Entropy at 0K

3. isolated system Surroundings Open System Closed System Matter & energy
Isothermal (T)
Surroundings
Isobaric (P)
energy
Isochoric (V)
Closed
System
adiabatic
no heat transfer
isolated
system

4. ENERGY OF SYSTEM E = K + V + U internal energy kinetic energy (½mv2)
potential energy
depends on the force field applied to the system. E.g. gravity, electrostatic, etc.
internal energy
U = Utr + Uvib + Urot + Uel + Uint + Unuc
Assuming K and V are not changing …… DE = DU

5. Thermodynamic Properties
Intensive
Extensive P T n V U
H, S, G
isobaric
isothermal
isochoric
V/n or Vm
U/n = Um, etc

6. The 1st Law of Thermodynamics
Conservation of Energy – The energy of the universe is constant.
The energy of a system can change if an equal amount of energy is transferred
to/from the surroundings in the form of heat or work.
DU = q + w
For a given change in U, the values
of q and w will depend on how the
process is carried out; q & w are not
state functions.
heat (q)
work (w)
State Function A property of a system that is indicative of the state of the system at the current time. P, V, T, U, H, G, S
Change of State – DT, DU, DP, etc. ….
The values of DT, DU, etc. are independent of path.
Equation of State – PV = nRT; G = H –TS;
indicates the relationship between the values of various state functions with
respect to other state functions

7. work = force applied over a distance
Work (w)
work = force applied over a distance
extension work
dw = Fx dx and w = ∫ Fx dx
gravitational work F = ma = mg (constant)
dw = mg dh or w = mgh
electrical work w = EIt
Volts (E) WA-1 or J s-1 A-1
current (I) Ampere (A)
Time (t) s

8. w =  dw =  Fx dx if Fx is cst …
Work derivations
dw = Fx dx …..
w =  dw =  Fx dx if Fx is cst …
= Fx (x2 - x1)
2.1
dw = Fx dx: for gas P = F/A (area) & F = PA
dw = PA dx = P  Adx = P dV
dw = - P dV (- due to sign convention)
expansion w is (-) compression w is (+)

9. PV (expansion) work
system must allow for DV.
P = F/A & Fx = PA

10. w = - Fx dx = - PA dx = -  P dV
PV (expansion) work
system must allow for DV.
Fx = PA
w = - Fx dx = - PA dx = -  P dV

11. Reversible vs. Irreversible
Pext = Pf = constant

12. Reversible vs. Irreversible
Pext is not constant

13. w = -RT ln (2) = -1729 J Work for reversible, isothermal, IG expansion
w = -  P dV =
-  nRT dV/V =
-nRT ln (V2/V1)
w = -RT ln (2) = J
PV Isotherm
P (Pa)
V (m3)

14. w = -P (V2 - V1) = -1247 J w = -  P dV = - P  dV = -P (V2 – V1)
Work for irreversible,
isothermal, IG expansion
n = 1.00
w = -  P dV
= - P  dV
= -P (V2 – V1)
P (Pa)
w = -P (V2 - V1) = J
V (m3)
2.4

15. Work - Solid/liquid examples
Solid/liquid volume changes are small compared to gas therefore work is much less.
Work - Solid/liquid examples
w = - ∫ P dV applies universally – not just to gases
however, do not try to substitute P = nRT/V for solids/liquids!
Work done during phase changes
(e.g. - side walk cracks from Duluth winters) See water data on page 37
How much work is done when 18.0 g of water freezes at 0 ºC?
density: ice = g ml-1, water = g ml-1
W = J
V (m3) = 1/(g ml-1) • 1 x 10-6 m3/cm3 • #g
Vm (m3) = 1/(g ml-1) • 1 x 10-6 m3/cm3 • FW g/mol
How much work is done heating 18.0 g of water from 0→100 ºC?
density: water (0ºC) = g ml-1, water (100ºC) = g ml-1
W = J

16. Special case – Expansion into a vacuum an irreversible process
What is Pext?
What is w?
w = -P • (V2 – V1) = 0
Assign 2.14 and 2.16

