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).
Calculating Critical Nucleus Size in Solidification
This problem involves calculating the critical radius ($r^*$) for the homogeneous nucleation of a spherical solid nucleus in an undercooled liquid. The calculation relies on the balance between the volume free energy change ($\Delta G_v$) and the solid-liquid interfacial energy ($\gamma$).
Homogeneous Nucleation Theory
For homogeneous nucleation, the total Gibbs free energy change ($\Delta G$) associated with forming a spherical nucleus of radius $r$ is:
$ \Delta G = \frac{4}{3}\pi r^3 \Delta G_v + 4\pi r^2 \gamma $
The critical radius ($r^*$) is the radius at which $\Delta G$ reaches its maximum value. This occurs when the derivative of $\Delta G$ with respect to $r$ is zero ($\frac{d(\Delta G)}{dr} = 0$).
Deriving the Critical Radius Formula
Differentiating $\Delta G$ with respect to $r$ and setting it to zero yields the formula for the critical radius:
$ \frac{d(\Delta G)}{dr} = 4\pi r^2 \Delta G_v + 8\pi r \gamma = 0 $
Solving for $r$ gives the critical radius, $r^*$:
$ r^* = -\frac{2 \gamma}{\Delta G_v} $
Step-by-Step Calculation
- Identify Given Values:
- Gibbs free energy of solidification, $\Delta G_v = -0.5 \times 10^8$ J/m$^3$.
- Solid-liquid interfacial energy, $\gamma = 0.1$ J/m$^2$.
- Substitute Values into the Formula:
$ r^* = -\frac{2 \times (0.1 \text{ J/m}^2)}{(-0.5 \times 10^8 \text{ J/m}^3)} $
- Calculate the Radius in Meters:
$ r^* = \frac{0.2}{0.5 \times 10^8} \text{ m} $
$ r^* = \frac{2}{5 \times 10^8} \text{ m} $
$ r^* = 0.4 \times 10^{-8} \text{ m} $
$ r^* = 4 \times 10^{-9} \text{ m} $
- Convert to Nanometers (nm):
Since $1 \text{ nm} = 10^{-9} \text{ m}$, the critical radius is:
$ r^* = 4 \text{ nm} $
- State the Final Answer:
The critical nucleus size is required as an integer. The calculated value is 4 nm.
Final Answer: 4
Important Questions from Solidification Cooling Curve Analysis
- Critical value of the Gibbs energy of nucleation at equilibrium temperature is
- 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
- 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)?
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 = $1.88 \times 10^9 \text{ J} \cdot \text{m}^{-3}$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).
