All Exams Test series for 1 year @ ₹349 only
Question

Critical value of the Gibbs energy of nucleation at equilibrium temperature is

The correct answer is
infinite

Understanding Nucleation Gibbs Energy

Nucleation is the initial process where a new thermodynamic phase forms. This requires overcoming an energy barrier, known as the Gibbs energy of nucleation.

The change in Gibbs energy ($\Delta G$) for forming a spherical nucleus involves two main contributions:

  • A negative contribution from the bulk free energy change ($\Delta G_v$), which favors the new phase formation.
  • A positive contribution from the surface energy ($\gamma$), which increases with the surface area of the nucleus.

Critical Gibbs Energy Barrier

The nucleation process involves reaching a critical size where the nucleus becomes stable. The energy required to reach this point is the critical Gibbs energy barrier, denoted as $\Delta G^*$.

The formula for the critical Gibbs energy barrier is:

$ \Delta G^* = \frac{16 \pi \gamma^3}{3 (\Delta G_v)^2} $

where:

  • $ \gamma $ is the surface tension.
  • $ \Delta G_v $ is the difference in Gibbs free energy per unit volume between the parent and new phase.

Gibbs Energy at Equilibrium Temperature

The equilibrium temperature ($T_{eq}$) is the specific temperature where the parent phase and the new phase are in equilibrium. At this temperature, there is no net driving force for the phase transformation from the bulk perspective.

Mathematically, this means the bulk free energy change per unit volume is zero:

$ \Delta G_v = 0 $

Substituting this condition into the formula for the critical Gibbs energy barrier:

$ \Delta G^* = \frac{16 \pi \gamma^3}{3 (0)^2} $

As the denominator approaches zero, the value of $\Delta G^*$ tends towards infinity.

Therefore, the critical value of the Gibbs energy of nucleation at the equilibrium temperature is effectively infinite, representing an infinitely large barrier to initiate nucleation under these precise equilibrium conditions.

Was this answer helpful?

Important Questions from Solidification Cooling Curve Analysis

  1. During the solidification of a pure metal, it was found that dendrites are formed. Assuming that the liquid-solid interface is at the melting temperature, the temperature from the interface into the liquid
  2. Which one of the following schematics represents the variation of the rate of nucleation of solid from a pure liquid metal as a function of undercooling ($\Delta T = T_m - T$, where $T_m$ and $T$ are the freezing temperature and the liquid temperature, respectively)?
  3. A given volume of liquid is undercooled just below the melting temperature to form a spherical solid nucleus (consider homogeneous nucleation). The Gibbs free energy of solidification ($\Delta G_v$) is ($- 0.5 \times 10^8$) J/m$^3$. The solid-liquid interfacial energy ($\gamma$) is isotropic and its value is 0.1 J/m$^2$. 

    The critical nucleus size for a stable nucleus is __________ nm (answer in integer).

  4. During solidification of a pure metal, the radius of critical nucleus at an undercooling of 10 K is ________ $\times 10^{-9} \text{ m}$ (answer rounded off to 1 decimal place).
    Given: solid/liquid interface energy = $0.177 \text{ J} \cdot \text{m}^{-2}$,
    melting point of the metal = 1356 K and
    latent heat of fusion = $1.88 \times 10^9 \text{ J} \cdot \text{m}^{-3}$

  5. Consider homogeneous nucleation of a spherical solid in liquid. For a given undercooling, if surface energy of a nucleus increases by $20\%$, the corresponding increase (in percent) in the critical radius of the nucleus is: ___________(round off to nearest integer).

Need Expert Advice?

Start Your Preparation with Prepp Mobile App

Download the app from Google Play & App Store
Download the app from Google Play & App Store
Prepp Mobile App