THERMAL CYU 33

Transition from Endothermic to Exothermic

The process turns from endothermic ($dQ > 0$) to exothermic ($dQ < 0$) when $dQ = 0$. This corresponds to the point where the process curve is tangent to an adiabat.

Using the first law of thermodynamics: $dQ = dU + dW$. For a monoatomic gas:

$$dQ = \frac{3}{2}(pdV + Vdp) + pdV = \frac{5}{2}pdV + \frac{3}{2}Vdp$$

Setting $dQ = 0$:

$$\frac{5}{2}pdV + \frac{3}{2}Vdp = 0 \implies \frac{dp}{dV} = -\frac{5p}{3V}$$

The process $A \to B$ is a straight line. Let the equation of the line be $p = -\alpha V + p_{int}$. From the diagram, the x-intercept is $V_0$. The equation can be written as:

$$p = C (V_0 – V)$$

The slope of this line is constant: $\frac{dp}{dV} = -C$.

At the transition point, equate the slopes:

$$-C = -\frac{5p}{3V} \implies C = \frac{5}{3V} [C(V_0 – V)]$$ $$1 = \frac{5(V_0 – V)}{3V}$$ $$3V = 5V_0 – 5V$$ $$8V = 5V_0$$ $$V = \frac{5}{8}V_0$$