CHAPTER 3 SECOND LAW OF THERMODYNAMICS 65
Since each term has a logarithm, this equation reduces to:
or (db3.11)
Tfinalc Vinitial=Tinitialc Vfinal
If lnx=lnythen ln(x/y) =1 and x=yFor the Carnot cycle, the products are:
VAThotc =VDTcoldc and VCTcoldc =VBThotc (db3.12)
Dividing these two expressions gives:
VAThotc =VDTcoldc and VCTcoldc =VBThotc
(db3.13)Consequently, we can substitute these volume ratios into the expression for the heat flow
(eqn db3.4):
(db3.14)VT
VT
VT
VT
AhotcBhot
cDcoldcCcold= c orVV
V
V
V
A
BD
C=
T
T
V
V
nal
initialc
nal
initial⎛ 33
⎝
⎜⎜
⎞
⎠
⎟⎟ =
⎛
⎝
⎜⎜
⎞
⎠⎠
⎟⎟
and
or
qT
q
Thot
hotcold
cold−= 0
q
qnRTV
V
nRTV
V
hot T
coldhotb
acoldb
alnln=
−
=−hhot
TcoldqnRT
V
V
nRTV
V
cold cold d nR
ccold
a
b==ln ln =− TTV
cold V
b
alnqnRT
V
hot hot Vb
a= ln