Do Problem 40 in Chapter 1

Have class do handout

Assign problem 20 and 35 for Friday

Pvdw = RT/(Vm – b) – a/Vm2

Pid = RT/Vm Vm,id = RT/P

Add Vid into VdW equation.Adjust Vm to change P value until target P reached.

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

Thermodynamics – The study of the interactions between matter and energy.

What is the composition of the Universe?

ClosedSystem

Surroundings

OpenSystem

isolated system

energy

Matter & energy

adiabatic no heat transfer

Isothermal (T)

Isobaric (P)

Isochoric (V)

ENERGY OF SYSTEM

E = K + V + U

kinetic energy (½mv2)potential energydepends 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

Thermodynamic Properties

Intensive Extensive

nVUH, S, G

PT

isobaric

isothermal isochoric

V/n or Vm

U/n = Um, etc

The 1st Law of ThermodynamicsConservation of Energy – The energy of the universe is constant.

DU = q + w

heat (q) work (w)

State FunctionA 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

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.

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.

Work (w)

work = force applied over a distance

dw = Fx dx and w = ∫ Fx dx extension work

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

Work derivations

= Fx (x2 - x1)

dw = Fx dx ….. w = dw = Fx dx if Fx is cst …

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 (+)

2.1

PV (expansion) work system must allow for DV.

P = F/A & Fx = PA

w = - Fx dx = - PA dx = - P dV

Fx = PA

PV (expansion) work system must allow for DV.

Reversible vs. IrreversiblePext = Pf = constant

Reversible vs. IrreversiblePext is not constant

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V (m3)

P (Pa)PV Isotherm

w = - P dV = - nRT dV/V = -nRT ln (V2/V1)

w = -RT ln (2) = -1729 J

Work for reversible, isothermal, IG expansion n = 1.00

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w = -P (V2 - V1) = -1247 J

w = - P dV = - P dV = -P (V2 – V1)

Work for irreversible, isothermal, IG expansion n = 1.00

V (m3)

P (Pa)

2.4

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 = 0.915 g ml-1, water = 0.9999 g ml-1

How much work is done heating 18.0 g of water from 0→100 ºC? density: water (0ºC) = 0.9999 g ml-1, water (100ºC) = 0.9585 g ml-1

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

W = -0.17 J

W = 0.079 J

Solid/liquid volume changes are small compared to gas therefore work is much less.

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

ThTc

Heat (q)A measure of thermal energy transfer that can be measured by the change in T of the system.

TfTf

q = m1c1(Th -Tf) = m2c2(Tf -Tc)

c = specific heat: e.g. cal g-1 oC-1

q

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 24.4H2O 75.9Fe 24.8Pb 32.9Cg 8.5Cd 6.2

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?

DU = DUtr + DUvib + DUrot + DUel + DUint + DUnuc

Isothermal change of state for an ideal gas

DU = 0 any isoT, IG process

q = DU – w = -w

T causes these motions so if DT = 0 they don’t contribute to DU

0 if no l absorption and DT = 0

0 for chemical process0 for IG

DU = DUtr + DUvib + DUrot + DUel + DUint + DUnuc

Isothermal change of state for an ideal gas

DU = 0 any isoT, IG process

q = DU – w = -w

T causes these motions so if DT = 0 they don’t contribute to DU

0 if no l absorption and DT = 0

0 for chemical process0 for IG

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.

Equipartition Theorem (CM)

For a collection of particles at thermal equilibriumthe 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

Monatomic IG (no rotations/vibrations) e.g. He, Ne, Ar… CV,m = 1.5R = 12.47 J mol-1 K-1

Utr contributes 3/2 RT toward Um for IG (dU/dT) = CV,m = 3/2 R

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

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

DH = qP

Properties of Water

Specific heat

J g-1 K-1

CP,m

J K-1 mol-1

DHJ mol-1

Densityg cm-3

Densitykg m-3

Volumem3 mol-1

ice 2.113 38.07 0.915 915 1.97 x 10-5

Fusionmelting

6007

Water 0C

4.18 75.4 0.9999 999.9 1.80 x 10-5

Water 100C

4.18 75.4 0.9584 958.4 1.88 x 10-5

Boilingcondensation

40,660

gas 1.874 33.76 5.88 x 10-4 0.588 0.0306

Calculating q, w, DH and DU? melting vs. boiling (n = 1)

Is melting/boiling a Cst V or Cst P process?