17. Th Tc Heat (q) q Tf q = m1c1(Th -Tf) = m2c2(Tf -Tc)
A measure of thermal energy transfer that can be measured
by the change in T of the system.
q = m1c1(Th -Tf) = m2c2(Tf -Tc) Tf Th Tc q
c = specific heat: e.g. cal g-1 oC-1

18. Heat Capacity CP = dqP/dT (J K-1) or CP,m (J K-1 mol-1)
CV = dqV/dT (J K-1) or CV,m (J K-1 mol-1)
substance
CP (J mol-1 K-1) Cu
H2O Fe Pb Cg Cd
What would make the most efficient frying pan?
How much heat does it take to raise a 10carat diamond (1 carat = 0.2 g)
from 298K – 500K?

19. Isothermal change of state for an ideal gas
T causes these motions so if DT = 0 they don’t contribute to DU
0 for IG
0 for chemical process
DU = DUtr + DUvib + DUrot + DUel + DUint + DUnuc
0 if no l absorption and DT = 0
DU = 0 any isoT, IG process
q = DU – w = -w

20. Isothermal change of state for an ideal gas
T causes these motions so if DT = 0 they don’t contribute to DU
0 for IG
0 for chemical process
DU = DUtr + DUvib + DUrot + DUel + DUint + DUnuc
0 if no l absorption and DT = 0
DU = 0 any isoT, IG process
q = DU – w = -w

21. Heat Capacity - CV,m and DU
CV,m = dqv/dT J K-1mol-1
From DUsys = q + w ……. Derive DU = qV
assume that only expansion work is possible
assume constant volume
DU = qV (constant V heat)
Sub in dU for dq in CV,m expression: CV = (dU/dT)V
Multiply both sides by dT: dU = CV dT
Integrate both sides: ∫dU = DU = ∫CV dT
Assume CV is constant over T:
DU = CV ∫dT = CV (T2 – T1) = CV DT
This is independent of how the process is carried out.

22. Equipartition Theorem (CM)
For a collection of particles at thermal equilibrium
the average contribution of each ‘degree of freedom’
to the total energy is 1/2kT.
Accurate for translation, Very good for rotations,
Poor for vibrations.
Utr contributes 3/2kT toward U for IG (per particle)
since 3 dimensions of motion
Utr contributes 3/2 RT toward Um for IG (dU/dT) = CV,m = 3/2 R
Monatomic IG (no rotations/vibrations) e.g. He, Ne, Ar…
CV,m = 1.5R = J mol-1 K-1

23. Enthalpy (H) Begin with 1st Law DU = q + w (or dU = dq + dw)
cst P and expansion work only
By definition H = U + PV or dH = dU + d(PV)
dH = dqP – P dV + P dV + V dP = dqP
DH = qP
U is cst V heat (qV) while H is cst P heat (qP)
CP = dqP/dT = (dH/dT)P and dH = CP dT
DH =  dH =  CP dT = CP DT
if CP is independent of T
If CP,m = f(T) = a + bT + c/T2 then DH = qP =  CP dT ….
=  (a + bT + c/T2) dT

24. Properties of Water Calculating q, w, DH and DU?
melting vs. boiling (n = 1)
Properties of Water
Specific heat
J g-1 K-1
CP,m
J K-1 mol-1 DH
J mol-1
Density
g cm-3
kg m-3
Volume
m3 mol-1
ice
2.113
38.07
0.915
915
1.97 x 10-5
Fusion
melting
 6007
Water
0C
4.18
75.4
0.9999
999.9
1.80 x 10-5
100C
0.9584
958.4
1.88 x 10-5
Boiling
condensation
 40,660
gas
1.874
33.76
5.88 x 10-4
0.588
0.0306
Is melting/boiling a Cst V or Cst P process?
How can we calculate DU?
What are easiest variable(s) to determine from given data?
How can we calculate work?
w = -P (V2 – V1)

25. How different are DH and DU?
Solids or Liquids
H = U + PV & dH = dU + d(PV)
since solids/liquids are not very compressible
D(PV) ~ 0 (very small) & DH  DU
Ideal Gas
H = U + PV & PV = nRT
show DH = DU + Dn(g)RT

