C.V. Air. Assume a reversible, adiabatic process.
Energy Eq u2 - u1 = 0 − 1w2 ;
Entropy Eq s2 - s1 = ∫ dq/T + 1s2 gen = 0
Process: Adiabatic 1q2 = 0 Reversible 1s2 gen = 0
Properties: k = 1.393
With these two terms zero we have a zero for the entropy change. So this is a constant s (isentropic) expansion process.
P2 = P1( T2 / T1) k/k-1 = 2015 kPa
Using the ideal gas law to eliminate P from this equation leads

