Entropy is measure of randomness in a fluid state. During compression the fluids tend to decrease in volume and reduce temperature of a fluid while solids increase its volume and vice versa.
The reduction of temperature leads to lower excitation energy of molecules for varying control volume which become more stable and reduces entropy. The opposite phenomenon happens during expansion.
When considering an isentropic compression of an ideal gas with constant specific heat ratio gamma, from a pressure Po=120 kPa to a pressure P=1000 kPa one would have an increase in temperature from To=(29+273.16)K =302.16K to a temperature T=To(P/Po)^(1-(1/gamma)). In case of air for the given condition gamma=1.4 is a good approximation hence we would have T=1.833 To=553.8 K=(280.16+273.16)K=280.6 Celcius. As you observe a final temperature of 66 Celcius, there is a significant cooling of the gas during this compression. The process is actually quite close to an isothermal compression. The cooling explains the reduction of the entropy.
The reduction of volume due to compression reduces the entropy of the as if the temperature does not increase. Regards, Mico
From the second law of thermodynamics in an reversible process one has:
TdS=dU+p dV
where S is the entropy, T is the absolute temperature (in K), p the pressure and V the volume. For an ideal gas p=R T/V where R is the ideal gas constant. For an ideal gas the internal energy is only a function of temperature hence dU =Cv dT, where Cv is the specific heat at constant temperature. Also for an ideal gas R=Cp-Cv where Cp is the specific heat at constant pressure. Hence the second law of thermodynamics combined with the first law becomes:
dS/Cv=(dT/T)+ (R/Cv)(dV/V)
For an isothermal process dT=0 so that a compression dV/V