26. Calculations Find q, w, DU, and DH for …….
Isobaric heating/cooling of solids, liquids, or gases
data needed: CP,m (or CV,m for gases) and density
Phase changes -
data needed: DHtr - density
IG changes of state with specified conditions ….
isothermal (reversible/irreversible)
adiabatic (reversible)

27. Starting Points for w, q, DU, & DH calculations
Regardless of condition ….
Work
w = -  P dV
Internal energy
DU =  CV dT
DU = q + w
DH = DU + D(PV)
enthalpy
DH =  CP dT
CP,m = CV,m + R (ideal gas)

28. w = -  P dV DU =  CV dT DU = q + w DH =  CP dT DH = DU + P • DV
2.14
Gas – V1 = 377 ml to V2 = 119 ml with constant P = 1550 torr while removing J of heat.
DU = q + w q = P (Pa) = 1550/760 • = 2.07 x 105
w = -∫ P dV = - P (V2 – V1) = x 105 ( – ) = J
 DU = 53.4 – = J
What is DH?
DH = DU P • DV
/760 •
= -124 = qP.

29. w = -  P dV DU =  CV dT DU = q + w DH =  CP dT DH = DU + D(PV)
2.16 IG n = mol V1 = 1.0 L to V2 = 10.0 L T = 298K
a. reversibly b. irreversibly against cst P = 1.00 atm
w = -∫ P dV = - nRT ln(V2/V1) = • • 298 • ln(10): wrev = J
w = -∫ P dV = - P (V2 – V1) = • (0.010 – ): wirr = -912 J
You get more work output when the expansion is done reversibly. 
Calculate DH, DU, and q for each part above?
DH = 0 , DU = 0, and q = J (reversible)
q = +912 J (irreversible)

30. w = -  P dV DU =  CV dT DU = q + w DH =  CP dT DH = DU + D(PV)
Note that CP,m - CV,m = R
 2.16 IG n = mol V1 = 1.0 L to V2 = 10.0 L T = 298K
a. reversibly b. irreversibly against cst P = 1.00 atm
w = -∫ P dV = - nRT ln(V2/V1) = • • 298 • ln(10): wrev = J
w = -∫ P dV = - P (V2 – V1) = • (0.010 – ): wirr = -912 J
For n = 1, V = m3, T = 298 K — gas is He with CV,m = J mol-1 K-1
Gas is heated at constant volume to 400K.
Using IG law and equations above…. Calculate P1 and P2, w, DU and DH
P1 = 2.48 x 106 Pa P2 = 3.33 x 106 Pa w = 0
DU = 1272 J
What is CP,m?
= 20.8 J mol-1 K-1.
DH = 2122 J

31. isothermal, IG expansion
wrev = -  P dV =
-  nRT dV/V =
-nRT ln (V2/V1)
Work for reversible,
isothermal, IG expansion
wirrev = -  P dV =
- P (V2 – V1)
wrev = -RT ln (2) = J
PV Isotherm
P (Pa)
wirrev = (0.0125) = J
DU = 0 ― DH = 0 ― q = -w
V (m3)

32. Reversible, Adiabatic IG Process
q = 0
w = DU = -  PdV =  CVdT
note that P, V, & T are all changing.
Equate above two expressions & solve to get
-  RT/Vm dV =  CV,m dT
ln (V1/V2)R/Cv,m = ln (T2/T1)
- R  dV/Vm = CV,m  dT/T
(V1/V2)R/Cv,m = T2/T1
- R/CV,m ln(V2/V1) = ln(T2/T1)
T2/T1 = (V1/V2)R/Cv,m or P1V1 g = P2V2 g (g = CP,m/CV,m)
DH = DU + D(PV)

33. PV graph P (Pa) V (m3) isothermal adiabatic T↓ T2/T1 = (V1/V2)R/Cv
1 mole of an IG at P = 2.00 bar and T = 300. K is expanded adiabatically
to a final volume of m3. What is w, q, DU, and DH.
CV.m = 15.0 J mol-1 K-1
isothermal
P (Pa)
adiabatic T↓
V (m3)

