In astrophysics, what is referred to as "entropy" is actually the adiabatic constant derived as follows.
Using the first law of thermodynamics for a quasi-static, infinitesimal process for a hydrostatic system
d
Q
=
d
U
−
d
W
.
{\displaystyle dQ=dU-dW.}
For an ideal gas in this special case, the internal energy, U, is a function of only the temperature T; therefore the partial derivative of heat capacity with respect to T is identically the same as the full derivative, yielding through some manipulation
d
Q
=
C
v
d
T
+
P
d
V
.
{\displaystyle dQ=C_{\text{v}}dT+P\,dV.}
Further manipulation using the differential version of the ideal gas law, the previous equation, and assuming constant pressure, one finds
d
Q
=
C
p
d
T
−
V
d
P
.
{\displaystyle dQ=C_{\text{p}}dT-V\,dP.}
For an adiabatic process
d
Q
=
0
{\displaystyle dQ=0\,}
and recalling
γ
=
C
p
/
C
v
{\displaystyle \gamma ={C_{\text{p}}}/{C_{\text{v}}}\,}
, one finds
One can solve this simple differential equation to find
P
V
γ
=
constant
=
K
{\displaystyle PV^{\gamma }={\text{constant}}=K}
This equation is known as an expression for the adiabatic constant, K, also called the adiabat. From the ideal gas equation one also knows
P
=
ρ
k
B
T
μ
m
H
,
{\displaystyle P={\frac {\rho k_{\text{B}}T}{\mu m_{\text{H}}}},}
where
k
B
{\displaystyle k_{\text{B}}}
is the Boltzmann constant.
Substituting this into the above equation along with
V
=
[
g
]
/
ρ
{\displaystyle V=[\mathrm {g} ]/\rho \,}
and
γ
=
5
/
3
{\displaystyle \gamma =5/3\,}
for an ideal monatomic gas one finds
K
=
k
B
T
(
ρ
/
μ
m
H
)
2
/
3
,
{\displaystyle K={\frac {k_{\text{B}}T}{(\rho /\mu m_{\text{H}})^{2/3}}},}
where
μ
{\displaystyle \mu \,}
is the mean molecular weight of the gas or plasma; and
m
H
{\displaystyle m_{\text{H}}}
is the mass of the hydrogen atom, which is extremely close to the mass of the proton,
m
p
{\displaystyle m_{p}}
, the quantity more often used in astrophysical theory of galaxy clusters.
This is what astrophysicists refer to as "entropy" and has units of [keV⋅cm2]. This quantity relates to the thermodynamic entropy as
Δ
S
=
3
/
2
ln
K
.
{\displaystyle \Delta S=3/2\ln K.}