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Lecture 11
1
As we have already noted for any ideal gas U=U(T)
Å
depends on
temperature only.
So, for an isothermal process:
Δ
U=0 for an ideal gas.
This implies
q+w=0, or q=w, or  q= w
δ
rev
Furthermore, if the process is reversible:
 q
ln
rev
w
PdV
nRTd
V
δδ
==
−
=
−
2
rev
1
and:
q
ln
V
nRT
V
=
w
rev
<0 i.e. work done on surroundings
For V
2
>V
1
q
rev
>0 heat entered system to keep gas at constant temperature.
Now, consider an adiabatic
process to the same final volume.
We define adiabatic to be
δ
q=0.
If there is no heat flow and we go to the same final volume:
T
2
<T
1
(gas must cool!).
But by how much?
More Thermodynamics
(19.419.9)
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2
Fig 19.5
Isobaric:
Δ
P=0,
path D
Isothermal:
Δ
T=0, path A
Adiabatic:
δ
q=0,
path B
Note for adiabatic path
δ
q=0 which means dU=
δ
w.
Adiabatic:
dU=
δ
w
V
ext
nRT
CdT
PdV
dV
V
=−
Ideal gas
reversible
Ideal gas
22
11
ln
V
TV
V
dT
dV
CR
C
dT
R
∫∫
Lecture 11
3
Remember for anything other than a monatomic ideal gas, the heat
capacity has a temperature dependence.
()
2
1
21
12
ln
T
V
T
CT
VV
dT
R
=−
=
∫
V
3
for a monatomic gas, C
2
R
=
3/2
2
1
1
2
3
So:
ln
ln
or
2
TV
T
V
T
V
⎛⎞
==
⎜⎟
⎝⎠
For an adiabatic expansion
V
2
>V
1
so T
2
<T
1
; the gas cools!
5/2
2
2
1
1
Since
,
TT
P
P
T
nRT
V
PT
T
P
P
T
→
=
Question:
If argon expands from a pressure of 10
5
Pa (1 bar) to 50 mbar
and had an initial temperature of 298K, what is the final temperature?
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This note was uploaded on 04/01/2012 for the course CHEM 444 taught by Professor Gruebele,m during the Spring '08 term at University of Illinois, Urbana Champaign.
 Spring '08
 Gruebele,M
 Physical chemistry, pH

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