34. CP,m – CV,m = R for an ideal gas (proof)
Define: DH = n CP,m DT , DU = n CV,m DT
let n = 1, rearrange & sub into above
CP,m – CV,m = (DH/DT) - (DU/DT)
substitute DH = DU + D (PV)
= {DU + (D(PV)}/DT) - (DU/DT)
sub D (PV) = R DT (for n = 1)
= (DU + R DT - DU)/DT)
CP,m – CV,m = R
CP - CV = [(dU/dV)T + P] (dV/dT)P
Internal Pressure

35. CP – CV = R (IG) this is a helpful relationship for IG problems
Given: CP = (dH/dT)P CV = (dU/dT)V H = U + PV show that…
CP – CV = R (IG) this is a helpful relationship for IG problems
CP - CV
H = U + PV expand
U (T,V) expand to dU dTP sub above
Reduce & factor
CP - CV = [(dU/dV)T + P] (dV/dT)P
(dU/dV)T is called “internal pressure”
solve (dU/dV)P for IG

36. Joule Experiment: measure DT as gas expands into vacuum: mJ = (dT/dV)U find (dU/dV)T for real gas
Adiabatic Walls ― q = 0
but w = 0 also since Pext = 0
(dT/dU)V(dU/dV)T(dV/dT)U = -1
& (dU/dV)T = -mJCV
Result: DT was too small to measure with Joule’s apparatus

37. Joule-Thomson Experiment: (dT/dP)H = mJT
Adiabatic Walls: Measure T change
P2, V2 T2 P2 P1
P1, V1 T1 P1 P2
(dT/dH)P(dH/dP)T(dP/dT)H = -1
(dH/dP)T = -mJTCP
JT throttling is used to liquefy gases when mJT > 0 ( since dP < 0)
Inversion temperature: mJT > 0 below inv. T

38. Inversion temperature: mJT > 0 below inv. T
atm cm6 mol-2
cm3 mol-1
Inversion T
Gas
MW (kg/mol) a b
diameter (å) Ar
1.34E+06
32.21
2.88
723
CH4
2.26E+06
42.87
968
CO2
3.61E+06
42.83
3.34
1500 H2
2.45E+05
26.63
2.34
202 He
3.41E+04
23.65
1.90 40 Kr
2.29E+06
39.58
3.69
1090 N2
1.35E+06
38.62
3.15
621 Ne
2.09E+05
16.97
231 O2
1.36E+06
31.67
2.98
764
JT throttling is used to liquefy gases when mJT > 0 ( since dP < 0)

39. U = Utr + Uvib + Urot + Uel + Uint + Unuc
Dependent on bonds in molecules
atom separation
Energy
Separated atoms
Bond Energy – Uel
difference between energy of AOs & MOs
DUrx & DHrx
bond length

40. Standard Enthalpy of Reaction
DHo298 = Si ni Hom,298,i = Si ni Hom,298,i (products) - Si ni Hom,298,i (reactants)
CH O2  CO H2O
nCO2 = 1; nH2O = 2 nO2 = -2; nCH4 = -1
DHo298 = H˚CO2 + 2H˚H2O - H˚CH4 – 2H˚O2
Can’t know H˚i
U = Utr + Uvib + Urot + Uel + Uint + Unuc
Replace with DH˚f,298,i

41. DHo298 = Si vi DHof,298,i DHof,298,i = Standard Heat of Formation
The standard heat of formation of any element in its ‘natural’ state is 0.
This eliminates any concern over Unuc, and focuses on energy differences due to bond formation.
e.g. C(gr) + O2(g) → CO2(g)
This can be experimentally determined using a bomb calorimeter, and is the value listed in your thermodynamic tables = kJ mol-1.

