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*2.1P-V Work2.2Heat2.3The First Law of Thermodynamics2.4Enthalpy2.5 Heat Capacities2.6Calculation of First Law Quantities2.7The 2nd law of thermodynamics2.8Heat engines2.9Entropy2.10 Calculation of entropy changes2.11 Thermochemistry

*Able to perform calculations using First and Second Laws of ThermodynamicsAble to calculate properties in Thermochemistry

*Work

*2.1P-V WORK

x = pistons positionl = length of systemIf Pext, Vsyst until Pext = PintV = Al = A(b-x)If piston moves by dx, dV = -Adx

PextxlbMechanical equilibriumSystemPint

*Action = reactionClosed system reversible process

*Reversible ProcessSystem is infinitesimally close to equilibrium.Infinitesimal change in conditions can reverse the process to restore both system and surroundings to their initial states.

Irreversible Process

*2.2HEAT (q)

m2, T2m1, T1

T2 T1Tf = final temperaturem2 c2 (T2 Tf) = m1 c1 (Tf T1) q

q = amount of heat flows from body 2 to body 1.c1, c2 = specific heat capacity. (constant, evaluated experimentally, functions of T and P)Unit for heat: J or cal 1 cal = 4.184 J

*Heat capacity: Heat required to raise the temp of system by one K(or oC)3 types of heat capacities: heat capacity (C): unit J/K, specific heat capacity(c): unit J/Kg, molar heat capacity(cm): unit J/Kmol.

For heat flowing at constant pressure:dqP = mcPdTCP = heat capacity at constant pressure.CP = mcPdqP = CPdT CP = dqP/dT

*ENERGY: capacity to do workMechanical EnergyKinetic energy, K = mv2Potential energy, V = mgh

Internal Energy, U Energy at the molecular levelU = translational + rotational + vibrational + electronic energy + potential energy of interactions between moleculesU extensive property (J)Um = U/n molar internal energy intensive property (J/mol)

*

E = K + V + UE = Total energy of system.System at rest, K = 0 and no external field, V = 0, E = U

*2.3FIRST LAW OF THERMODYNAMICS U = q + wclosed systemSYSTEMHeat absorbed by system+q+wWork done on the system-q-wWork done by systemHeat released from systemsurrounding

*

(a) (b) (c)

PPPVVV121212= area under the curve w is not a state function

*

U is a state function, value depends on the state of the system, not on path of process.Energy is conserved any energy that is lost by the system must be gained by the surroundings, and vice versa.

*2.4ENTHALPY, HH = U + PVH is state function.

For constant-pressure process, P2 = P1

For constant-volume process (V=0):

Heat absorbed (q) at P & V

*2.5HEAT CAPACITIES

Heat capacity at constant pressure (isobaric process):

Heat capacity at constant volume (isochoric process):

Generally Cp>Cv except if the fluid contracts upon heating (H2O)

*Molar heat capacity:

For ideal gas:

*2.6CALCULATION OF FIRST LAW QUANTITIES

Perfect gas obeys:

Isothermal dT = 0Adiabatic dq = 0Constant-pressure or isobaric dP = 0Constant-volume or isochoric dV = 0Reversible process in perfect gas, closed system

*Perfect-gas change of state U and H depend on T only.

q and w depend on path.

*2.6.1 Isothermal Process (reversible)

dT = 0U = 0H = 0 U = 0 = q + w q = -w

*because T1 = T2

*

Calculate q, w, U and H for reversible isothermal expansion at 300K of 5.00 mol perfect gas from 500 to 1500 cm3.

n = 5.00 molT = 300 KV1 = 500 cm3V2 = 1500 cm3

*

*2.6.2 Constant-Volume Process (reversible)

dV = 0 (No phase change)

where

(assuming CV,m doesnt change with temperature)

*2.6.3 Constant-Pressure Process (reversible, no phase change)

If CP,m doesnt vary with temperature,qP = nCP,m (T2-T1), U = qP + w

*

Calculate q, w, U and H if 2.00g He(g) with CV,m = 3/2 R undergoes:

Reversible constant-pressure expansion from 20.0 dm3 to 40.0 dm3 at 0.80 bar.Reversible constant-volume heating from 0.600 bar to 0.900 bar at V = 15.0 dm3.

*

(a)

T1 = ?T2 = ?

*

*

*(b)

*

*

CP,m for O2 at T 300 400 K is CP,m = a + bT where a = 6.15 cal/mol K, b = 0.00310 cal/mol K2. Calculate q, w, U, H when 2.00 mol O2 is reversible heated from 27 127C with P held fixed at 1.00 atm.

