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An in-depth analysis of adiabatic exponents in the context of ionization zones. Adiabatic exponents describe how pressure, temperature, and specific volume change during an adiabatic process. In ionization zones, these exponents can differ from the ideal gas values due to the presence of free particles and ionization energy. the methodology for calculating adiabatic exponents, their values for ideal gases and radiation, and their significance in determining stellar pulsations. It also discusses the role of ionization zones in second ionization of He and the instability strip on the Hertzsprung-Russell diagram.
Typology: Exercises
Uploaded on 09/12/2022
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− 1
( )
ρ
= dE + Pd
1
ρ
⎛
⎝
⎜
⎞
⎠
⎟
ρ
2
d ρ
First law of thermodynamics
1
∂ ln ρ
ad
ρ
ad
2
2
ad
ad
ρ
3
− 1
( ) =^
ρ
− 1
( )
3
∂ ln ρ
ad
ρ
ad
ad
∂ ln ρ
ad
∂ ln ρ
ad
3
1
2
2
Ideal Gas:
1
2
3
Radiation:
1
2
3
3
is equivalent to the γ in
the γ - law equation of state
P = ( γ − (^1) ) ρ E
For mixtures of gas and radiation, as well as for cases where
chemical reaction happen, the adiabatic exponents can differ
from each other.
(relevant for dynamical processes like pulsations)
(relevant for determining whether convection takes place)
2
3
1
P = P gas
=
N A
k
μ
ρ T +
1
3
aT
4
E = E gas
=
3
2
N A
k
μ
T + a
T
4
ρ
=
N A
k
μ
T
V ρ
1
3
aT
4
=
3
2
N A
k
μ
T + aT
4
V ρ
Pressure
Specific Internal Energy
V ρ
ρ
T
ρ
ρ
A
3
ρ
4
( ) dV ρ
A
ρ
4
ρ
A
k
μ
ρ
aT
4
E =
3
2
N A
k
μ
T + aT
4
V ρ
A
k
2 μ
3
V ρ
dT +
A
k
μ
ρ
aT
4
dV ρ
A
k
2 μ
ρ
4
ρ
dT +
A
k
μ
ρ
aT
4
dV ρ
gas
rad
dT
gas
rad
( )
dV ρ
ρ
Adiabatic change
dQ = 0
gas
rad
2
2
gas
rad
( )
dT
gas
dV ρ
ρ
gas
rad
dT
gas
rad
( )
dV ρ
ρ
1
st Law of
Thermodynamics
Equation
of State
Ax + By = 0
Cx + Dy = 0
⎫
⎬
⎪
⎭
⎪
→
A
C
=
B
D
gas
rad
2
2
gas
( (^) rad)
gas
rad
gas
gas
rad
P gas
= β P
P rad
= (^) ( 1 − β) P
β P + (^4) ( 1 − β) P −
2
2
β P + (^12) ( 1 − β) P
β P + (^4) ( 1 − β) P
β + (^4) ( 1 − β) −
2
2
β + 12 ( 1 − β)
β + (^4) ( 1 − β)
2
2
2
2
( 4 −^3 β) =^ ( −^ β) 12 −^
2
−
2
2
( 4 −^3 β) =^ −^12 β^ +^
2
2
2
( 4 −^3 β) =^16 −^12 β^ −^
2 →
2
2
2
2
2
2
2
2
2
Thermodynamics of partially ionized gas is different because
n
= yn I
n e
= yn I
n
0
= (^) ( 1 − y ) n I
⎫
⎬
⎪
⎭
⎪
y
2
1 − y
=
1
n I
2 π m e
kT
h
2
⎛
⎝
⎜
⎞
⎠
⎟
3 2
e
− χ kT
n I
= N A
ρ
Assuming a pure H ideal gas
=
A
N A
ρ
T
3 2
e
− χ kT
y
2
1 − y
=
A
N A
V ρ
T
3 2
e
− χ kT
P = n
0
n
n e
E =
3
2
n
0
n
n e
kT
ρ
n
χ
ρ
A
ρ kT →
3
2
N A
kT + yN A
χ
Pressure
Specific Internal Energy
I
yn I
yn I
⎡ ⎣
⎤ ⎦
kT
I
kT
I
3
2
kT
ρ
yn I
ρ
χ
A
k
T
V ρ
V ρ
, y
ρ
T , y
ρ
T , V ρ
= (^) ( 1 + y ) N A
ρ
dT − (^) ( 1 + y ) N A
ρ
2
ρ
A
ρ
Equation
of State
P = (^) ( 1 + y ) N A
k
T
V ρ
E = (^) ( 1 + y )
3
2
N A
kT + yN A
χ
dP = (^) ( 1 + y ) N A
k
ρ
dT
dV ρ
ρ
dy
1 + y
= (^) ( 1 + y ) N A
ρ
− (^) ( 1 + y ) N A
ρ
ρ
ρ
ρ
dP = P
dT
dV ρ
ρ
dy
1 + y
V ρ
ρ
T
ρ
( 1 − y )
2
2 y (^) ( 1 − y ) + y
2
A
ρ
1 2
− χ kT
3 2
− χ kT
2
A
3 2
− χ kT
ρ
Saha Equation
f (^) ( y ) =
y
2
1 − y
=
A
N A
V ρ
T
3 2
e
− χ kT
( 1 − y )
2
2 y (^) ( 1 − y ) + y
2
A
ρ
3 2
− χ kT
2
ρ
ρ
(^1 −^ y )
2
2 y ( 1 − y ) + y
2
2
ρ
ρ
= D (^) ( y )
ρ
ρ
ideal
dT
dV ρ
ρ
kT
dy
1 + y
dP = P
dT
dV ρ
ρ
dy
1 + y
1
st Law of
Thermodynamics
Equation
of State
Saha
ideal
dT
dV ρ
ρ
kT
D ( y )
kT
dT
dV ρ
ρ
1
st Law of
Thermodynamics
ideal
dT
dV ρ
ρ
kT
dy
1 + y
ideal
kT
D ( y )
kT
dT
kT
D ( y )
dV ρ
ρ
ideal
1 + D ( y )
kT
2
⎡
dT
1 + D ( y )
kT
dV ρ
ρ
1
st Law
Equation
of State
2
2
kT
dT
dV ρ
ρ
1 + D ( y )
kT
2
⎡
dT
1 + D ( y )
kT
dV ρ
ρ
2
2
kT
1 + D (^) ( y )
kT
2
D ( y ) − 1
1 + D (^) ( y )
kT
2
2
= 1 + D (^) ( y )
kT
(^ D^ (^ y )^ −^1 ) 1 +^ D^ (^ y )^
kT
2
⎡
1 + D ( y )
kT
2
2
= 1 + D (^) ( y )
kT
(^ D^ (^ y )^ −^1 ) 1 +^ D^ (^ y )^
kT
2
⎡
1 + D (^) ( y )
kT
2
2
kT
kT
2
⎡
1 + D (^) ( y )
kT
D (^) ( y ) =
y (^) ( 1 − y )
(^1 +^ y ) ( 2 −^ y )
…algebra…
3
2 + 2 D (^) ( y )
kT
3 + 2 D (^) ( y )
kT
2
Similarly:
