Proof of the formula for a reversible adiabatic change

We will assume that we are dealing with an ideal gas and therefore the first law of thermodynamics becomes:

dQ = dE + dW

where dQ is the heat energy input, dE is the increase in internal energy of the gas and dW is the work done by the gas.

But dE = C_VdT where C_V is the molar specific heat capacity of the gas at constant volume.

This gives:

dQ = C_VdT + P dV

For an adiabatic change dQ = 0 and so C_VdT + P dV = 0.

Let the gas expand from V to V + dV at constant pressure and let the temperature fall from T to T- dT.

Then: C_VdT + PdV = 0

But PV = nRT. Therefore for one mole: PV = RT and so C_VdT/T + RdV/V = 0

But since C_P - C_V = R for an ideal gas: C_VdT/T + (C_P - C_V) dV/V =0

Writing C_P/C_V = γ we have: dT/T + (γ- 1)dV/V = 0

which when integrated gives:TV^(γ-1) = constant

From this the other versions of the adiabatic equation may be obtained using PV = RT for one mole of the gas.