For water at 100°C and 1 bar,
$$\Delta_{vap}H - \Delta_{vap}U = \_ \times 10^2$$ 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]

We start with the relation that connects the enthalpy change and the internal-energy change for any process:

$$\Delta H \;=\;\Delta U \;+\;\Delta(pV)$$

Therefore, for vaporisation of one mole, the quantity asked is

$$\Delta_{\text{vap}}H \;-\;\Delta_{\text{vap}}U \;=\;\Delta(pV).$$

Next we evaluate the change in the $$pV$$ term between liquid water and its vapour at the boiling point. The given instructions allow us to neglect the volume of the liquid in comparison with the gas, and to treat the vapour as an ideal gas.

For one mole of an ideal gas the equation of state is

$$pV = RT.$$

Consequently, inside the difference $$pV_{\text{gas}} - pV_{\text{liquid}}$$ we keep only the gas contribution:

$$\Delta(pV) \;=\; p_{\text{gas}}V_{\text{gas}} \;-\; p_{\text{liquid}}V_{\text{liquid}} \;\approx\; RT \;-\; 0 \;=\; RT.$$

At the boiling point of water the temperature is $$100^{\circ}\text{C} = 373\;\text{K}$$. Using the gas constant $$R = 8.31\;\text{J mol}^{-1}\text{K}^{-1}$$, we obtain

$$\Delta(pV) = RT = 8.31\;\text{J mol}^{-1}\text{K}^{-1} \times 373\;\text{K}.$$

We now perform the multiplication step by step:

$$8.31 \times 373 = (8 \times 373)\;+\;(0.31 \times 373).$$

First term:

$$8 \times 373 = 2984.$$

Second term:

$$0.31 \times 373 = 0.3 \times 373 + 0.01 \times 373 = 111.9 + 3.73 = 115.63.$$

Adding both parts,

$$2984 + 115.63 = 3099.63\;\text{J mol}^{-1}.$$

Hence

$$\Delta_{\text{vap}}H - \Delta_{\text{vap}}U \;=\; 3099.63\;\text{J mol}^{-1}.$$

The question asks us to express this value as $$\_\times 10^{2}\;\text{J mol}^{-1}$$ and round to the nearest integer. Dividing by $$10^{2}$$ gives

$$\frac{3099.63}{100} = 30.9963 \approx 31.$$

So, the answer is $$31$$.