Across a first order phase transition, the free energy is
Options:
(A) proportional to the temperature
(B) a discontinuous function of the temperature
(C) a continuous function of the temperature but its first derivative is discontinuous
(D) such that the first derivative with respect to temperature is continuous
Official Answer Key:
B
Step-by-Step Solution & Explanation
### **Official Solution & Step-by-Step Derivation (GATE PH 2013 Q.5)**
**1. Thermodynamics of Phase Transitions (Ehrenfest Classification)**:
- According to Ehrenfest's classification, an $n$-th order phase transition is characterized by:
- Continuity of the Gibbs free energy $G(T, P)$ and all derivatives up to order $(n-1)$.
- A discontinuity in the $n$-th order partial derivatives of $G(T, P)$.
**2. First-Order Phase Transition Behavior**:
- The Gibbs free energy $G(T, P)$ itself is strictly **continuous** across the phase boundary ($G_1 = G_2$).
- The first derivatives of $G$ represent entropy $S$ and volume $V$:
$S = -\left( \frac{\partial G}{\partial T} \right)_P, \quad V = \left( \frac{\partial G}{\partial P} \right)_T$
- In a first-order phase transition (e.g. solid-liquid melting, liquid-gas vaporization), latent heat $L = T \Delta S \neq 0$ is absorbed or released, causing a finite discontinuity in entropy:
$\Delta S = S_2 - S_1 = \frac{L}{T} \neq 0$
- Similarly, a finite density change produces a discontinuity in volume $\Delta V = V_2 - V_1 \neq 0$.
- Hence, the free energy $G$ is **continuous**, while its first derivatives are **discontinuous**.
**3. Official Verification**:
The verified official IIT Bombay Master Answer is **Option (B)**.
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