Consider that an ideal gas `(n mol es)` is expanding in a process given by `P=f(V)`, which passes through a point `(P_(0), V_(0))`. Show that the gas is absorbing heat at `(P_(0), V_(0))` if the slope of the curve `P=f(V)` is larger than the slope of the adiabat passing through `(p_(0), V_(0))`.

Explanation

The slope of `P= f(V) "curve at" (P_(0), V_(0))`
The slope of an adiabat passing through `(P_(0), V_(0)) = -(gammaP_(0))/(V_(0))`
If `dQ` is the amount of heat absorbed in the process, `P=f(V)`, then
`dQ=dU+dW`
or `dQ=nC_(v)dT+PdV`….(i)
From `PV= nRT, T=((1)/(nR))PV=((1)/(nR))f(V).V [ :. P=f(V)]`
`:. dT((1)/(nR))[f(V)dV+V f' (V)dV]`....(ii)
Using (ii) in (i), we get
`(dQ)/(dV)=(nC_(v))/(dV)[((1)/(nR)){f(V)dV+Vf'(V)dV}]+(PdV)/(dV)`
`=(C_(v))/R[f(V)+Vf'(V)]+P`
At `(P_(0), V_(0)), (dQ)/(dV)= 1/(gamma-1)[f(V_(0))+V_(0)f'(V_(0))]+P_(0) (As (R)/(C_(v))=gamma-1)`
As `f(V_(0))=P_(0)`, therefore, at `(P_(0), V_(0))`
`(dQ)/(dV)=(V_(0)f'(V_(0)))/(gamma-1)+(1/(gamma-1)+1)P_(0)=(V_(0)f'(V_(0)))/(gamma-1)+(gammaP_(0))/(gamma-1)`
`=(1/(gamma-1))[V_(0)f'(V_(0))+gammaP_(0)]`
The gas will absorb heat at `(P_(0), V_(0))`, when at `(P_(0), V_(0))`
`(dQ)/(dV) gt 0 i.e., 1/(gamma-1)[V_(0)f'(V_(0))+gammaP_(0)] gt 0`
or `V_(0)f'(V_(0))+gammaP_(0) gt 0 or f'(V_(0)) gt (-gammaP_(0))/(V_(0))`
which is the required result.