C.V. Air. Assume a reversible, adiabatic process.

Energy Eq u_{2} - u_{1} = 0 − _{1}w_{2} ;

Entropy Eq s_{2} - s_{1} = ∫ dq/T + 1s_{2} gen = 0

Process: Adiabatic _{1}q_{2} = 0 Reversible _{1}s_{2} 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.

P_{2} = P_{1}( T_{2} / T_{1}) ^{k/k-1 }= 2015 kPa

Using the ideal gas law to eliminate P from this equation leads