Question

Prove that `"PV"^gamma="constant"`. ?

Answer 1

We know from the 1st law of thermodynamics `dQ = dU+dW`

For an adiabatic process, we can say `dQ =0`,

so, `dU+dW =0=dU+PdV`......1

We know, `dU = nC_v dT`.....2

For an ideal gas, `PV= nRT`

Or, `T =(PV)/(nR)`

or, `dT = (PdV + VdP)/(nR)`

Putting in 2,we get,

`dU = C_v/R (PdV +VdP)`

Putting this value of `dU` in 1 we get,

`(dP)/P = -(dV)/V (C_v +R)/C_v`

or, `(dP)/P = -(dV)/V *gamma` (as `gamma = (C_p)/(C_v)=(C_v +R)/C_v`)

or, `int (dP)/P = -gamma int (dV)/V`

or, `ln P = -gamma ln V +c`

or, `ln P + gamma ln V = k`

or, `ln P + ln V ^gamma =k`

or, `ln(PV^gamma) = k`

so, `PV^gamma` = constant