CHM2132 Tutorial notes for Oct. 9
th
, 13
th
1) When 3.00 mol of a gas at 230 K and 150 kPa is subjected to isothermal compression,
its entropy decreases by 15.0 J K
1
.
Calculate (a) the final pressure of the gas and (b)
Δ
G
for the compression.
We are given:
n
=3.00 mol,
T
1
=230K,
p
1
= 150 kPa,
∆
S
= 15.0 J K
1
,
∆
T
= 0
a) What is
p
2
?
We are given entropy and we have been told that this is an isothermal compression.
The equation for this entropy change is:
!
S
=
nR
ln
V
2
V
1
"
#
$
%
&
’
Since the process is isothermal,
p
1
V
1
=
p
2
V
2
and;
!
S
=
nR
ln
p
1
p
2
"
#
$
%
&
’
Since we have been given
p
1
,
n
and
Δ
S
we can rearrange to find
p
2
:
ln
p
1
p
2
!
"
#
$
%
&
=
’
S
nR
ln
p
2
=
(
’
S
nR
+
ln
p
1
=
(
(
15 J K
1
mol
1
3.00 mol
(
)
8.314 J K
1
mol
1
(
)
+
ln 150
(
)
=
5.61
p
2
= 274 kPa
b) What is
Δ
G
?
For isothermal processes, the pressure dependence of Gibbs is given by:
!
G
!
p
"
#
$
%
&
’
T
=
V
We can rearrange this to isolate the variables to solve for Gibbs:
dG
=
Vdp
Then integrate both sides:
dG
G
1
G
2
!
=
Vdp
p
1
p
2
!
Since the volume of the gas changes as a function of pressure we must cast
V
in terms of
pressure by substituting in the ideal gas equation:
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dG
G
1
G
2
!
=
nRT
p
dp
p
1
p
2
!
=
nRT
dp
p
p
1
p
2
!
!
G
=
nRT
ln
p
2
p
1
"
#
$
%
&
’
This is the Gibbs energy equation for the isothermal expansion of a gas. We can now
substitute in the given values:
!
G
=
3.00 mol
(
)
8.314 J K
1
mol
1
(
)
230 K
(
)
ln
274 kPa
150 kPa
"
#
$
%
&
’
=
3.45 kJ
Note that since
!
G
=
!
H
"
T
!
S
and
Δ
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 Fall '08
 GIORGI
 Entropy, Gibbs, isothermal expansion

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