Diamonds are formed from graphite under high pressure in coal mines. Calculate the equilibrium pressure (in atm) at which graphite is converted to diamonds at `25^(@)C` (assumed constant) given densities of `rho_("graphite")=2g//c c & rho_("diamond")=3g//c c (DeltaG_(f)^(@))` for diamonds is `"3 k J m"^(-1)` from graphite

Answer

To calculate the equilibrium pressure at which graphite is converted to diamonds at \(25^\circ C\), we can use the relationship between the change in Gibbs free energy (\(\Delta G\)), the change in volume (\(\Delta V\)), and pressure (\(P\)). The relevant equation is:

\[ dG = V dP - S dT \]

Since the temperature is constant (\(dT = 0\)), we can simplify this to:

\[ dG = V dP \]

### Step 1: Calculate \(\Delta G\)

Given that \(\Delta G_f^\circ\) for diamonds from graphite is \(3 \, \text{kJ/mol}\), we convert this to Joules:

\[ \Delta G_f^\circ = 3 \, \text{kJ/mol} = 3000 \, \text{J/mol} \]

### Step 2: Calculate the change in volume \(\Delta V\)

We need to find the molar volumes of graphite and diamond using their densities:

- Density of graphite, \(\rho_{\text{graphite}} = 2 \, \text{g/cm}^3\)
- Density of diamond, \(\rho_{\text{diamond}} = 3 \, \text{g/cm}^3\)

The molar volume (\(V_m\)) can be calculated using the formula:

\[ V_m = \frac{M}{\rho} \]

Where \(M\) is the molar mass. For carbon, \(M = 12 \, \text{g/mol}\).

Calculating the molar volumes:

\[ V_m(\text{graphite}) = \frac{12 \, \text{g/mol}}{2 \, \text{g/cm}^3} = 6 \, \text{cm}^3/\text{mol} = 6 \times 10^{-6} \, \text{m}^3/\text{mol} \]

\[ V_m(\text{diamond}) = \frac{12 \, \text{g/mol}}{3 \, \text{g/cm}^3} = 4 \, \text{cm}^3/\text{mol} = 4 \times 10^{-6} \, \text{m}^3/\text{mol} \]

Now, we can find \(\Delta V\):

\[ \Delta V = V_m(\text{diamond}) - V_m(\text{graphite}) = (4 - 6) \times 10^{-6} \, \text{m}^3/\text{mol} = -2 \times 10^{-6} \, \text{m}^3/\text{mol} \]

### Step 3: Substitute into the equation

Now we substitute \(\Delta G\) and \(\Delta V\) into the equation:

\[ \Delta G = \Delta V \cdot P \]

Rearranging gives us:

\[ P = \frac{\Delta G}{\Delta V} \]

Substituting the values:

\[ P = \frac{3000 \, \text{J/mol}}{-2 \times 10^{-6} \, \text{m}^3/\text{mol}} = -1.5 \times 10^9 \, \text{Pa} \]

### Step 4: Convert to atm

To convert from Pascals to atmospheres, we use the conversion factor \(1 \, \text{atm} = 1.013 \times 10^5 \, \text{Pa}\):

\[ P = \frac{-1.5 \times 10^9 \, \text{Pa}}{1.013 \times 10^5 \, \text{Pa/atm}} \approx -14800 \, \text{atm} \]

Since pressure cannot be negative, we take the absolute value:

\[ P \approx 14800 \, \text{atm} \]

### Final Answer

The equilibrium pressure at which graphite is converted to diamonds at \(25^\circ C\) is approximately **14800 atm**.

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