Since I can't resist:
from the first law dU=dw + dq...since the sum of heat and work only depends on the final and initial state although the type of work and the amount of heat used in the process are dependent upon the path taken to the final and initial state the total energy change remains constant..
so,
I can assume all the change is entirely adiabatic so the 'q' term disappears for an adiabatic change since U is a state function therefore I can use the equation for isothermal work since if 'dq'= 0 then work done adiabatically =work done isothermally wad=wIsT and since w= -pdV and adiabatic work = CvdT [or the product heat capacity at constant volume and the infinitesimal temperature change].
Equating the two terms wad =wIsT we have,
CvdT =-Pdv
From the ideal gas law p=nRT/V and dividing both sides by 'T'
we have CvdT/T =-nRdV/V
integrating both sides between final and initial state of 'T and V'
we have Cv
∫ dT/T =-nR
∫dV/V
therefore,
Cv*ln[Tf/Ti ]=-nR* ln[Vi/Vf]
since the ratio 'gamma' is given by Cp/Cv and Cp-Cv =nR; it follows that gamma = Cv/nR which is thus obtained by dividing both sides by -nR
finally,
we have gamma*ln [Tf/Ti ]= ln[Vi/Vf] raising both sides to 'e' which subsequently cancels; one obtains the adiabatic gas law
[Tf/Ti]^gamma =[Vi/Vf]
this equation is used to describe the relationship between pressure [since it can be related to T and V], volume and temperature during adiabatic processes...and so the mysterious 'gamma' is finally revealed...
Here's a good thermodynamics lecture I found
Happy New Year PPRuNe....
