Melting point of a metal is $1356 \ K$. When the liquid metal is undercooled to $1256 \ K$, the free energy change for solidification, $\Delta G^{L \to S} = -1000 \ J \ mol^{-1}$. On the other hand, if the liquid metal is undercooled to $1200 \ K$, the free energy change (in $J \ mol^{-1}$) for solidification is ________.
Understanding Free Energy Change During Solidification
The question relates the free energy change ($\Delta G$) for solidification to temperature and melting point ($T_m$). We are given:
- Melting Point, $T_m = 1356 \ K$
- Temperature 1, $T_1 = 1256 \ K$
- Free Energy Change at $T_1$, $\Delta G_1 = -1000 \ J \ mol^{-1}$
- Temperature 2, $T_2 = 1200 \ K$
- We need to find the Free Energy Change at $T_2$, $\Delta G_2$.
The change in free energy upon solidification at a temperature $T$ below the melting point $T_m$ can be approximated using the latent heat of fusion ($\Delta H_f$):
$ \Delta G \approx \frac{\Delta H_f (T_m - T)}{T_m} $
This formula assumes $\Delta H_f$ is constant and $\Delta G = 0$ at $T_m$. At temperatures close to $T_m$, $\Delta G$ is approximately proportional to the undercooling ($T_m - T$).
Calculating Latent Heat of Fusion
First, we use the data at $T_1$ to estimate the latent heat of fusion, $\Delta H_f$. The undercooling at $T_1$ is:
$ \Delta T_1 = T_m - T_1 = 1356 \ K - 1256 \ K = 100 \ K $
Now, substitute the values into the approximation formula:
$ \Delta G_1 \approx \frac{\Delta H_f \Delta T_1}{T_m} $
$ -1000 \ J \ mol^{-1} \approx \frac{\Delta H_f \times 100 \ K}{1356 \ K} $
Solving for $\Delta H_f$:
$ \Delta H_f \approx \frac{-1000 \ J \ mol^{-1} \times 1356 \ K}{100 \ K} $
$ \Delta H_f \approx -13560 \ J \ mol^{-1} $
Calculating Free Energy Change at $T_2$
Next, we calculate the undercooling at $T_2$:
$ \Delta T_2 = T_m - T_2 = 1356 \ K - 1200 \ K = 156 \ K $
Using the estimated $\Delta H_f$ and the undercooling $\Delta T_2$, we can find $\Delta G_2$:
$ \Delta G_2 \approx \frac{\Delta H_f \Delta T_2}{T_m} $
$ \Delta G_2 \approx \frac{-13560 \ J \ mol^{-1} \times 156 \ K}{1356 \ K} $
$ \Delta G_2 \approx -10 \ J \ mol^{-1} \times 156 $
$ \Delta G_2 \approx -1560 \ J \ mol^{-1} $
This calculated value of $-1560 \ J \ mol^{-1}$ falls within the specified correct answer range.
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)?
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).
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 = $1.88 \times 10^9 \text{ J} \cdot \text{m}^{-3}$
