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
δ
rev
Furthermore, if the process is reversible:
- q
ln
rev
w
PdV
nRTd
V
δ
δ
=
= −
= −
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
(19.4-19.9)

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Lecture 11
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
C dT
P dV
PdV
dV
V
= −
= −
= −
Ideal gas
reversible
Ideal gas
2
2
1
1
ln
ln
V
T
V
V
T
V
dT
dV
C
R
T
V
C
d
T
d
V
R
= −
= −
∫
∫