Asaresultofthedifferenceintheirinitial

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ‐
The
Second
Law
of
Thermodynamics
 98
 
 Entropy
Changes
for
the
System
during
Specific
Reversible
Processes:
 
 1)
 Isothermal
Reversible
Expansion
or
Compression
of
an
Ideal
Gas
 dU
=
δw
+
δq
=
CVdT
=
0

δq
=
‐δw
=
PextdV
=
P
dV
=
nRT
(dV/V)
 dS
=
δq
/
T
=

nR
(dV/V)
 Note
dU
=
0
because
the
process
is
isothermal
for
an
ideal
gas.
 Note
Pext
=
P
because
the
process
is
reversible.
 Note
dS
=
dδ
/
T
because
the
process
is
reversible.
 Note
when
a
process
is
reversible,
it
is
mechanically
reversible
(Pext
=
P)
and
 it
is
thermally
reversible
(Tsurr
=
T)
 Therefore,
ΔS
=
nR
ln
(Vf
/
Vi)
 2)
 Reversible
Adiabatic
Process
(all
materials)
 If
a
process
is
reversible,
then
dS
=
δq
/
T
 If
the
process
is
adiabatic
then
δq
=
0.
Therefore,
for
a
reversible
adiabatic
 process,
dS
=
0.
 One
says
that
a
reversible
adiabatic
process
is
also
isentropic
(constant
 entropy).
 
 Marand’s
Notes:
Chapter
3
‐
The
Second
Law
of
Thermodynamics
 99
 3)
 Reversible
Heating
at
Constant
Pressure
(all
materials)
 dS
=
δq
/
T
=
δqP
/
T
=
dH
/
T
=
CP
dT
/
T
 Therefore,
ΔS
=
CP
ln(Tf
/
Ti)
 
 4)
 Reversible
Heating
at
Constant
Volume
(all
materials)
 dS
=
δq
/
T
=
δqV
/
T
=
dU
/
T
=
CV
dT
/
T
 Therefore,
ΔS
=
CV
ln(Tf
/
Ti)
 
 5)
 Reversible
Phase
Transformations:
 If
a
phase
transformation
takes
place
under
equilibrium
conditions,
then

 dS
=
δq
/T
=
dH
/
T
 since
during
a
phase
transformation,
the
temperature
(as
well
as
the
 pressure,
see
above
δq
=
dH)
are
constant.
Then
we
can
integrate
dS
to
 obtain:
 ΔS
=
ΔH...
View Full Document

This note was uploaded on 01/26/2014 for the course CHEM 3615 taught by Professor Aresker during the Spring '07 term at Virginia Tech.

Ask a homework question - tutors are online