The Reversible Carnot Cycle

The Reversible Carnot Cycle
I. Reversible Isothermal Expansion (1 2)
II. Reversible Adiabatic Expansion (2 3)
III. Reversible Isothermal Compression (3 4)
IV. Reversible Adiabatic Compression (4 1)
Processes I II III IV
State 1 State 2 State 3 State 4 State 1
T1 T1 T3 T3 T1
V1 V2 V3 V4 V1
P1 P2 P3 P4 P1
Processes I II III IV
Notice that since I & III are isothermal processes, T2 has been replaced by T1 and T4 has been replaced by
T3. Additionally, above and below the chart, the Roman numerals for the Processes have been placed
along the borders between the states so that the cycle can be more clearly visualized.
States 1 2 3 4 1
Process I Process II Process III Process IV
q -wI 0 -wIII 0
w -nRT1 ln (V2/V1) ΔUII -nRT3 ln (V4/V3) ΔUIV
ΔU 0 nCv(T3-T1) 0 nCv(T1-T3)
ΔH 0 nCp(T3-T1) 0 nCp(T1-T3)
States 1 2 3 4 1
In this chart, the method of calculation of the various thermodynamic variables have been inserted. (Note
that here the Arabic numerals of the States have been placed along the “borders’ of the Processes to help
with the visualization.)
Since this is a cycle, the sum of the thermodynamic state variables must be zero. An examination of the
equations for ΔU and ΔH illustrates this quite nicely: (T3 – T1) is the negative of (T1 – T3) so ΔUII must
equal -ΔUIV and ΔHII must equal -ΔHIV and their sums must equal zero.
Using the pair of adiabatic processes to analyze system
Process II (2 ! 3) Process IV (4 ! 1)
q II = 0 so dU II = w II q IV = 0 so dU IV = w IV
3
2 ln T " % 3
#
$
&
' = - ln V " % 3
#
$
&
' 3
2 ln T " % 1
#
$
&
' = - ln V " % 1
#
$
&
'
T 2
V 2
But T 1 = T 2 & T 3 = T 4 so
3
2 ln T ! $ 4
"
#
%
& = - ln V ! $ 3
"
#
%
& 3
2 ln T ! $ 1
"
#
%
& = - ln V ! $ 1
"
#
%
&
T 1
V 2
T 4
T 4
V 4
V 4

and that means ln V ! $ 3
"
#
%
& = - ln V ! $ 1
"
#
%
&
V 2
which rearranges to V ! $ 3
"
#
%
& = V ! $ 4
"
#
%
& ' V ! $ 1
"
#
%
& = V ! $ 4
"
#
%
& or V ! $ 2
"
#
%
& = V ! $ 3
"
#
%
&
V 2
V 1
Since the total work is {-nRT1 ln (V2/V1)} + {-nRT3 ln (V4/V3)} & ln (V2/V1) = - ln (V4/V3)
Then
-nR T1 ln V ' ! $ * 2
)
"
# V1 %
& , + -nR T3 ln
(
+
V ' ! $ * 4
)
"
# V3 %
& , = -nR T1 ln
(
+
V ' ! $ * 2
)
"
# V1 %
& , + +nR T3 ln
(
+
V ' ! $ * 2
)
"
# V1 %
& ,
(
+
work total = T 3 - T 1
'
)
(
V 2
!
"
#
V 4
( ) nR ln V 2
and because T1 ≠ T3. Thus, the net total work for the cycle cannot be ZERO
HOMEWORK PROBLEM
V1 = n RT1 =
P1 (1.000)(8.31447)(300.0)
! V1 = 24.94341 Liters
100.0
Processes I II III IV
State 1 State 2 State 3 State 4 State 1
Temperature (K) 300.0 300.0 100.0 100.0 300.0
Volume (Liters) 24.94 V2 V3 V4 24.94
Pressure (kPa) 100.0 P2 P3 P4 100.0
Processes I II III IV
States 1 2 3 4 1
Process I Process II Process III Process IV
q 2.500 0 -wIII 0
w -2.500 ΔUII -nRT3 ln (V4/V3) ΔUIV
ΔU 0 nCv(T3-T1) 0 nCv(T1-T3)
ΔH 0 nCp(T3-T1) 0 nCp(T1-T3)
States 1 2 3 4 1
Just like above, use the adiabatic process to identify the states (starting with IV)
Process IV (4 ! 1)
q IV = 0 so dU IV = w IV " 3
2
# 100.0 &
ln
$
%
300.0'
( = - ln 24.94341 # &
$
%
'
(
V 4
V 1
V 3
$ *
%
& ,
+
V 1
V 4

ln 24.94341 ! $
! 24.94341$
"
#
%
& = -1.647918 &
"
#
%
& = 0.192645 & V4 = 129.60976 Liters
V 4
P 4 = n RT 4
V 4
= (1.000)(8.31447)(100.0)
129.60976
Now the first table looks like this
V 4
' P 4 = 6.41500 kPa
Processes I II III IV
State 1 State 2 State 3 State 4 State 1
Temperature (K) 300.0 300.0 100.0 100.0 300.0
Volume (Liters) 24.93 V2 V3 129.6 24.93
Pressure (kPa) 100.0 P2 P3 6.415 100.0
Processes I II III IV
Since wI = -2.500 kJ , wI = - n RT1 ln V ! $ 2
"
#
%
& and V1 = 24.9434, V2 = 67.95719 Liters
P 2 = n RT 2
V 2
V 1
= (1.000)(8.31447)(300.0)
67.95719
! P 2 = 36.70459 kPa
The above proof shows that ln V ! $ 3
"
#
%
& = - ln V ! $ 1
"
#
%
& that means V ! $ 3
"
#
%
& = V ! $ 4
"
#
%
And finally P 3 = n RT 3
V 3
V 2
V 4
= (1.000)(8.31447)(100.0)
353.1305
V 2
! P 3 = 2.35450 kPa
V &
1
& V3 = 353.1305 Liters
Processes I II III IV
State 1 State 2 State 3 State 4 State 1
Temperature (K) 300.0 300.0 100.0 100.0 300.0
Volume (Liters) 24.93 67.96 353.1 129.6 24.93
Pressure (kPa) 100.0 36.70 2.355 6.415 100.0
Processes I II III IV
States 1 2 3 4 1
Process I Process II Process III Process IV
q 2.500 0 -0.83334 0
w -2.500 -2.494 0.83334 2.494
ΔU 0 -2.494 0 2.494
ΔH 0 -4.157 0 4.157
States 1 2 3 4 1