The densities of graphite and diamond at `298K` are `2.25` and `3.31gcm^(-3)` , respectively. If the standard free energy difference `(DeltaG^(0))` is equal to `1895Jmol^(-1)` , the pressure at which graphite will be transformed into diamond at `298K` is

Options

A. `9.92xx106Pa`
B. `9.92xx105Pa`
C. `9.92xx108Pa`
D. `9.92xx107Pa`

Explanation

To solve the problem, we need to find the pressure at which graphite will be transformed into diamond at 298 K, given the standard free energy difference (ΔG°) and the densities of both forms of carbon. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the relationship between ΔG°, pressure (P), and volume change (ΔV)
The relationship is given by the equation:
\[ \Delta G^{0} = P \Delta V \]
where ΔG° is the standard free energy change, P is the pressure, and ΔV is the change in volume during the transformation.

### Step 2: Calculate the volume of graphite and diamond
The volume (V) can be calculated using the formula:
\[ V = \frac{m}{\rho} \]
where \(m\) is the mass and \(\rho\) is the density.

Assuming we take 1 mole of each substance (which is approximately 12 grams for carbon):
- For graphite:
- Density (\(\rho_{graphite}\)) = 2.25 g/cm³ = 2250 kg/m³
- Volume (\(V_{graphite}\)) = \(\frac{12 \, \text{g}}{2.25 \, \text{g/cm}^3} = \frac{12 \times 10^{-3} \, \text{kg}}{2250 \, \text{kg/m}^3} = 5.33 \times 10^{-6} \, \text{m}^3\)

- For diamond:
- Density (\(\rho_{diamond}\)) = 3.31 g/cm³ = 3310 kg/m³
- Volume (\(V_{diamond}\)) = \(\frac{12 \, \text{g}}{3.31 \, \text{g/cm}^3} = \frac{12 \times 10^{-3} \, \text{kg}}{3310 \, \text{kg/m}^3} = 3.62 \times 10^{-6} \, \text{m}^3\)

### Step 3: Calculate the change in volume (ΔV)
\[ \Delta V = V_{diamond} - V_{graphite} = 3.62 \times 10^{-6} \, \text{m}^3 - 5.33 \times 10^{-6} \, \text{m}^3 = -1.71 \times 10^{-6} \, \text{m}^3 \]

### Step 4: Rearrange the equation to find pressure (P)
Substituting ΔG° and ΔV into the equation:
\[ P = \frac{\Delta G^{0}}{\Delta V} \]
Given that ΔG° = 1895 J/mol and ΔV = -1.71 × 10^{-6} m³, we find:
\[ P = \frac{1895 \, \text{J/mol}}{-1.71 \times 10^{-6} \, \text{m}^3} \]

### Step 5: Calculate the pressure
\[ P = -1.107 \times 10^{9} \, \text{Pa} = 1107 \, \text{MPa} \]
Converting to units of 10^8 Pa:
\[ P \approx 110.7 \times 10^{6} \, \text{Pa} = 110.7 \times 10^{8} \, \text{Pa} \]

### Conclusion
The pressure at which graphite will be transformed into diamond at 298 K is approximately \(110.7 \times 10^{8} \, \text{Pa}\).