The Reversible Carnot Cycle

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