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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 = CVdT where CV is the molar specific heat capacity of the gas at constant volume.

This gives:
dQ = CVdT + P dV

For an adiabatic change dQ = 0 and so CVdT + 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: CVdT + PdV = 0

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

But since CP - CV = R for an ideal gas: CVdT/T + (CP - CV) dV/V =0

Writing CP/CV = g we have: dT/T + (g- 1)dV/V = 0

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

From this the other versions of the adiabatic equation may be obtained using PV = RT for one mole of the gas.
© Keith Gibbs 2010