Question

Consider that an ideal gas (n moles) is expanding in a process given by P = f(V), which passes through a point (V₀, P₀). 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 adiabat passing through (P₀, V₀).

Answer 1

 Slope of the graph at `(V_0, P_0) = ((dP)/(dV))_((V_0, P_0)`

We know that for the adiabatic process `PV^γ = K`

⇒ `P = K/V^γ`

⇒ `(dP)/(dV) = K(-γ) V^(-γ-1)`

⇒ `(dP)/(dV) = - γPV^γ V^-γ V^-1`

⇒`(dP)/(dV) = (-γP)/V`

⇒ `((dP)/(dV))_((V_0"," P_0)) = (-γP_0)/V_0`

Now, heat absorbed in the process `P = f(V)`,

∴ `dQ = dU + dW`

⇒ `dQ = nC_VdT + PdV`  ......(i)

We know that `PV = nRT`. Therefore, we get

⇒ `T = (PV)/(nR)`

⇒ `T = V/(nR) f(V)`  .....(Given, `P = f(V)`)

⇒ `(dT)/(dV) = 1/(nR)`  .....`[f(V) + Vf^' (V)]`

Now, continue solving equation (i)

⇒ `(dQ)/(dV) = nCv (dT)/(dV) + P(dV)/(dV)`

⇒ `(dQ)/(dV) = (nCv)/(nR) [f(V) + Vf^' (V)] + P`

⇒ `((dQ)/(dV))_(V = V_0) = (Cv)/R [f(V_0) + V_0f^' (V_0)] + f(V_0)`

⇒ `((dQ)/(dV))_(V = V_0) = f(V_0) [(Cv)/R + 1] + V_0f^' (V_0) (Cv)/R`

We know that `C_P - C_V = R`

⇒ `(C_P)/(C_V) - 1 = R/(C_V)`

⇒ `γ - 1 = R/(C_V)`

⇒ `C_V = R/(γ - 1)`

⇒ `C_V/R = 1/(γ - 1)`

⇒ `((dQ)/(dV))_(V = V_0) = f(V_0) [1/(γ - 1) + 1] + V_0f^' (V_0) 1/(γ - 1)`

⇒ `((dQ)/(dV))_(V = V_0) = f(V_0) [(1 + γ - 1)/(γ - 1)] + (V_0f^' (V_0))/(γ - 1)`

⇒ `((dQ)/(dV))_(V = V_0) = γ/((γ - 1)) f(V_0) + V_0 (f^' (V_0))/(γ - 1)`

⇒ `((dQ)/(dV))_(V = V_0) = 1/((γ - 1)) [γf (V_0) + V_0f^' (V_0)]`

⇒ `((dQ)/(dV))_(V = V_0) = 1/((γ - 1)) [γP_0 + V_0f^' (V_0)]`  .....`(∵ f(V_0) = P_0)`

If γ > 1, so `(1/(γ - 1))` is positive.

Then, heat is absorbed where `((dQ)/(dV))_(V = V_0) > 0` when gas expands.

Hence, `γP_0 + V_0f^' (V_0) > 0`

⇒ `V_0f^' (V_0) > (- γ P_0)`

⇒ `f^' (V_0) > ((- γP_0)/V_0)`