# Isentropic Equation

Deduce the equation of the Isentropic (Adiabatic) transformation starting from the first principle. Assume an ideal gas.

From first principle one has

$d Q=d U+d W$   (Clausius sign convention:work done by the system on exterior is $>0$)

$d Q=N c_v d T+P d V=0$ (since the process is isentropic)

The general equation of gases is

$P V=N R T$ which differentiated is

$P d V+V dP=N R dT$ or $d T=(P dV+V dP)/NR$

Back into first principle:

$0=N c_v*(P dV+V dP)/NR+P dV=P dV(c_v/R+1)+c_v/R* V dP$  or equivalent

$0=(c_v+R)P dV+c_v V dP$

Rearranging we get

$(c_v+R)*P dV=-c_v*V dP$ or $(c_v+R)d V/V=-c_v*d P/P$

But we know that by definition

$γ=c_p/c_v$    and $c_p-c_v=R$
So that

$c_p*dV/V=-c_v*dP/P$     or $γ*dV/V=-dP/P$
which integrated gives $γ*ln⁡(V_1/V_2 )=-ln⁡(P_1/P_2 )$

and finally

$P_2/P_1 =(V_1/V_2 )^γ or P_1*V_1^γ=P_2*V_2^γ$   (1)

If we get back to the general equation of gases above we can write
$(P_1*V_1 )*V_1^{γ-1}=(P_2*V_2 )*V_2^{γ-1}$     (2)
or equivalent $T_1*V_1^{γ-1}=T_2*V_2^{γ-1}$

Equations (1) and (2) can be rewritten as

$(P_2/P_1 )^{1/γ}=V_1/V_2$     and $(T_1/T_2 )^{1/(γ-1)}=V_2/V_1$
which gives

$(P_2/P_1 )^{1/γ}=(T_2/T_1 )^{1/(γ-1)}$

or equivalent $T_2/T_1 =(P_2/P_1 )^{(γ-1)/γ}$