9 th Week Chap(12-13) Thermodynamics and Spontaneous Processes Standard State: The Thermodynamically...
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9 th Week Chap(12-13) Thermodynamics and Spontaneous Processes • Standard State: The Thermodynamically stable state for pure liquids(Hg) and solid(graphite), for gases ideal gas behavior, for solutions 1.0 molar concentration of the dissolved species at P= 1.0 atm and some specified T in each case • Reversible and Irreversible processes: Processes that occur through a series of equilibrium states are reversible. Adiabatic(q=0) paths are reversible • Thermodynamic Universe= System + Surrounding Isolated : No Energy and Matter can go in or out Adiabatic: No heat goes in or out • Entropy (S): measure of disorder Absolute Entropy S=k B ln Number of Available Microstate ~ # of quantum states~size • Second Law of Thermodynamics: Heat cannot be transferred from cold to hot without work S ≥ q/T In-quality of Clausius
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Transcript of 9 th Week Chap(12-13) Thermodynamics and Spontaneous Processes Standard State: The Thermodynamically...
9th Week Chap(12-13) Thermodynamics and Spontaneous Processes• Standard State: The Thermodynamically stable state for
pure liquids(Hg) and solid(graphite), for gases ideal gas behavior, for solutions 1.0 molar concentration of the dissolved species at P= 1.0 atm and some specified T in each case
• Reversible and Irreversible processes:Processes that occur through a series of equilibrium states are reversible. Adiabatic(q=0) paths are reversible
• Thermodynamic Universe= System + Surrounding Isolated : No Energy and Matter can go in or outAdiabatic: No heat goes in or out
• Entropy (S): measure of disorderAbsolute Entropy S=kBln Number of Available Microstate ~ # of quantum states~size
• Second Law of Thermodynamics: Heat cannot be transferred from cold to hot without workS ≥ q/T In-quality of ClausiusS = q/T for reversible (Isothermal) processes S > q/T for irreversible processes
• Third Law of Thermodynamics: S as T for a pure substance
• Reversible Processes: S =∫qrev/T but not IsothermalFrom the 2nd Law ST=0∫T qrev/T Absolute Entropy
Thermodynamic Processes no reactions/phase Transitions
isothermPext
T>T
T
T<T
Isotherms T>T>TP = nRT/V for T =const
Isothermal Paths are always reversible, e.g. AB and BA
Whereas ABC and ADB are not reversible since they occur ina finite number of steps.
Though both paths must result in the same changes in U, U=UB – UA, since U is a state function.
Thermodynamic Processes no reactions/phase Transitions
UAC = qin + wAC qin= n cP(TC – TA) > 0 and wAC = - PextV
UCB = qout + wCB qout= n cV(TC – TB) < 0 and wCB = - PV=0
UAB = UAC + UCB = n cP(TC – TA) - PextV + n cV(TC – TB)
isotherm
qin>0
qout<0
Pext
T>T
T
T<T
Thermodynamic Processes no reactions/phase Transitions
isotherm
qin>0
qout<0
Pext
wAC= - PextVAC
work=-(area)Under PV curve
wAC= - PextVAC
work=-(area)Under PV curve
Area under A-C-B (-)work done by the systemor (+)work done on the system
Thermodynamic Processes no reactions/phase Transitions
isotherm
qin>0
qout<0
Pext
wDB= - PextVAC
work=-(area)Under PV curve
wDB= - PextVAC
work=-(area)Under PV curve
Area under A-D-C (-)work done by the systemor (+)work done on the systemIn the irreversible path A-D-B-C-A more work isdone on the system than by the system!Whereas the reversible path AB the work is the same
Fig. 12-14, p. 506
Hess’s Law applies to all State Function
A
B
D
C
AD (1)AB (2)BC
(3)CD
Example of Hess’s LawC(s,G) + O2(g) CO2(g) = -393.5 kJ CO2((g ) CO(g) + ½O2(g) = + 283 kJC(s,G) + O2(g) CO(g) + ½ O2(g) = ?
