Molar enthalpy change for vapourisation of $1.0 \mathrm{~mol}$ of water at $1.0 \mathrm{bar}$ and $100^{\circ} \mathrm{C}$ is $41.0 \mathrm{~kJ} \mathrm{~mol}^{-1}$. If water vapour is assumed to be an ideal gas, the internal energy change for $1.0 \mathrm{~g}$ of water in $\mathrm{kJ}$ is

• A

  $1.053$
• B

  $4.212$
• C

  $3.159$
• Correct answer
  D

  $2.106$

Hence, the correct answer is option: 'D'.