Calculate the entropy change when one mole of hydrogen is mixed with two moles of oxygen at room temperature assuming that the gas gas behaves ideally
Answers
Explanation:
About
28.8
35
.
J/K
−−−−−−−−−− , assuming
C
P
stays constant throughout the rather large temperature range. This is usually a good approximation for a gas in general, particularly at high temperatures.
If you really want to be exact, look up the Shomate equation for argon, and integrate that instead:
Δ
S
(
const. P
)
≡
∫
T
2
T
1
C
P
(
T
)
T
d
T
=
∫
T
2
T
1
[
20.78600
T
+
2.825911
×
10
−
10
−
1.464191
×
10
−
13
T
+
1.092131
×
10
−
17
T
2
−
3.661371
×
10
−
2
T
3
]
d
T
≈
∫
T
2
T
1
20.78600
T
d
T
when ignoring the higher-order terms. The most exact answer is
28.8
155
.
J/K
, and the approximated answer is
28.8
35
J/K
. Pretty good!
DERIVING THE RELATIONSHIP BETWEEN ENTROPY AND HEAT CAPACITY
Well, the influence on the entropy due to a change in temperature at constant pressure is given by:
(
∂
S
∂
T
)
P
,
the partial derivative of
S
(entropy) with respect to
T
(temperature) at constant pressure (
P
).
Recall that the differential entropy is given by:
d
S
=
δ
q
r
e
v
T
(
1
)
where
rev
indicates the process is reversible, i.e. that is done infinitesimally slowly.
At constant pressure, the differential heat flow
δ
q
is equal to differential change in enthalpy,
d
H
, because from the first law of thermodynamics...
d
H
=
Internal Energy
d
E
+
d
(
P
V
)
=
First Law of Thermodynamics
δ
q
r
e
v
+
δ
w
r
e
v
+
P
d
V
+
V
d
P
=
δ
q
r
e
v
−
P
d
V
+
P
d
V
+
V
d
P
=
δ
q
r
e
v
+
V
d
P
0
at const. pressure,
where
w
r
e
v
is the reversible work done from the perspective of the system.
Furthermore, by definition, for a reversible process, the constant-pressure heat capacity
C
P
is given by:
C
P
=
(
∂
H
∂
T
)
P
=
(
∂
q
r
e
v
∂
T
)
P
(
2
)
Therefore, we use
(
2
)
to modify
(
1
)
to get:
T
(
∂
S
∂
T
)
P
=
(
∂
q
r
e
v
∂
T
)
P
=
C
P
(
3
)
As a result, we have that:
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
∣
∣
∣
∣
(
∂
S
∂
T
)
P
=
C
P
T
∣
∣
∣
∣
−−−−−−−−−−−−−−−−−−−−
at constant pressure
−−−−−−−−−−−−−−−−− .
(If you have done the derivation before, you can feel free to just memorize this result.)
FINDING THE CHANGE IN ENTROPY
Keeping that in mind, separate the variables and integrate as follows:
∫
(
2
)
(
1
)
d
S
=
∫
T
2
T
1
C
P
T
d
T
Assuming that the constant-pressure heat capacity
C
P
(
T
)
(normally a function of temperature) stays constant in the temperature range, we can pull it out of the integral to get:
Δ
S
≈
C
P
ln
(
T
2
T
1
)
≈
n
¯¯¯
C
P
ln
(
T
2
T
1
)
(where
¯¯¯
C
P
=
C
P
n
)
≈
1 mol
⋅
20.8 J/
mol
⋅
K
⋅
ln
(
1200 K
300 K
)
≈
28.8 J/K
−−−−−−−−