Transcript pchemc3-2.ppt

Chapter 3
The Second Law
Unit 2 Carnot Cycle
Spring 2009
http://galileoandeinstein.physics.virginia.edu/more_stuff/flashlets/carnot.htm
Figure 5.1
A schematic depiction of a heat engine is shown. Changes in temperature of the
working substance brought about by contracting the cylinder with hot or cold
reservoirs generate a linear motion that is mechanically converted to a rotary
motion, which is used to do work.
Carnot Cycle
1. Reversible isothermal expansion from A to B
at Th; the entropy change is qh/Th,where qh is the
energy supplied to the system as heat from the
hot source.
2. Reversible adiabatic expansion from B to C.
No energy leaves the system as heat, so the
change in entropy is zero. In the course of this
expansion, the temperature falls from Th to Tc,the
temperature of the cold sink.
3. Reversible isothermal compression from C to
D at Tc. Energy is released as heat to the cold sink;
the change in entropy of the system is qc/Tc; in
this expression qc is negative.
4. Reversible adiabatic compression from D to A.
No energy enters the system as heat,so the
change in entropy is zero. The temperature rises
from Tc to Th.
Sec
ond
La
w
of
The
rm
ody
na
mic
s
Carnot Cycle
Step 1. Reversible isothermal expansion from A to B at Th
U  0, q   w
H  0
VB
w   nRTh ln( )
VA
VB
q  nRTh ln( )
VA
VB
S  nR ln( )
VA
Carnot Cycle
Step 2. Reversible adiabatic expansion from B to C.
q  0, S  0,
U  w
U  nCV , m (Tc  Th )
w  nCV , m (Tc  Th )
H  nC p , m (Tc  Th )
Carnot Cycle
Step 3. Reversible isothermal expansion from C to D at Tc
U  0, q   w
H  0
VD
w   nRTc ln( )
VC
VD
q  nRTc ln( )
VC
VD
S  nR ln( )
VC
Carnot Cycle
Step 4. Reversible adiabatic expansion from D to A.
q  0, S  0,
U  w
U  nCV , m (Th  Tc )
w  nCV , m (Th  Tc )
H  nC p , m (Th  Tc )
Carnot Cycle
Step
q
w
U
H
S
A→B
nRTh ln(VB/VA)
-nRTh ln(VB/VA)
0
0
nR ln(VB/VA)
B→C
0
+nCv (Tc - Th)
C→D
nRTc ln(VD/VC)
-nRTc ln(VD/VC)
D→ A
0
+nCv (Th - Tc)
Totals
nRTh ln(VB/VA) [-nRTh ln(VB/VA)]
+
+
nRTc ln(VD/VC) [-nRTc ln(VD/VC)]
+nCv (Tc - Th) +nCp (Tc - Th)
0
0
+nCv (Th - Tc) +nCp (Th - Tc)
0
0
0
nRln(VD/VC)
0
nR ln(VB/VA)
+
nRln(VD/VC)
Carnot Cycle
• Step 2
c
VB Th
• Step 4
c
 VC Tc
VB VD Thc Tcc  VC V A Tcc Thc
VB VD  VC V A
V A VD

VB VC
VD Tcc  VA Thc
Carnot Cycle
• Step1: A → B
q A B
VB
 nRTh ln( )
VA
• Step3:C → D
qC  D
VD
 nRTc ln( )
VC
VA
 nRTc ln( )
VB
VB
 nRTc ln( )
VA
q h q A B

q c qC  D
VB
VB
nRTh ln( )
nRTh ln( )
Th
VA
VA



VD
VB
Tc
nRTc ln( )  nRTc ln( )
VC
VA
Carnot Cycle
qh
Th
qh
qc
 

qc
Tc
Th
Tc

q h qc
dS 

0
Th Tc
Effeciency, e
In Carnot cycle |w|=qh + qc
q h  qc
qc
Tc
e

 1
 1
qh
qh
qh
Th
w
 Tc
w  e q h  qh 1 
 Th




Clausius Inequality
U  q  w  qrev  wrev
U is a state function
w in a reversible process
Is the maximum work
wrev  w
q rev  q  wrev  w  0
q rev  q
q rev q

T
T
q
dS 
T
Spontaneous Cooling
Consider the transfer of energy as heat from one system—the
hot source—at a temperature Th to another system—the cold
sink—at a temperature Tc.
When |dq| leaves the hot source (so dqh < 0),the Clausius
inequality implies that dS ≥ dqh/Th.
When |dq| enters the cold sink the Clausius inequality implies
that dS ≥ dqc/Tc (with dqc > 0).