What are easiest variable(s) to determine from given data?

How can we calculate work? w = -P (V2 – V1)

How can we calculate DU?

Ideal Gas

H = U + PV & PV = nRT

show DH = DU + Dn(g)RT

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

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)

Starting Points for w, q, DU, & DH calculations

Regardless of condition ….

w = - P dV

DU = CV dT

DH = CP dT

DU = q + w

DH = DU + D(PV)

Work

Internal energy

enthalpy

CP,m = CV,m + R (ideal gas)

w = - P dV DU = CV dT

DH = CP dT

DU = q + w

DH = DU + P • DV -70.6 1550/760 • 101325 -0.000258

= -124 = qP.

2.14Gas – V1 = 377 ml to V2 = 119 ml with constant P = 1550 torr while removing 124.0 J of heat. DU = q + w q = -124.0 P (Pa) = 1550/760 • 101325 = 2.07 x 105

w = -∫ P dV = - P (V2 – V1) = - 2.07 x 105 (0.000119 – 0.000377) = + 53.4 J DU = 53.4 – 124.0 = -70.6 J

What is DH?

w = - P dV DU = CV dT

DH = CP dT

DU = q + w

DH = DU + D(PV) 2.16 IG n = 1.000 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) = -1.000 • 8.314 • 298 • ln(10): wrev = -5700 J w = -∫ P dV = - P (V2 – V1) = -101325 • (0.010 – 0.0010): 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 = +5700 J (reversible) q = +912 J (irreversible)

w = - P dV DU = CV dT

DH = CP dT

DU = q + w

DH = DU + D(PV)

 2.16 IG n = 1.000 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) = -1.000 • 8.314 • 298 • ln(10): wrev = -5700 J w = -∫ P dV = - P (V2 – V1) = -101325 • (0.010 – 0.0010): wirr = -912 J

For n = 1, V = 0.00100 m3, T = 298 K — gas is He with CV,m = 12.5 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

What is CP,m?

P1 = 2.48 x 106 Pa P2 = 3.33 x 106 Pa w = 0 DU = 1272 J

DH = 2122 J= 20.8 J mol-1 K-1.

Note that CP,m - CV,m = R

Reversible, Adiabatic IG Process

q = 0

note that P, V, & T are all changing. Equate above two expressions & solve to get ......

T2/T1 = (V1/V2)R/Cv,m or P1V1 g = P2V2

g (g = CP,m/CV,m)

DH = DU + (D PV)

- RT/Vm dV = CV,m dT

- R dV/Vm = CV,m dT/T

- R/CV,m ln(V2/V1) = ln(T2/T1)

ln (V1/V2)R/Cv,m = ln (T2/T1)

(V1/V2)R/Cv,m = T2/T1

w = DU = - PdV = CVdT

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PV graph

isothermal

adiabatic

T↓

P (Pa)

V (m3)

1 mole of an IG at P = 2.00 bar and T = 300. K is expanded adiabatically to a final volume of 0.0250 m3. What is w, q, DU, and DH.

T2/T1 = (V1/V2)R/Cv

CV.m = 15.0 J mol-1 K-1

Define: DH = n CP,m DT , DU = n CV,m DT let n = 1, rearrange & sub into above

CP,m – CV,m = R for an ideal gas (proof)

CP - CV = [(dU/dV)T + P] (dV/dT)P

Internal Pressure

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

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

Joule-Thomson Experiment: (dT/dP)H = mJT

P1, V1

T1

P1 P2P2, V2

T2

P2P1

Adiabatic Walls: Measure T change

(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

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 0.03995 1.34E+06 32.21 2.88 723

CH4 0.01604 2.26E+06 42.87 968CO2 0.04401 3.61E+06 42.83 3.34 1500H2 0.00202 2.45E+05 26.63 2.34 202He 0.00400 3.41E+04 23.65 1.90 40Kr 0.08380 2.29E+06 39.58 3.69 1090N2 0.02801 1.35E+06 38.62 3.15 621Ne 0.02018 2.09E+05 16.97 231O2 0.03200 1.36E+06 31.67 2.98 764

JT throttling is used to liquefy gases when mJT > 0 ( since dP < 0)