Example of Hess’s LawC(s,G) + O2(g) CO2(g) = -393.5 kJ CO(g) + ½O2(g) CO2((g) = - 283 kJ===========================C(s,G) + O2(g) CO(g) + ½ O2(g) == -110.5 kJ
The Standard State:Elements in their most stables form are assigned a standard heat of Formation H°f= 0
The standard state of solids/liquids is the pure phase @1 atm
Gases it’s the ideal gas @ 1 atm
For species in solution the Standard State is the concentration of 1.0 Molar@ P=1.0 atm H°f= 0 for H+ ions solution
For compounds the H°f is defined by formation of one moleof the compound from its elements in their Standard States @ 1 atm, 1 molar and some T
f = 0
C(s,G) + O2(g) CO2(g) H°f = -393 kJmol-1 @25 °C
In General for a reaction, with all reactants and products at a partial pressure of one 1 atm and/or concentration of 1 Molar
aA + bB fF + eE
The Standard Enthalpy Change at some specified Temperature
(rxn) f(prod) -
f(react)
(rxn) = f f(F) + e
f(E) – {af(A) + b
f(B)}energy
Elements in their standard states f(elements)=0Elements in their standard states f(elements)=0
f(reactants)
(rxn
f(product)
For Example consider the reaction:
1/2 O2(g) O(g) f[O(g)]
(atomization energy/bond enthalpy) the formation of Ozone3/2O2(g) O3(g)
f[O3(g)]energy
f[O2(g)]=0f[O2(g)]=0
f(reactants)
(rxn)
f[O3]
O(g)
f[O(g)]>0
O(g) + O2(g) O3(g) =f[O3] -
f[O] - f[O2]
O3
Standard Heat of rxn for
For example consider the combustion rxn(not balanced):
CH4(g) + O2(g) CO2(g) + H2O(l) = standard ?
energy
f[O2(g)]=0f[O2(g)]=0
f[CH4] +
f[O2]
(rxn)
= f[CO2] +
f[H2O] - f[CH4] -
f[O2]
CH4(g) + O2(g)
CO2(g) + H2O(l)
f[CO2] +
f[H2O]
The combustion rxn(balanced):
CH4(g) + 2O2(g) CO2(g) +2H2O(l) = standard ?
energy
f[O2(g)]=0f[O2(g)]=0
f[CH4] +
f[O2]
(rxn)
= f[CO2] +2
f[H2O] - f[CH4] - 2
f[O2]
CH4(g) + O2(g)
CO2(g) + H2O(l)
f[CO2] +
f[H2O]
For example consider the combustion rxn:
CH4(g) + O2(g) CO2(g) + H2O(l) = standard ?
= f[CO2] +
f[H2O] - f[CH4] -
f[O2]
energy
f[O2(g)]=0f[O2(g)]=0
f[CH4] +
f[O2]
(rxn)
CH4(g) + O2(g)
CO2(g) + H2O(l)
f[CO2] +
f[H2O]
Consider the Reversible Isothermal Expansion of an Ideal Gas:So T=const for the system. U = q + w = 0
q = -w = (Pext V)
Heat must be transferred from the Surroundings to the System And the System does work on the surroundings
If the process is reversible, Pext ~ P=nRT/VWhich means done slowly and changes are infinitesimalV dV: The gas expands from V1 to V2 then
q ~ PdV= (nRT/V)dV = (nRT) dlnV since dlnV=dV/V
Integrating from V1 to V2 q = nRT[lnV2 - lnV1]=nRTln(V2/V1)
(q) Heat transferred in the Reversible Isothermal Expansion of an Ideal Gas
q = nRTln(V2/V1)
Reversible
Irreversibleq = nRTln(V2/V1)
Carnot will use both Isothermal and Adiabatic q = 0
processes to define the entropy change S
For an Ideal gasU = q + w =w
U=ncV T = -Pext VReversible Process Pext ~ P=nRT/V
T dT and VdV
ncV dT = - P dV
Carnot will use both Isothermal and Adiabatic q = 0
processes to define the entropy change S
Reversible Process Pext ~ P=nRT/V
T dT and VdV
ncv dT = - P dVncv(dT/T)=-nRdV/V
cvdln(T) =-R dln(V)
cvln(T2/T1)=Rln(V1/V2) T2
T1
V1 V2
cvln(T2/T1)=-Rln(V1/V2)
A simpler form cab be obtained
cvln(T2/T1)= ln(T2/T1)Cv =Rln(V1/V2)= ln(V1/V2)R
(T2/T1)Cv = (V1/V2)R
Recall that cp = cv + R
(T2/T1)Cv = (V1/V2)R = (V1/V2)Cp-Cv let =cp/cv
(T2/T1) = (V1/V2) -1 this relationship is very important foradiabatic processes in engines
Carnot considered a cyclical process involving the
adiabatic/Isothermal Compression/Expansion of an Ideal Gas:
The so called Carnot cycle
Which Yields (qh/Th) –(ql/Tl) = 0 and therefore S = q/T and that S is a State Function
Since q = nRTln(V2/V1) then S = q/T= nRln(V2/V1)
Th
Tl
qh;Th
ad ql:: Tl
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