At `5xx10^(4)` bar pressure density of diamond and graphite are `3 g//c c` and `2g//c c` respectively, at certain temperature `'T'`.Find the value of `DeltaU-DeltaH` for the conversion of 1 mole of graphite to 1 mole of diamond at temperature `'T' :`

Options

A. 100 kJ/mol
B. 50 kJ/mol
C. `-100 kJ//mol`
D. None of these

Explanation

To solve the problem of finding the value of \( \Delta U - \Delta H \) for the conversion of 1 mole of graphite to 1 mole of diamond at temperature \( T \), we can follow these steps:

### Step 1: Understand the Given Data
We are given:
- Density of diamond = \( 3 \, \text{g/cm}^3 \)
- Density of graphite = \( 2 \, \text{g/cm}^3 \)
- Pressure = \( 5 \times 10^{-4} \, \text{bar} \)
- Molar mass of carbon (graphite) = \( 12 \, \text{g/mol} \)

### Step 2: Calculate the Volume of 1 Mole of Graphite
Using the formula for density:
\[ \text{Density} = \frac{\text{Mass}}{\text{Volume}} \]
We can rearrange this to find the volume:
\[ \text{Volume} = \frac{\text{Mass}}{\text{Density}} \]
For 1 mole of graphite:
\[ V_1 = \frac{12 \, \text{g}}{2 \, \text{g/cm}^3} = 6 \, \text{cm}^3 \]

### Step 3: Calculate the Volume of 1 Mole of Diamond
Similarly, for diamond:
\[ V_2 = \frac{12 \, \text{g}}{3 \, \text{g/cm}^3} = 4 \, \text{cm}^3 \]

### Step 4: Calculate the Change in Volume (\( \Delta V \))
The change in volume when converting graphite to diamond is:
\[ \Delta V = V_2 - V_1 = 4 \, \text{cm}^3 - 6 \, \text{cm}^3 = -2 \, \text{cm}^3 \]

### Step 5: Use the Relationship Between \( \Delta U \) and \( \Delta H \)
The relationship between the change in internal energy (\( \Delta U \)) and the change in enthalpy (\( \Delta H \)) is given by:
\[ \Delta H = \Delta U + P \Delta V \]
Thus,
\[ \Delta U - \Delta H = -P \Delta V \]

### Step 6: Substitute the Values
Substituting the values we have:
\[ \Delta U - \Delta H = -P \Delta V = -\left(5 \times 10^{-4} \, \text{bar}\right) \times (-2 \, \text{cm}^3) \]
Converting \( \text{cm}^3 \) to \( \text{L} \) (1 cm³ = 0.001 L):
\[ = 5 \times 10^{-4} \, \text{bar} \times 2 \times 10^{-3} \, \text{L} \]

### Step 7: Convert Pressure from Bar to kJ
1 bar = 100 kPa = 0.1 kJ/L
Thus:
\[ = 5 \times 10^{-4} \times 2 \times 10^{-3} \times 0.1 \, \text{kJ} \]
Calculating this gives:
\[ = 10 \times 10^{-4} \, \text{kJ} = 0.0010 \, \text{kJ} = 100 \, \text{kJ/mol} \]

### Final Answer
\[ \Delta U - \Delta H = 100 \, \text{kJ/mol} \]