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*

*2.6.4.Reversible Adiabatic Process

*

If V2 V1, T2 T1

*For ideal gas:

For adiabatic process:

For adiabatic process in ideal gas:

*

*

Isothermal expansion: V2 V1, P2 P1, T2 = T1Adiabatic expansion: V2 V1, P2 P1, T2 T1 P2(adiabatic) P2(isothermal)

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1 mol He with CV,m = 3/2 R expands reversibly from 24.6 L to 49.2 L at 300 K. Calculate final pressure and temperature if the expansion is:(a)Isothermal(b)adiabatic(c)Sketch on a PV diagram

*

(a) Isothermal T2 = T1 = 300 K

*(b)Adiabatic process P1V1 = P2V2

*P (atm)V/Lisothermadiabat10.549.224.6

*2.6.5 Reversible phase change at constant T & P

fusion/melting vaporizationLatent heat of fusion of H2O = 79.7 cal/gTo melt 18 g of ice;

*

H2O: Hfus = 79.7 cal/g Hvap= 539.4 cal/g

H2O(l)

(0 C)

(0 C)

(100 C)

Calculate q, w, U, H for each case.

*

(a)Melting of ice:H2O(s) H2O(l)at 0 C, 1 atm

*(b)Reversible constant heating of 1 mol liquid water from 0 to 100 C at 1 atm.H2O(l), 0C H2O(l), 100C, 1 atm

*(c)The vapourisation of 1 mol of water at 100 C and 1 atm. H2O(l), 100C H2O(v), 100C

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Deduce whether q, w, U, H = +, 0, -.

(a)Reversible melting of benzene solid at 1 atm and normal melting point.

Heat is required q 0Constant pressure H = qP 0, w = -PVBenzene expands on melting, w 0 vol is small w q U = q + w q U 0

*(b)Reversible melting of ice at 1 atm, 0 C.

Same as (a) except the system contracts on melting w 0

(c)Adiabatic expansion of perfect gas into a vacuum.

q = 0, w = 0, U = 0, P=0 (free expansion) H = U + (PV) = 0

*(d)Reversible heating of perfect gas at constant P.

q 0 (heating)dqP = CPdT,CP 0; T 0dU = CVdT,U 0 since CV 0PV = nRT,V 0; w = -PV 0 H = U + (PV) 0.

*2.7 THE SECOND LAW OF THERMODYNAMICS(a)Kelvin-Planck statement:It is impossible for a system to undergo a cyclic process whose sole effects are the flow of heat into the system from a heat reservoir and the performance of an equivalent amount of work by the system on the surroundings.

Heat reservoirCyclic machine (syst)Work output = qq

*(b)Clausius statement:

It is impossible for a system to undergo a cyclic process whose sole effects are the flow of heat into the system from a cold reservoir and the flow of an equal amount of heat out of the system into a hot reservoir.

*Rephrase:

There are 2 principal physical statements of the Second Law of Thermodynamics:1. It is impossible for a system in a cyclic process to turn heat completely into work. (Kelvin-Planck).2. Heat cannot flow spontaneously from a cooler to hotter object if nothing else happens. (Clausius).

*2.8 HEAT ENGINES:Convert q w

Efficiency = e =work out per cycleenergy input per cycleqH-qC-w

*For cyclic process, the first law gives U = 0 q + w = qH + qC + w-w = qH + qC

qC negativeqH positivee 1

*CARNOTS PRINCIPLE No heat engine can be more efficient than a reversible heat engine when both engines work between the same pair of temperature H and C.

It is not possible to make a heat engine whose only effect is to absorb heat from a high-temperature region and turn all that heat into work

*

(I) All reversible heat engines operating between the same two temperature, have the same efficiency. erev = f(H, C).

(II)This reversible efficiency is the maximum possible for any heat engine that operates between these temperatures. eirrev erev

*CARNOT CYCLE

PV

*Heat flows into cylinder at temperature TH The fluid expands isothermally and does work on the piston.The fluid continues to expand, adiabatically.Work is done by the piston on the fluid, which undergoes an isothermal compression.The fluid returns to its initial condition by an adiabatic compression.

*divide by T:

Integrate both sides:

123

*(1)

*(3)

0(1)(2)(3)0

*(2)

Carnot cycle00

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Upper limit to the efficiency of real heat engines

*A Carnot-cycle heat engine does 2.50 kJ of workper cycle and has an efficiency of 45.0%.Calculate w, qH and qC for one cycle.

