The molar heat capacities of Iodine vapour and solid are `7.8` and 14 cal/mol respectively. If enthalpy of iodine is 6096 cal/mol at `200^(@)C`, then what is `DeltaU` (internal energy change) at `250^(@)C` in cal/mol

Options

A. 5360
B. 4740
C. 6406
D. None of these

Explanation

To solve the question, we need to find the change in internal energy (ΔU) of iodine when it transitions from solid to vapor at different temperatures. We are given the molar heat capacities of iodine vapor and solid, the enthalpy of iodine at 200°C, and we need to find ΔU at 250°C.

### Step-by-Step Solution:

1. **Identify Given Data**:
- Molar heat capacity of iodine vapor (Cₚ,vapor) = 7.8 cal/mol·K
- Molar heat capacity of iodine solid (Cₚ,solid) = 14 cal/mol·K
- Enthalpy of iodine at 200°C (H₁) = 6096 cal/mol
- Temperature change from 200°C to 250°C (ΔT) = 250°C - 200°C = 50°C

2. **Calculate ΔCₚ**:
\[ \Delta Cₚ = Cₚ,vapor - Cₚ,solid = 7.8 - 14 = -6.2 \text{ cal/mol·K} \]

3. **Calculate Enthalpy at 250°C (H₂)**:
Using the formula:
\[ H₂ = H₁ + \Delta Cₚ \cdot \Delta T \]
Substituting the values:
\[ H₂ = 6096 + (-6.2) \cdot 50 \]
\[ H₂ = 6096 - 310 = 5786 \text{ cal/mol} \]

4. **Calculate Δn (Change in Moles)**:
In the reaction of solid iodine to vapor, we consider:
- Moles of products (gaseous iodine) = 1
- Moles of reactants (solid iodine) = 0
\[ \Delta n = 1 - 0 = 1 \]

5. **Use the Relation between ΔH and ΔU**:
The relation is given by:
\[ \Delta H = \Delta U + \Delta n \cdot R \cdot T \]
Here, R (gas constant) in cal/mol·K = 2 cal/mol·K and T must be in Kelvin:
\[ T = 250 + 273 = 523 \text{ K} \]

6. **Substituting Values to Find ΔU**:
Rearranging the equation for ΔU:
\[ \Delta U = \Delta H - \Delta n \cdot R \cdot T \]
Substituting the known values:
\[ \Delta U = 5786 - 1 \cdot 2 \cdot 523 \]
\[ \Delta U = 5786 - 1046 = 4740 \text{ cal/mol} \]

### Final Answer:
\[ \Delta U = 4740 \text{ cal/mol} \]