Thermodynamics and Statistical Mechanics Review for Quiz 1.
Transcript of Thermodynamics and Statistical Mechanics Review for Quiz 1.
Thermodynamics and Statistical Mechanics
Review for Quiz 1
Thermo & Stat Mech - Spring 2006 Class 11
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Laws of Thermodynamics
First law: đQ – đW = dU
Energy is conserved
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Laws of Thermodynamics
Second Law: The entropy of an isolated system increases in any irreversible process and is unaltered in any reversible process. This is the principle of increasing entropy.
S 0
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Laws of Thermodynamics
Third Law: The entropy of a true equilibrium state of a system at a temperature of absolute zero is zero.
Equivalent to: It is impossible to reduce the temperature of a system to absolute zero using a finite number of processes.
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Second Law Variations
No series of processes is possible whose sole result is the absorption of heat from a thermal reservoir and the complete conversion of this energy to work.
There are no perfect engines!
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Second Law Variations
No series of processes is possible whose sole result is the transfer of heat from a reservoir at a given temperature to a reservoir at a higher temperature.
There are no perfect refrigerators!
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Zeroth Law
If two systems are separately in thermal equilibrium with a third system, they are in thermal equilibrium with each other.
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Work done by a gas
f
i
V
VPdVW
PdVdW
AdsAF
dW
FdsdW
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Ideal gas law
Ideal gas law:PV = nRT
In terms of molar volume, v = V/n, this becomes:
Pv = RT, or P = RT/v
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van der Waals equation of state
RTbvva
P
va
bvRT
P
2
2 or , Then,
This equation has a critical value of T which suggests a phase change. The next slide shows graphs for several values of T .
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Thermal Expansion
Expansivity or Coefficient of Volume Expansion, .
TVTTV
V
PT
Tv
vTV
V
P
PP
),(
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Compressibility
Volume also depends on pressure.
Isothermal Compressibility:
PVPPV
V
PTPV
V
T
T
),( 1
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Cyclical Relation
1
0
PVT
VTP
VTP
VT
TP
PV
TP
PV
TV
TP
PV
TV
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Carnot Cycle
A Carnot cycle is an idealized reversible cycle that operates between two heat reservoirs at temperatures T1 and T2, where T2 > T1. It can operate as a heat engine, or a refrigerator.
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Thermal Efficiency ()
2
12
2
1
2
1
2
12
2
1
1
TTT
TT
Q
Q
Q
Q
W
If T1 = 0, = 1 (100%)
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For a Carnot Engine
TQTT
Q
Q
1
2
1
2
1
2
1
2
TT
QQ or 0
2
2
1
1 TQ
TQ
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Entropy
0 TQd
TQd
i i
i
dSTQd
For reversible processes.Entropy is a state variable.
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First and Second Laws
First Law: dU = đQ – đW
First law, combined with the second law:
dU = TdS – PdV
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Tds Equations
dPc
dvv
cdP
P
Tcdv
v
TcTds
dPTvdTcdPT
vTdTcTds
dvT
dTcdvT
PTdTcTds
vP
vv
PP
PP
P
vv
v
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Ideal Gas
RccTPv
PT
Tvcc
Tvcc
vp
vp
vp
2
2
1
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Properties
From first law: TdS = dU + PdV, or
Internal Energy dU = TdS – PdV U(S, V)
Enthalpy: H = U + PV
dH = TdS + VdP H(S, P)
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New Potentials
Helmholtz Function:
F = U – TS
Gibbs Function:
G = U – TS + PV
G = H – TS
G = F + PV
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All Four
dU = TdS – PdV U(S, V)
dH = TdS + VdP H(S, P)
dF = – PdV – SdT F(V, T)
dG = – SdT + VdP G(T, P)
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Maxwell Relations
PTPS
VTVS
T
V
P
S
S
V
P
T
T
P
V
S
S
P
V
T
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Clausius-Clapeyron Equation
liquid-Solid )(
vapor-Solid )(
vapor-Liquid )(
12
12
13
13
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vvTdT
dP
vvTdT
dP
vvTdT
dP