CHM2132  Tutorial Notes for Oct. 20
th
, 23
rd
1) Atkins & dePaula, Problem 3.7
We need to determine the entropy of NH
3
(g) at 100
o
C.
We can find the standard entropy
of NH
3
(g) at 298 K from Table 2.6 in the textbook  all we have to calculate is the
amount of entropy that arises from increasing the temperature of this gas by an additional
75
o
C:
!
S
m
o
373 K
( )
=
!
S
m
o
298 K
( )
+
!
S
m
o
298 K
"
373 K
( )
where
!
S
m
o
298 K
"
373 K
( )
is the entropy change of raising the temperature.
In order to calculate the second entropy term we must start from the general equation for
entropy:
!
S
=
"
q
rev
T
#
As this is a constant pressure heating process we can start with the general equation for
constant pressure heat capacity to find the reversible heat:
!
H
!
T
"
#
$
%
’
p
=
C
p
,
m
Rearranging into a form to isolate for enthalpy (which is the reversible heat):
dH
=
C
p
,
m
dT
=
!
q
rev
However, in this case we are told that the heat capacity is temperature dependent
according to:
C
p
,
m
=
a
+
bT
+
c
T
2
Therefore we must substitute this equation into the equation for reversible heat:
q
rev
=
a
+
bT
+
c
T
2
"
#
$
%
’
dT
Now this expression for reversible heat can be substituted back into the equation for
entropy:
!
S
=
C
p
,
m
dT
T
"
=
a
+
bT
+
c
T
2
#
$
%
’
(
T
"
dT
=
a
T
+
b
+
c
T
3
#
$
%
’
(
"
dT
Now using the following general integration formulae:
x
n
!
dx
=
1
n
+
1
x
n
+
1
+
c
and
dx
x
!
=
ln
x
( )
+
c
We can solve for entropy:
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S
=
a
ln
T
+
bT
"
c
2
T
2
T
1
T
2
=
a
ln
T
2
T
1
#
$
%
’
(
+
b
!
T
"
c
2
1
T
2
2
"
1
T
1
2
#
$
%
’
(
Now we can find the values for
a
,
b
and
c
for NH
3
(g) and plug into this equation.
However, since we know the absolute entropy at 298 K we will only need to use this
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 Fall '08
 GIORGI
 Thermodynamics, Entropy, Gibbs, mole fractions

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