The latent heat of vapourisation of a liquid at `500K` and `atm` pressure is `10kcal mol^(-1)`. What will be change in internal energy of `3mol` of liquid at same temperature?

Options

A. `13.0 kcal`
B. `-13.0 kcal`
C. `27.0 kcal`
D. `-27.0 kcal`

Explanation

To find the change in internal energy (ΔU) of 3 moles of a liquid at 500 K when it vaporizes, we can use the relationship between the change in enthalpy (ΔH), internal energy (ΔU), and the change in the number of moles of gas (Δn) during the phase transition.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Latent heat of vaporization (ΔH) = 10 kcal/mol
- Number of moles (n) = 3 mol
- Temperature (T) = 500 K
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