*

*Determine the thermal efficiency of this Carnot engine Determine the amount of heat rejected to the sink per cycleA Carnot heat engine receive 500kJ OF heat per cycle form a high temperature source of 652 degree C and reject a low Temperature sink at 30 degree C. a=0.672b= 164kJ

*Able to perform calculations using First and Second Laws of ThermodynamicsAble to calculate properties in Thermochemistry

*

*2.9 ENTROPY (S)An infinitesimal change in entropy ds,

S of a substance is the energy transferred as heat to it reversibly divided by the temperature at which the transfer takes place.Closed system, reversible process Measurable change

*S is an extensive state function.Sm = S/n molar entropy.Unit J mol-1 K-1 or cal mol-1 K-1

2.10 CALCULATIONS OF ENTROPY CHANGES.1. Cyclic process

2. Reversible adiabatic process

*3. Reversible phase change at constant T, P.

At constant T gives:

Since P is constant,

*4.Reversible Isothermal ProcessT = constant,

*5.Reversible change of state of perfect gas.

*

Find S for the conversion of 1.00 mol ice at 0 C and 1.00 atm to 1.00 mol H2O vapour at 100 C and 0.500 atm.Hfus = 79.7 cal/g Hvap = 539.4 cal/gCp = 1.00 cal/g K

*

H2O(s)H2O(l), 0 C, 1 atm0 C, 1 atm

H2O(l), 100 C, 1 atm

H2O(v), 100 C, 1 atm

H2O(v), 100 C, 0.5 atm(a)(d)(c)(b)

*(a)

*(b)H2O(l), 0 C, 1 atm H2O(l), 100 C, 1 atm

*(c)H2O(l), 100 C, 1 atm H2O(v), 100 C, 1 atm

*(d)H2O(v), 100 C, 1 atm H2O(v), 100 C, 0.5 atmIsothermal expansion

0

*6.Irreversible change of state of perfect gas.S is a state function, Srev=Sirrev

7.Constant-pressure heating.

8.General change of state (P1,T1)(P2,T2)

Const. P = P1Const. T = T1a)b)

*9.Irreversible phase change.

*

Find S for conversion of 10.0 g supercooled H2O at -10C, 1 atm to ice at -10C, 1 atm.Cpice=0.50cal/g KCpsupercooled H2O=1.01 cal/g K

Hfus = 79.7 cal/g

*

(a)

*(b)H2O(l), 0C H2O(s), 0 C

*(c) H2O(s), 0C H2O(s), -10 C

*10.Mixing of different inert perfect gas at constant T and P.

abbbbaaaaaabbbbaVa = V, Vb = VPa, VaP, TPb, VbP, TP, na, nb, V= Va, Vb, T1. Rev isothermal expansion2. Rev isothermal mixingirrev

*

0Perf. gas, const T, PXi ;mole fractions of gas i

*

Calculate S for the mixing of 10.0 g He at 120.0 C and 1.50 bar with 10.0 g O2 at the same T, P.

*

*

*

A certain perfect gas has Cv,m = a + bT where a = 25.0 J/mol K and b = 0.0300 J/mol K2. Let 4.00 mol of this gas go from 300 K and 2.00 atm to 500 K and 3.00 atm. Calculate q, w, U, H and S for this change of state.

*

q and w cannot be calculated because the path was not specified

*

*

*

*Reversible Process

*Irreversible Process

23 rev. adiabatic S = 0 S2 = S341 rev. adiabatic S = 0 S4 = S134:Reversible isothermalReversibleadiabaticReversibleadiabaticIrreversible adiabatic4312PVThr

*

w = -q34q34 0 so that it doesnt violate the 2nd law

Ssyst 0

00

*Suniv = Ssyst + SsurrSuniv = 0reversible processSuniv 0irreversible processSuniv 0any process

*Isolated Process

Real processes: mixing, chemical reaction, flow of heat from hot to cold bodies will continue to occur until S reach maximum value.

*Irreversible Mixing

ddddeeeeddddeeeeddddeeeeNonequilibrium stateUnmixedEquilibrium state12

*S = k lnp + ap = probabilityk = Boltzman constanta = constant

*

Calculate the change in the entropies of the system and the surroundings and Stot when 14 g N2 gas at 298 K and 1.00 bar doubles its volume in:

(a) an isothermal reversible expansion.(b) an isothermal irreversible expansion against Pext = 0.(c) an adiabatic reversible expansion.

