Isentropic Process

Process where $d S = 0$

Requirements

Closed system: $d m = 0$
Adiabatic: $δ Q = 0$
Reversible: $d S = \frac{δ Q}{T} = 0$

Special Case: Liquids and Solids

Isentropic = Isothermal: $d s = C \frac{d T}{T} \to d T = 0$

Recall:

Ratio of specific heats:
$γ = \frac{C_{P}}{C_{V}}$
$C_{P} = C_{V} + \frac{R}{w}$
$C_{V} = \frac{1}{γ - 1} \frac{R}{W}, C_{P} = \frac{γ}{γ - 1} \frac{R}{W}$

Table of Specific Heats:

Material	Specific Heat Capacity ( $γ$ )
Air	$\approx$ 1.4

Entropy Expressions:

d s = C_{v} \frac{d T}{T} + \frac{R}{W} \frac{d v}{v} = 0 ⟶ T^{\frac{1}{γ - 1}} v = c o n s t

d s = C_{p} \frac{d T}{T} - \frac{R}{W} \frac{d P}{P} = 0 ⟶ T^{\frac{γ}{γ - 1}} P^{- 1} = c o n s t

P v^{γ} = c o n s t