Page
1
of
12
CHEM 3615
Problem Set #4 –Real Gases
Due September 18, 2007
Finish reading Chapter 2.
Problems 1 through 7 are due at the start of class and will be graded.
1.
In the last problem set, you plotted Z values for methane as a function of temperature expressed in terms of
nT
B
for the two-term virial expansion for a van der Waal's gas:
RT
p
RT
a
b
1
RT
V
p
Z
⎟
⎠
⎞
⎜
⎝
⎛
−
+
=
=
Plot the derivative of the compression factor with respect to pressure at constant temperature as a function
of temperature (100K < T < 6000K) and make a second plot where the temperature varies between (100K <
T < 1x10
6
K) using a logarithmic scale.
In both cases, pick your y axis in such a way as to clearly see any
maximum in temperature.
At what temperature (T
max
) does the slope maximize in terms of the Boyle
temperature?
Prove this mathematically.
Finally, what is the value of the maximum slope in terms of the
van der Waal's constants a and b?
{Units for T
max
are K, the maximum slope is in units of Pa
-1
}
Solution:
RT
p
)
RT
a
b
(
1
Z
−
+
=
, and
2
2
T
T
R
a
RT
b
p
Z
−
=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
∂
∂
(this is the slope of the Z - p plot).
where
2
6
mol
m
Pa
227
.
0
a
−
•
•
=
and
1
3
5
mol
m
10
28
.
4
b
−
−
•
×
=
So we can get:
2
6
T
T
00328
.
0
T
10
15
.
5
p
Z
−
×
=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
∂
∂
−
The derivative of the compression factor with respect to pressure at constant temperature as a function of
temperature is plotted as following: the first of which has the temperature of 100K~6000K, while the second
has 100 K ~ 1
×
10
6
K.
3
2
2
T
T
R
a
2
RT
b
p
Z
T
+
−
=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
∂
∂
∂
∂
and
4
2
3
T
2
2
T
R
a
6
RT
b
2
p
Z
T
−
=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
∂
∂
∂
∂
then,
at T=T
max
, where
0
T
R
a
2
RT
b
p
Z
T
3
max
2
2
max
T
=
+
−
=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
∂
∂
∂
∂
so,
B
max
T
2
Rb
/
a
2
T
=
=
.
To show it is a maximum, at T = T
max
= 2 T
B
0
a
b
R
8
1
)
Rb
/
a
2
(
R
a
6
)
Rb
/
a
2
(
R
b
2
T
R
a
6
RT
b
2
p
Z
T
3
4
2
4
2
3
4
2
3
T
2
2
<
−
=
−
=
−
=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
∂
∂
∂
∂

This
** preview**
has intentionally

**sections.**

*blurred***to view the full version.**

*Sign up*
Page
2
of
12
The maximum slope is
a
4
b
a
4
b
a
2
b
T
R
a
RT
b
2
2
2
2
max
2
max
=
−
=
−
For methane,
2
6
mol
m
Pa
227
.
0
a
−
•
•
=
and
1
3
5
mol
m
10
28
.
4
b
−
−
•
×
=
So maximum slope is 2.02
×
10
-9
Pa
-1
. This only refers to the initial slope.
This means big gas molecules with weak attractions have large positive maximum slopes while small molecules
with strong attractions have small positive maximum slopes.
Slope vs Temperature (100K~6000K)
2.5
2.0
1.5
1.0
0.5
0.0
-0.5
-1.0
(
∂
Z/
∂
p)
T
x 10
9
/Pa
-1
6000
4000
2000
0
T /K
Figure 1A
T
B
T = 2T
B