42. P + K 25oC 25oC Bomb Calorimeter R + K P + K ignite DU = 0 25 + DToC
experiment
step #1
1. Reaction heats water – DT measured
2. Cool to original T1
Requirements:
~ 100% Products
exothermic
no side reactions
Power
Supply
w = EIt
3. T1 → T2; w = EIt = qrx = DUrx
ignite
R + K
25oC
P + K
25oC
DUrx (298)
desired information
Uel
experiment
step #2

43. Conversion from DUrx to DHrx
DHrx = DUrx + D(PV)
Only gases contribute significantly to D(PV)
Let D(PV) = DngRT …….
DHrx = DUrx + DngRT
Examples:
C(gr) + O2(g) → CO2(g) Dn = 0
CO(g) + ½ O2(g) → CO2(g) Dn = -½

44. Cg + ½ O2(g) → COg Cg + O2(g) → CO2(g) COg + ½ O2(g) → CO2
This can’t be done in a bomb calorimeter
Cg + O2(g) → CO2(g)
COg + ½ O2(g) → CO2
These can be done in a bomb calorimeter
DH˚rx = DH˚f,CO2 - DH˚f,CO
DH˚f,CO = DH˚f,CO2 - DH˚rx
Using similar procedures the values for DH˚f has been
determined for most common compounds and the results
can be found in standard thermodynamic tables.

45. For a reaction at T  298K ……. dH = ∫ CP dT ~ CP DT
DHT – DH298 = ∫298T DCP,rx dT
DCPrx = Si ni DCPof,298,i
DHT = DH298 + DCP,rx (T – 298)
Note: you must keep units consistent: convert DCP,rx to kJ

46. DHT = DH298 + DCP,rx (T – 298) Ethanol combustion at 298 K at 500 K
Compound
DH°
DG° S°
Cp°
kJ mol-1
J mol-1 K-1
C(graphite)
5.74
8.527
C(diamond)
1.897
2.9
2.377
6.115
CH4 (g)
-74.81
-50.72
35.309
CO (g)
29.116
CO2 (g)
213.74
37.11
C2H2 (g)
226.73
209.2
200.94
43.93
C2H4 (g)
52.26
68.15
219.56
43.56
C2H6 (g)
-84.68
-32.82
229.6
52.63
C3H8 (g)
-23.37
270.02
73.51
C6H6 (g)
82.93
129.7
269.31
81.67
glucose
-910.1
212.1
218.8
ethanol(l)
160.67
Cl2 (g)
33.907
F2 (g)
202.78
31.3
H2 (g)
28.824
HD (g)
0.318
-1.464
29.196
D2 (g)
144.96
HBr (g)
-36.4
-53.45
29.142
HCl (g)
29.12
HF (g)
-271.1
-273.2
29.133
HN3 (g)
294.1
328.1
238.97
43.68
H2O (l)
69.91
75.21
H2O (g)
33.577
H2O2 (l)
109.6
89.1
H2S (g)
-20.63
-33.56
205.79
34.23
N2 (g)
191.61
29.125
NH3 (g)
-46.11
-16.45
192.45
35.06
NO (g)
90.25
86.55
29.844
NO2 (g)
33.18
51.31
240.06
37.2
N2O4 (g)
9.16
97.89
304.29
77.28
O2 (g)
29.355
SO2 (g)
248.22
39.87
PCl3 (g)
-287
-267.8
311.78
71.84
PCl5 (g)
-374.9
-305
364.58
112.8
Ethanol combustion
at 298 K
at 500 K
DHT = DH298 + DCP,rx (T – 298)
Note: you must keep units consistent: convert DCP,rx to kJ

47. AVG Bond Energy in kJ mol-1
If DHf,m,i is not in table …..
DHrx,298 can be estimated from tables of averaged bond energies.
DHº298 = Si -ni • bond energy (note sign is opposite)
C – C C = C H – H
C – Cgr C = O O – H
C – H C = N N - H
C – O N = N
C – N O = O O – O 143
C – F N – N 159
C – Cl C ≡ C Cl – Cl 243
C – Br C ≡ N F – F 158
C – S N ≡ N
AVG Bond Energy in kJ mol-1
Ethanol combustion
Errors  as resonance energy  - particularly aromatic cpds