Consider that an ideal gas (n moles) is expanding in a process given by p = f (V), which passes through a point `(V_0, p_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 adiabatic passing through `(p_0,V_0)`.

Explanation

Slope of graph P=f(V) at point `(V_0, p_0)=f(V_0)`
For adiabatic process `PV^(gamma)`=K (Constant)
`therefore` At point `(V_0,P_0)P_0 V_0^gamma=k`
`therefore P_0=k/V_0^gamma`
`therefore dP_0=k(-gamma)V_0^(-gamma-1). dV_0`
`therefore (dP_0)/(dV_0)=-gamma k/V_0^gamma xx 1/V_0`
`therefore` Slope =`-gamma P_0/V_0`....(1) `[because k/V_0^gamma =P_0]`
Now dQ=dU+dW
`=nC_V dT + PdV`...(2)
and PV =nRT
`therefore T=(PV)/(nR)=(f(V)V)/(nR)`
`therefore dT=1/(nR)[f(V)dV+Vf.(V)dV]`
`=1/(nR) [f(V)+Vf.(V)]dV`...(3)
`rArr` Equ. (2) `dQ=nC_V xx1/(nR)[f(V)+Vf.(V)]dV`
`therefore ["dQ"/"dV"]_(V=V_0) =C_V/R[f(V_0)+V_0f.(V_0)]+f(V_0)` ...(4)
Now, `C_P-C_V=R`
`therefore C_P/C_V-1=R/C_V`
`gamma-1=R/C_V`
`therefore C_V/R=1/(gamma-1)`...(5)
Putting value of (5) in (4),
`((dQ)/(dV))_(V=V_0)=(C_V/R+1)f(V_0)+V_0f.(V_0)`
`=[1/(gamma-1)+1]f(V_0) +(V_0f.(V_0))/(gamma-1)`
`=gamma/(gamma-1)P_0+V_0/(gamma-1)f.(V_0) [ because f(V_0)=P_0]`
When gas expended then `(dQ)/(dV) gt 0` and heat is absorbed.
Now `(dQ)/(dV) gt 0` hence , `(gammaP_0)/(gamma-1)+V_0/(gamma-1) f.(V_0) gt0`
`therefore gammaP_0+V_0f.(V_0) gt 0`
`therefore f.(V_0)=-(gammaP_0)/V_0`