For water $$\Delta_{vap}H = 41$$ kJ mol$$^{-1}$$ at 373 K and 1 bar pressure. Assuming that water vapour is an ideal gas that occupies a much larger volume than liquid water, the internal energy change during evaporation of water is _________ (kJ mol$$^{-1}$$):
[Use: R = 8.3 J mol$$^{-1}$$ K$$^{-1}$$]

We are given that the molar enthalpy change of vaporisation of water at 373 K and 1 bar is $$\Delta_{\text{vap}}H = 41\;\text{kJ mol}^{-1}.$$ We have to find the corresponding molar internal-energy change $$\Delta_{\text{vap}}U.$$

First, we recall the thermodynamic relation that connects enthalpy change and internal-energy change for any process carried out at constant pressure:

$$\Delta H = \Delta U + \Delta(PV).$$

Rearranging this formula for the required quantity gives

$$\Delta U = \Delta H - \Delta(PV).$$

During vaporisation, one mole of liquid water turns into one mole of water vapour. Liquid water occupies a negligibly small volume compared with the vapour, so the change in the product $$PV$$ is effectively the value for the vapour itself:

$$\Delta(PV) \approx P_{\text{vap}}\,V_{\text{vap}} - P_{\text{liq}}\,V_{\text{liq}} \approx P_{\text{vap}}\,V_{\text{vap}}.$$

At 1 bar the vapour behaves ideally, so we can use the ideal-gas equation for one mole:

$$P_{\text{vap}}\,V_{\text{vap}} = RT.$$

Substituting this into the earlier expression, we obtain

$$\Delta U = \Delta H - RT.$$

Now we insert the numerical values. The universal gas constant in kJ units is

$$R = 8.3\;\text{J mol}^{-1}\text{ K}^{-1} = 0.0083\;\text{kJ mol}^{-1}\text{ K}^{-1}.$$

Hence

$$RT = 0.0083\;\text{kJ mol}^{-1}\text{ K}^{-1} \times 373\;\text{K}.$$

Multiplying gives

$$RT = 3.0959\;\text{kJ mol}^{-1} \approx 3.1\;\text{kJ mol}^{-1}.$$

Finally, we substitute the values of $$\Delta H$$ and $$RT$$ into $$\Delta U = \Delta H - RT$$:

$$\Delta_{\text{vap}}U = 41\;\text{kJ mol}^{-1} - 3.1\;\text{kJ mol}^{-1}.$$

Carrying out the subtraction,

$$\Delta_{\text{vap}}U = 37.9\;\text{kJ mol}^{-1}.$$

Rounding to the appropriate number of significant figures, we have

$$\Delta_{\text{vap}}U \approx 38\;\text{kJ mol}^{-1}.$$

So, the answer is $$38$$.