To solve the problem, we need to find the pressure at which graphite transforms into diamond at 298 K using the given densities and the standard free energy difference.
### Step 1: Write down the given data
- Density of graphite, \( \rho_g = 2.25 \, \text{g/cm}^3 \)
- Density of diamond, \( \rho_d = 3.31 \, \text{g/cm}^3 \)
- Standard free energy difference, \( \Delta G^{0} = 1895 \, \text{J/mol} \)
### Step 2: Calculate the volume of 1 mole of each form of carbon
The molar mass of carbon (C) is approximately \( 12 \, \text{g/mol} \).
1. **Volume of graphite**:
\[
V_g = \frac{\text{mass}}{\text{density}} = \frac{12 \, \text{g}}{2.25 \, \text{g/cm}^3} = 5.33 \, \text{cm}^3
\]
2. **Volume of diamond**:
\[
V_d = \frac{\text{mass}}{\text{density}} = \frac{12 \, \text{g}}{3.31 \, \text{g/cm}^3} = 3.63 \, \text{cm}^3
\]
### Step 3: Calculate the change in volume (\( \Delta V \))
\[
\Delta V = V_d - V_g = 3.63 \, \text{cm}^3 - 5.33 \, \text{cm}^3 = -1.70 \, \text{cm}^3
\]
### Step 4: Convert \( \Delta V \) to liters
\[
\Delta V = -1.70 \, \text{cm}^3 \times 10^{-3} \, \text{L/cm}^3 = -1.70 \times 10^{-3} \, \text{L}
\]
### Step 5: Use the relationship between \( \Delta G^{0} \), pressure (\( P \)), and change in volume (\( \Delta V \))
The equation relating these quantities is:
\[
\Delta G^{0} = -P \Delta V
\]
Rearranging gives:
\[
P = -\frac{\Delta G^{0}}{\Delta V}
\]
### Step 6: Substitute the values and calculate pressure
Substituting the values:
\[
P = -\frac{1895 \, \text{J/mol}}{-1.70 \times 10^{-3} \, \text{L}} \times 101.3 \, \text{J/(L \cdot atm)}
\]
\[
P = \frac{1895 \times 101.3}{1.70 \times 10^{-3}} \, \text{atm}
\]
Calculating this gives:
\[
P \approx 10.93 \times 10^{3} \, \text{atm}
\]
### Step 7: Convert pressure to Pascals
1 atm = \( 1.01 \times 10^{5} \, \text{Pa} \)
\[
P \approx 10.93 \times 10^{3} \, \text{atm} \times 1.01 \times 10^{5} \, \text{Pa/atm} \approx 11.07 \times 10^{8} \, \text{Pa}
\]
### Final Answer
The pressure at which graphite will be transformed into diamond at 298 K is approximately \( 11.07 \times 10^{8} \, \text{Pa} \).
---
