Homework 2: Due at the beginning of class 09/26/2011
Problem 1.
In class, we learn about the reversible isothermal transformation of ideal gas.
There is another important path of the volume change where the system is
thermally
isolated
from the surrounding. i.e.
dq
= 0. This is called the “
adiabatic
” path. Let’s
derive the expression of a reversible adiabatic expansion of ideal gas.
(a)
Begin with
dw
dq
dU
+
=
to get the relation between
dT
and
dV
. (
Hint
) Replace
dU
with
C
V
dT
where
C
V
is
the
heat
capacity
at
constant
volume
(remember
dT
C
dU
V
=
for ideal gas in general conditions). We only consider PV
work here, thus
PdV
dw
−
=
. Replace
P
with
V
nRT
.
(b)
Integrate both sides from (T
i
,V
i
) to (T
f
, V
f
) to show that
f
i
c
i
f
V
V
T
T
=
Ο
Ο
Π
Ξ
Μ
Μ
Ν
Λ
where
c
is a constant to make the expression simpler. Express
c
with the constants
described above
(c) Both at the initial (i) and final (f) states above, you can apply ideal gas equations
to change the above result to the relation between (P
i
,V
i
) to (P
f
, V
f
). Show that
P
f
P
i
=
V
i
V
f
!
"
#
#
$
%
&
&
γ
What is
γ
? Express it with
c
.
(d) With the above results, obtain the following relation.
γ
γ
Ο
Ο
Π
Ξ
Μ
Μ
Ν
Λ
=
Ο
Ο
Π
Ξ
Μ
Μ
Ν
Λ
−
f
i
i
f
T
T
P
P
1
Problem 2.
The results in Prob.1 imply significant differences between the reversible
adiabatic process and the reversible isothermal one.
Assume all the processes are done
reversibly.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview.
Sign up
to
access the rest of the document.
 Fall '08
 LIM
 Physical chemistry, Thermodynamics, pH, Adiabatic process

Click to edit the document details