UTEG COLLEGE
The Second Law
Brief review of last class (adiabatic processes)
The ideal gas and entropy
The second law
The Carnot cycle
A new function of state - entropy
Heat capacity
Using the first law, it is easily shown that:
Q
CV
Always true
V
For an ideal gas, U = f n( ) only. Therefore,
dU
CV
and U U U 0 CV d CV 0
0
d
Enthalpy, H = U + PV, therefore:
dH = dU + PdV + VdP = Q + VdP
Ideal gas:
dH
CP
dT
Q
CP
Always true
P
and H H H 0 C P d C P 0
0
Calculation of work for a reversible process
(1)
1. Isobaric (P = const)
(2)
2. Isothermal (PV = const)
3. Adiabatic (PV = const)
(3)
4. Isochoric (V = const)
(4)
V
For a given reversible path, there is some associated physics.
W PdV area under curve; U Q + W
Configuration Work on an ideal gas
W PdV 0
Isochoric
W P dV P V f Vi
Isobaric
V f
dV
W PdV nR
nR ln
Isothermal
Vi V
Vi
Note: for an ideal gas, U = U(), so W = Q for isothermal processes.
It is also always true that, for an ideal gas,
Vf
U CV f i
and
H C P f i
Adiabatic processes: Q = 0, so W = U, also PV = constant.
1
W CV f i
Pf V f PV
i i
1
3R
5R
cP 5
Monatomic:
c
;
c
V
P
2
2
cV 3
5R
7R
cP 7
Diatomic: cV
; cP
;
2
2
cV 5
Chapter 2
100% Conversion of Heat to Work
2
Q
M
W=Q
Heat in equals heat out; energy is conserved! However,
common sense tells us this will not work (or it will in a
while).
100% transfer of heat to from cold to hot body
2 > 1
Something is
clearlyMmissing
Q
from the first
law!
Q1
1 < 2
Heat in equals heat out; energy is conserved! But we know
this never happens in the real world.
The Second Law of Thermodynamics
Clausius statement: It is impossible to construct a device
that operates in a cycle and whose sole effect is to transfer
heat from a cooler body to a hotter body.
Kelvin-Planck statement: It is impossible to construct a
device that operates in a cycle and produces no other effect
than the performance of work and the exchange of heat
from a single reservoir.
Carnots theorem: No engine operating between two
reservoirs can be more efficient than a Carnot engine
operating between the same two reservoirs.
Conversion of Heat to Work (a heat engine)
Efficiency (*):
Heat reservoir at
temperature 2 > 1
W
W
output
Q2 Q2
input
Q2 Q1
Q1 *
1
Q2
Q2
Q1
Q2
Q2
Heat
Engine
Q heat
W work
both in Joules
W Q2 Q1 *
Q1
Cold reservoir at
temperature 1 < 2
Q2 Q1
*Be careful with the
signs for heat!
The Carnot Cycle
ab isothermal expansion
bc adiabatic expansion
cd isothermal compression
da adiabatic compression
1.
2.
3.
4.
2
1
1.
2.
3.
4.
W2 > 0, Q2 > 0 (in)
W' > 0, Q = 0
W1 < 0, Q1 < 0 (out)
W'' < 0, Q = 0
1
1
2
Stirlings engine is a good approximation to Carnots cycle.
The Carnot Cycle
1.
2.
3.
4.
ab isothermal expansion
bc adiabatic expansion
cd isothermal compression
da adiabatic compression
1.
2.
3.
4.
W2 > 0, Q2 > 0 (in)
W' > 0, Q = 0
W1 < 0, Q1 < 0 (out)
W'' < 0, Q = 0
T1
1
T2
Stirlings engine is a good approximation to Carnots cycle.
Pressure
The absolute temperature (Kelvin) scale
T(K) = T(oC) + 273.15
Triple point
of water:
273.16 K
Based on the
ideal gas law
0
100
200
300
400
Temperature (K)
The Second Law of Thermodynamics
Clausius statement: It is impossible to construct a device
that operates in a cycle and whose sole effect is to transfer
heat from a cooler body to a hotter body.
Kelvin-Planck statement: It is impossible to construct a
device that operates in a cycle and produces no other effect
than the performance of work and the exchange of heat
from a single reservoir.
Carnots theorem: No engine operating between two
reservoirs can be more efficient than a Carnot engine
operating between the same two reservoirs.
