Derivation of equations for adiabatic processes in ideal gas from purely thermodynamic reasoning:
1.
Adiabatic free expansion: Allow a gas to expand freely in an adiabatic container. No heat enters; no work is done. Hence the internal energy U must be constant.
2.
In general, U
=
( , ) . Measure temperature change in afe—if negligible, supports suggestion that U
=
( ) for ideal gas.
3.
Experiment:
•
For monatomic gases (He, Ar, Ne, Hg, …) one finds that C
V
=
3
2
R
•
For many diatomic gases (N
2
, O
2
, NO, CO, H
2
(approximately) C
V
=
5
2
R
There are other possibilities, of course. We might expect that for polyatomic gases,
C
V
= 3 R
, or even higher if vibrational degrees of freedom are present. All we really need, as we will see in the following derivation, is that
C
V
= constant.
4.
First Law: dU
= dQ
+ dW
=
C dT
V
− pdV
Note, of course, that C
V
=
⎛
⎝ dQ dT
⎞
⎠
V
= dU dT
. for a gas.
Now, use the ideal gas law:
PV
= nRT
PdV
+
VdP
= nRdT
PdV
= nRdT
−
VdP whence the first law becomes dQ
=
C dT
V
+ pdV
=
C dT
V
V
+
+ nRdT
−
VdP
) −
VdP whence
C
P
=
⎛ dQ
⎞
⎝ dT
⎠
P
=
C
V
+ nR for any ideal gas.
1

5.
Now, consider quasistatic adiabatic processes: dQ
=
C dT
V
+ pdV dQ
=
C dT
P
−
VdP
But dQ = 0 for an adiabat; hence, letting
γ ≡
C
P
C
V
C
P
C
V
γ dV
V
= − dP
P
V dP
P dV from which it follows at once that
PV
γ = constant
Note again that this equation applies to quasistatic adiabatic processes in ideal gases.
It does not apply to an adiabatic free expansion—be sure that you understand why!
Exercises:
•
Note that Schroeder, page 26, defines
γ
as ( f +2)/ f where f is (loosely) the number of degrees of freedom, and more precisely, the number of quadratic terms in the energy. Show that these two definitions are equivalent.
•
Use our result, PV
γ = constant , along with the ideal gas law, to show how P and T , and V and T , are related in adiabatic processes. See G&T, Eqs. 2.44,
2.46, and 2.47.
2