Consider that an ideal gas (n moles) is expanding in a process given by p= f(V), which passes through a point (P0, V0). Show that the gas is absorbing heat at (P0, V0) if the slope of the curve p= f(V) is larger than the slope of the adiabatic passing through (P0, V0).

Hint: Use the first law of thermodynamics.
According to the question, the slope of the curve = f(V) where V is volume.
 The slope of p=f(V) curve at (P0, V0)=f(V0)
Step 1: Find the slope of the adiabatic curve.
 The slope of adiabatic at (P0, V0)=k(-γ)V0-1-γ=-γ P0/V0
Step 2: Find the heat absorbed in the process.
Now heat absorbed in the process p = f(V),
dQ=dU+dW=nCvdT+pdV                                 ...(i)
As pV=nRTT=1nRpV
T=1nRV×f(V)
dT=1nR[f(V)+V×f'(V)]dV                              ...(ii)
Now from Eq. (i), dQdV=nCvdTdV+pdVdV=nCVdTdV+p
                         =nCvnR×[f(V)+V×f'(V)]+p                          [from Eq. (ii)]
                        =CVR[f(V)+V×f'(V)]+ f(V)                          [p =f(V)]
           dQdVV=V0=CVR[f(V0)+V0×f'(V0)]+f(V0)
                               =f(V0)CVR+1+V0×f'(V0)CVR
                      CV=Rγ-1 CVR=1γ-1
      dQdvv=v0=1γ-1+1f(V0)+V0×f'(V0)γ-1
                          =γγ-1P0+V0γ-1×f'(V0)
Step 3: Find the condition when the heat is absorbed.
Heat is absorbed where dQdV> 0 when the gas expands.
Hence, γP0+V0×f'(V0)>0 or f'(V0)>-γpoVo