Consider that an idela gas (n mole) is espanding in a process gives by `p=f(v)`, which passes throught a point `(V_(0),p_(0))`. Shown that the gas is absorbing heat at `(p_(0),v_(0))` if the slope of the curve `p=f(v)` is large than the slope of the adabatic passing throught `(p_(0), v_(0))`.

Explanation

According to question , slope of the curve = f(V) where V is volume.
`therefore` `" "` Slpoe of p = f(v) curve at `(V_(0),p_(0))` = f(v)
Slope of addiabtic at `(V_(0),p_(0))` = `k(-gamma )v_(0)^(1-gamma-1) = - gamma p_(0)//v_(0)`
Now heat absorbed in the process `p=f(v)`
`dQ = dU + dW = nC(v) dT + pdV " "......(i)`
`because``" " pV = nRT rArr T= ((1)/(nR))pV `
`rArr" "T= ((1)/(nR)) V f(V)`
`rArr " "dT= ((1)/(nR))[f(v) + Vf(V) ]dV" ".........(ii)`
Now from Eq. (i) `(dQ)/(dV) = nC_(v) (dT)/(dV) + p(dV)/(dV) = nC_(V) (dT)/(dV)+ p`
`= (nC_(V))/(nR) xx [f(V) + Vf'(V) ] +p " "["from Eq. (ii)"]`
`" "=(C_(V))/(R)[f(V)+Vf'(V)]+f(V)" "[becausep=f(V)]`
`rArr" "[(dQ)/(dV)]_(V=V_(0))=(C_(V))/(R)[f(V_(0))+V_(0)f'](V_(0))]+f(V_(0))` ltBrgt `" "=f(V_(0))[(C_(V))/(R)+1]+V_(0)f'(V_(0))(C_(V))/(R)`
`because" "C_(V)=(R)/(gamma-1)rArr(C_(V))/(R)=(1)/(gamma-1)`
`rArr" "[(dQ)/(dv)]_(V=V_(0))=[(1)/(gamma-1)+1]f(V_(0))+(V_(0)f'(V_(0)))/(gamma-1)`
`" "(gamma)/(gamma-1)p_(0)+(V_(0))/(gamma-1)f'(V_(0))`
Heat is absorbed where `(dQ)/(dV)gt0`, when gas expands
Hence, `gamma p_(0)+V_(0)f'(V_(0))gt0orf'(V_(0))gt(-gamma(p_(o))/(V_(o)))`