U = Utr + Uvib + Urot + Uel + Uint + Unuc

Dependent on bonds in molecules

Bond Energy – Uel difference between energy of AOs & MOs

atom separation

EnergySeparated atoms

bond lengthDUrx & DHrx

Standard Enthalpy of Reaction

DHo298 = Si ni Ho

m,298,i = Si ni Hom,298,i (products) - Si ni Ho

m,298,i (reactants)

CH4 + 2O2 CO2 + 2H2O

nCO2 = 1; nH2O = 2 nO2 = -2; nCH4 = -1

DHo298 = H˚CO2 + 2H˚H2O - H˚CH4 – 2H˚O2

Can’t know H˚i

Replace with DH˚f,298,i

U = Utr + Uvib + Urot + Uel + Uint + Unuc

DHo298 = Si vi DHo

f,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 = -393.5 kJ mol-1.

Bomb Calorimeter

ignite

R + K 25oC

Uel

experiment step #2

P + K 25oC

DUrx (298)

desired information

PowerSupplyw = EIt

P + K 25 + DToC

DU = 0experiment step #1

1. Reaction heats water – DT measured

2. Cool to original T1

3. T1 → T2; w = EIt = qrx = DUrx

Requirements: ~ 100% Products exothermic no side reactions

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

Cg + ½ O2(g) → COg

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 calorimeterDH˚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.

For a reaction at T 298K …….

dH = ∫ CP dT ~ CP DT

DHT – DH298 = ∫298T DCP,rx dT

DCPrx = Si ni DCPof,298,i

DHT = DH298 + DCP,rx (T – 298)Note: you must keep units consistent: convert DCP,rx to kJ

CompoundDH° DG° S° Cp°

kJ mol-1 kJ mol-1 J mol-1 K-1 J mol-1 K-1

C(graphite) 0 0 5.74 8.527C(diamond) 1.897 2.9 2.377 6.115

CH4 (g) -74.81 -50.72 186.264 35.309CO (g) -110.525 -137.168 197.674 29.116CO2 (g) -393.509 -394.359 213.74 37.11C2H2 (g) 226.73 209.2 200.94 43.93C2H4 (g) 52.26 68.15 219.56 43.56C2H6 (g) -84.68 -32.82 229.6 52.63C3H8 (g) -103.85 -23.37 270.02 73.51C6H6 (g) 82.93 129.7 269.31 81.67glucose -1274.4 -910.1 212.1 218.8

ethanol(l) -276.98 160.67 -174.14 -180.92Cl2 (g) 0 0 223.066 33.907F2 (g) 0 0 202.78 31.3H2 (g) 0 0 130.684 28.824HD (g) 0.318 -1.464 143.801 29.196D2 (g) 0 0 144.96 29.196

HBr (g) -36.4 -53.45 198.695 29.142HCl (g) -92.307 -95.299 186.908 29.12HF (g) -271.1 -273.2 173.799 29.133HN3 (g) 294.1 328.1 238.97 43.68H2O (l) -285.83 -237.129 69.91 75.21H2O (g) -241.818 -228.572 188.825 33.577H2O2 (l) -187.78 -120.35 109.6 89.1H2S (g) -20.63 -33.56 205.79 34.23N2 (g) 0 0 191.61 29.125

NH3 (g) -46.11 -16.45 192.45 35.06NO (g) 90.25 86.55 210.761 29.844NO2 (g) 33.18 51.31 240.06 37.2N2O4 (g) 9.16 97.89 304.29 77.28

O2 (g) 0 0 205.138 29.355SO2 (g) -296.83 -300.194 248.22 39.87PCl3 (g) -287 -267.8 311.78 71.84PCl5 (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

If DHf,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)

Errors as resonance energy - particularly aromatic cpds

AVG Bond Energy in kJ mol-1

C – C 344 C = C 615 H – H 436C – Cgr 717 C = O 725 O – H 463C – H 415 C = N 615 N - HC – O 350 N = N 418C – N 292 O = O 498 O – O 143C – F 441 N – N 159C – Cl 328 C ≡ C 812 Cl – Cl 243 C – Br 276 C ≡ N 890 F – F 158C – S 259 N ≡ N 946

AVG Bond Energy in kJ mol-1

Ethanol combustion