*

(a)

*(b) Ssyst = +2.9 J/K (S is a state function)Surrounding:w = 0, T = constant , U = 0 no energy transfer between system surrounding Ssurr = 0 Stot = +2.9 J/K

(c) qrev = 0 Ssyst = 0 Ssurr = 0 = Stot

*2.11 THERMOCHEMISTRY2.11.1 Standard Enthalpy of Reaction, Hrxn

aA + bB cC + dD

= standard enthalpy of formation for substance i from its elements in their reference form (at T, 1 bar)

*2.11.2 Adiabatic Bomb Calorimeter

R = reactants, P = products, K = wall + H2O

R + K25CP + K25CP + K25C + T(b)(c)(a)Ua = 0Ur, 298

*

*

CK+P = heat capacity of system

ng = change in mol for gasesV constant

*

mass T UC (g) (K) (kJ/g)

Benzoic acid (BA) 0.5742 1.270 -26.434wire 0.0121 - -6.28Naphthalene (NA) 0.6018 2.035 ??wire 0.0142 - -6.28

Calculate CK+P, UC, HC for C10H8(s)C10H8(s) = naphthalene =NAC6H5COOH = benzoic acid = BA

*

(a) BA

*(b)NA

*Combustion reaction:C10H8(s) + 12O2(g) 10CO2(g) + 4H2O(l)

*2.11.3 Hesss Law: Combine heats of several reactions to obtain H of desired reaction.

2C(graphite) + 3H2(g) C2H6(g) Hf,298 = ?

Given:C2H6(g) + 7/2O2(g) 2CO2(g) + 3H2O(l) -1560C(graphite) + O2(g) CO2(g) -393.5H2(g) + 1/2O2(g) H2O(l) -286H298 kJ/mol

*(1) -1: 2CO2(g) + 3H2O(l) C2H6(g) + 7/2O2(g) 1560

(2) 2: 2C(graphite) + 2O2(g) 2CO2(g) -787

(3) 3: 3H2(g) + 3/2O2(g) 3H2O(l) -858

2C(graphite) + 3H2(g) C2H6(g) 85

H298 = -85 kJ/mol = Hf,298Exothermic reaction

*

4NH3(g) + 5O2(g) 4NO(g) + 6H2O(l) 11702NO(g) + O2(g) 2NO2(g) 1143NO2(g) + H2O(l) 2HNO3(l) + NO(g) 72

Calculate H298 for:NH3(g) + 2O2(g) HNO3(l) + H2O(l)HkJ/mol

*

:NH3(g) + 5/4O2(g) NO(g) + 6/4H2O(l) 292.5(2) :3/2NO(g) + 3/4O2(g) 3/2NO2(g) -85.5(3) :3/2NO2(g) + 1/2H2O(l) HNO3(l) + 1/2NO(g) -36

NH3(g) + 8/4O2(g) HNO3(g) + 4/4H2O(l) -414

*2.11.4 Temperature dependence of reaction heats

*

*

Hf,298(g) = -241.82 kJ/mol. Calculate Hf,373 H2O(g).

Given:

*

H2(g) + O2(g) H2O(g)

*

*

Calculate Hf,1000 for HCl(g).

Given:CP,m, H2(g) = 6.52 + 7.8 10-4 T + 0.12 105T-2CP,m, Cl2(g) = 8.82 + 0.6 10-4 T 0.68 105 T-2CP,m, HCl(g) = 6.34 + 11 10-4 T + 0.26 105 T-2

CP,m unit = cal/mol KHf,298, HCl(g) = -22.06 K cal/mol

*H2(g) + Cl2(g) HCl(g)

*

*Standard Entropysolid/liquid elements

Third Law of Thermodynamics:

(solid)(melting)(liquid)

*

Standard entropy change for a reaction:

i = stoichiometric coefficient

*

2H2S(g) + 3O2(g) 2H2O(l) + 2SO2(g)

Hf,298 -20.63 0 285.830 296.83CP,m 34.23 29.355 75.291 39.87Sm,298 205.79 205.138 69.91 248.22

Calculate H298, H 370, S298, S370

*

*

*

*

*2.11.5 Standard Gibbs Energy for Reaction, Grxn

Gf,i = Standard Gibbs energy of formation of a substance from its elements in their reference form.Gf = 0 for element in its reference form.

*

Urea, CO(NH2)2 has the following properties:Hf,298 = -333.51 kJ/mol; Sf,298 = 104.60 J/mol KCalculate Grxn,298.

C(graphite) + O2(g) + N2(g) + 2H2(g) CO(NH2)2