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 $T_2/T_1 =(P_2/P_1 )^{(γ-1)/γ}$


valentin68

Leave a Reply