Consider two heat engines
Carnot
Hypothetical
T2 > T1
T2 > T1
Q2
Q2'
W = Q2 Q1
Q1
M'
W' = Q2' Q1'
Q1'
W' W
Q2' Q2
T1 < T2
T1 < T2
Efficiency =
Efficiency = ' >
Connect M' to M and run M in reverse
W = W'
T2 > T1
Q2'
M'
M
Q2
W'
Q1'
Q1
T1 < T2
Therefore, using:
W' W
,
Q2' Q2
Q2 Q2'
&
Q1 Q'1
M is a Carnot engine, so we are entitled to run it in reverse
This would be equivalent to:
T2 > T1
Q1
M
Q1
T1 < T2
This violates Clausius statement of the 2nd law!
Conversion of Heat to Work (a heat engine)
T2 > T1
T2 > T1
Q1
Q2 Q1
Q2
M'
Q1
Q1
T1 < T2
This violates Kelvin-Planck statement of the 2nd law!
The Clausius Inequality and the 2nd Law
Efficiency (*):
W
W
output
Q2 Q2
input
Q2 Q1
Q2
Q1
1
Q2
T1
1
T2
Heat reservoir at
temperature T2 > T1
or
T1 Q2Q1
Q1
T
Q2
Q2
Heat
2
Engine
Q Q
2
W Q2 Q1
T2 QT
11
Q2 Q1
orHeat reservoir
at 0
T2
T1
temperature T1 < T2
The Clausius Inequality and the 2nd Law
Divide any reversible cycle into a
series of thin Carnot cycles, where
the isotherms are infinitesimally
short:
Q2
Q1
We have proven that the entropy, S, is a state variable,
since the integral of the differential entropy (dS = dQ/T),
around a closed loop is equal to zero, i.e. the integration
of differential entropy is path independent!
The Clausius Inequality and the 2nd Law
Divide any reversible cycle into a
series of thin Carnot cycles, where
the isotherms are infinitesimally
short:
Q2
Q2 Q1
0
T2
T1
Q1
Qi
Qr
0
i T
T
i
There is one major caveat: the cycle must be reversible.
In other words, the above assumes only configuration
work (PdV) is performed.
If the cycle additionally includes dissipative work, it is
not clear how to include this in the above diagram.
Thermodynamic cycles for designing ideal engines
and heat pumps
http://auto.howstuffworks.com/engine1.htm
Engine process:
P (1.013 x 105) Pa
Pf
Work of engine :
B
Heat input to system : Q Qin Qout
Efficiency :
Pi
Vi
Vf
Weng W
Weng
Qin
Examples process by an ideal gas:
P (1.013 x 105) Pa
Pf
AB
Q
W
Pi
Eint
BC
Vf
DA
Vi ( Pf Pi )
Pf (V f Vi )
V f ( Pf Pi )
-Pi (V f Vi )
-1
-1
-1
-1
-Pf(Vf-Vi)
Pi(Vf-Vi)
Vi ( Pf Pi )
Pf (V f Vi )
V f ( Pf Pi )
-Pi (V f Vi )
-1
-1
-1
-1
Efficiency :
Vi
CD
Weng
Qin
Pi V f Vi
QAB QBC
Example from homework
Efficiency :
Weng
Qin
Pi V f Vi
QAB QBC
Also : W Weng Q Qin Qout
Qin Qout
Qout
1
Qin
Qin
QCD QDA
QAB QBC
Most efficient thermodynamic cycle -- Carnot
Sadi Carnot 1796-1832
Carnot cycle:
AB
BC
CD
DA
Isothermal at Th
Adiabatic
Isothermal at Tc
Adiabatic
Efficiency of Carnot cycle
Qin Qout
Qin
Tc
1
Th
Qout
Qin
iclicker exercise:
We discussed the efficiency of an engine as
Qin Qout
Qin
Qout
Qin
Is this result
A. Special to the Carnot cycle
B. General to all ideal thermodynamic cycles
iclicker exercise:
We discussed the efficiency of an engine running
with hot and cold reservoirs as
T
1 c
Th
Is this result
A. Special to the Carnot cycle
B. General to all ideal thermodynamic cycles
Note that for a Carnot cycle :
Qout
Qin
WAB
WCD
nRTc ln(VC / VD )
nRTh ln(VB / VA )
For adiabatic process
ThVB 1 TcVC 1
ThVA 1 TcVD 1
VC / VD VB / VA
Qout Tc
Qin
Th
iclicker exercise:
Why should we care about the Carnot
cycle?
A. We shouldnt
B. It approximately models some
heating and cooling technologies
C. It provides insight into another
thermodynamic variable -- entropy
The Otto cycle
Theoretical efficiency :
V2
1
V1
PHY 113 A Fall 2012 -- Lecture 32
11/19/2012
30
The Diesel cycle
Theoretical efficiency :
1 TD TA
1
TC TB
31
Engine vs heating/cooling designs
