Answer :
Certainly! Let's break down the problem and solve it step-by-step to determine the number of Schottky defects per cubic meter in potassium chloride (KCl) at 500°C.
### Step 1: Given Values and Constants
- Energy per Schottky defect (Q): [tex]\(2.6 \, \text{eV}\)[/tex]
- Density of KCl at 500°C (ρ): [tex]\(1.955 \, \text{g/cm}^3\)[/tex]
- Temperature (T): [tex]\(500°C\)[/tex]
- Boltzmann constant (k_B): [tex]\(8.617333262 \times 10^{-5} \, \text{eV/K}\)[/tex]
- Avogadro's number (N_A): [tex]\(6.022 \times 10^{23} \, \text{mol}^{-1}\)[/tex]
- Molar mass of KCl (M): [tex]\(74.55 \, \text{g/mol}\)[/tex]
### Step 2: Converting Temperature to Kelvin
The temperature in Kelvin (T_K) can be calculated as:
[tex]\[ T_K = 500\, ^\circ\text{C} + 273.15 = 773.15 \, \text{K} \][/tex]
### Step 3: Convert Density to Proper Units
Convert the density from [tex]\( \text{g/cm}^3 \)[/tex] to [tex]\( \text{kg/m}^3 \)[/tex]:
[tex]\[ \rho_{\text{KCl}} = 1.955 \, \text{g/cm}^3 \times 1000 \, \text{kg/m}^3/\text{g/cm}^3 = 1955 \, \text{kg/m}^3 \][/tex]
### Step 4: Calculate the Molar Volume
The molar volume [tex]\(V_m\)[/tex] is the volume occupied by one mole of KCl:
[tex]\[ V_m = \frac{M}{\rho_{\text{KCl}}} \][/tex]
Substitute the values:
[tex]\[ V_m = \frac{74.55 \, \text{g/mol}}{1955 \, \text{g/L}} = \frac{74.55 \times 10^{-3} \, \text{kg/mol}}{1955 \times 10^{-3} \, \text{kg/L}} = \frac{74.55}{1955} \, \text{m}^3/\text{mol} \approx 0.03814 \, \text{m}^3/\text{mol} \][/tex]
### Step 5: Number of Formula Units per Cubic Meter
Calculate the total number of formula units (N_total) in one cubic meter:
[tex]\[ N_{\text{total}} = \frac{N_A}{V_m} = \frac{6.022 \times 10^{23} \, \text{mol}^{-1}}{0.03814 \, \text{m}^3/\text{mol}} \approx 1.5792 \times 10^{25} \, \text{units/m}^3 \][/tex]
### Step 6: Calculate the Number of Schottky Defects
Using the formula for the number of Schottky defects [tex]\( N_s \)[/tex]:
[tex]\[ N_s = N_{\text{total}} \times \exp \left( -\frac{Q}{2k_B T_K} \right) \][/tex]
Substitute the values:
[tex]\[ N_s = 1.5792 \times 10^{25} \times \exp \left( -\frac{2.6 \, \text{eV}}{2 \times 8.617333262 \times 10^{-5} \, \text{eV/K} \times 773.15 \, \text{K}} \right) \][/tex]
Simplifying the exponent:
[tex]\[ N_s \approx 5.3014 \times 10^{16} \][/tex]
### Final Answer
- Number of formula units per cubic meter [tex]\(N_{\text{total}}\)[/tex]: [tex]\(1.5792 \times 10^{25} \, \text{units/m}^3\)[/tex]
- Number of Schottky defects per cubic meter [tex]\(N_s\)[/tex]: [tex]\(5.3014 \times 10^{16} \, \text{defects/m}^3\)[/tex]
These values represent the density of atomic formula units and the density of Schottky defects in potassium chloride at 500°C respectively.
### Step 1: Given Values and Constants
- Energy per Schottky defect (Q): [tex]\(2.6 \, \text{eV}\)[/tex]
- Density of KCl at 500°C (ρ): [tex]\(1.955 \, \text{g/cm}^3\)[/tex]
- Temperature (T): [tex]\(500°C\)[/tex]
- Boltzmann constant (k_B): [tex]\(8.617333262 \times 10^{-5} \, \text{eV/K}\)[/tex]
- Avogadro's number (N_A): [tex]\(6.022 \times 10^{23} \, \text{mol}^{-1}\)[/tex]
- Molar mass of KCl (M): [tex]\(74.55 \, \text{g/mol}\)[/tex]
### Step 2: Converting Temperature to Kelvin
The temperature in Kelvin (T_K) can be calculated as:
[tex]\[ T_K = 500\, ^\circ\text{C} + 273.15 = 773.15 \, \text{K} \][/tex]
### Step 3: Convert Density to Proper Units
Convert the density from [tex]\( \text{g/cm}^3 \)[/tex] to [tex]\( \text{kg/m}^3 \)[/tex]:
[tex]\[ \rho_{\text{KCl}} = 1.955 \, \text{g/cm}^3 \times 1000 \, \text{kg/m}^3/\text{g/cm}^3 = 1955 \, \text{kg/m}^3 \][/tex]
### Step 4: Calculate the Molar Volume
The molar volume [tex]\(V_m\)[/tex] is the volume occupied by one mole of KCl:
[tex]\[ V_m = \frac{M}{\rho_{\text{KCl}}} \][/tex]
Substitute the values:
[tex]\[ V_m = \frac{74.55 \, \text{g/mol}}{1955 \, \text{g/L}} = \frac{74.55 \times 10^{-3} \, \text{kg/mol}}{1955 \times 10^{-3} \, \text{kg/L}} = \frac{74.55}{1955} \, \text{m}^3/\text{mol} \approx 0.03814 \, \text{m}^3/\text{mol} \][/tex]
### Step 5: Number of Formula Units per Cubic Meter
Calculate the total number of formula units (N_total) in one cubic meter:
[tex]\[ N_{\text{total}} = \frac{N_A}{V_m} = \frac{6.022 \times 10^{23} \, \text{mol}^{-1}}{0.03814 \, \text{m}^3/\text{mol}} \approx 1.5792 \times 10^{25} \, \text{units/m}^3 \][/tex]
### Step 6: Calculate the Number of Schottky Defects
Using the formula for the number of Schottky defects [tex]\( N_s \)[/tex]:
[tex]\[ N_s = N_{\text{total}} \times \exp \left( -\frac{Q}{2k_B T_K} \right) \][/tex]
Substitute the values:
[tex]\[ N_s = 1.5792 \times 10^{25} \times \exp \left( -\frac{2.6 \, \text{eV}}{2 \times 8.617333262 \times 10^{-5} \, \text{eV/K} \times 773.15 \, \text{K}} \right) \][/tex]
Simplifying the exponent:
[tex]\[ N_s \approx 5.3014 \times 10^{16} \][/tex]
### Final Answer
- Number of formula units per cubic meter [tex]\(N_{\text{total}}\)[/tex]: [tex]\(1.5792 \times 10^{25} \, \text{units/m}^3\)[/tex]
- Number of Schottky defects per cubic meter [tex]\(N_s\)[/tex]: [tex]\(5.3014 \times 10^{16} \, \text{defects/m}^3\)[/tex]
These values represent the density of atomic formula units and the density of Schottky defects in potassium chloride at 500°C respectively.
