For water at `100^(@)C` and 1 bar, `Delta_("vap") H - Delta_("vap")U =` _______ `xx 10^(2) J mol^(-1)`.
(Round off to the Nearest Integer)
[Use : `R = 8.31 J mol^(-1) K^(-1)`]
[Assume volume of `H_(2)O(l)` is much smaller than volume of `H_(2)O(g)`. Assume `H_(2)O(g)` treated as an ideal gas]

Answer

To solve the problem, we need to find the difference between the enthalpy of vaporization (ΔvapH) and the internal energy of vaporization (ΔvapU) for water at 100°C and 1 bar. We will use the relationship between these quantities and the ideal gas law.

### Step-by-Step Solution:

1. **Understand the relationship between ΔvapH and ΔvapU:**
The relationship between the change in enthalpy (ΔH) and the change in internal energy (ΔU) is given by the equation:
\[ \Delta H = \Delta U + \Delta n \cdot R \cdot T \]
where:
- Δn = change in the number of moles of gas
- R = universal gas constant (8.31 J mol⁻¹ K⁻¹)
- T = temperature in Kelvin

2. **Identify Δn:**
In the case of vaporization of water, we have:
- 1 mole of water (liquid) converts to 1 mole of water vapor (gas).
- Therefore, the change in moles (Δn) is:
\[ \Delta n = 1 - 0 = 1 \]

3. **Convert temperature to Kelvin:**
The temperature given is 100°C. To convert this to Kelvin:
\[ T = 100 + 273 = 373 \text{ K} \]

4. **Substitute values into the equation:**
Now we can rearrange the equation to find ΔvapH - ΔvapU:
\[ \Delta H - \Delta U = \Delta n \cdot R \cdot T \]
Substituting the values:
\[ \Delta H - \Delta U = 1 \cdot 8.31 \cdot 373 \]

5. **Calculate the right-hand side:**
\[ \Delta H - \Delta U = 8.31 \cdot 373 = 3103.03 \text{ J/mol} \]

6. **Express the result in the required format:**
We need to express this result in the form of \(xx \times 10^2 \text{ J/mol}\):
\[ 3103.03 \text{ J/mol} = 31.03 \times 10^2 \text{ J/mol} \]
Rounding off to the nearest integer gives:
\[ 31 \times 10^2 \text{ J/mol} \]

### Final Answer:
\[ \Delta_{\text{vap}} H - \Delta_{\text{vap}} U = 31 \times 10^2 \text{ J/mol} \]