1 g of water, of volume 1 ~cm^3 at 100^ C is converted into steam...

Given, mass of water, $\mathrm{m}=\mathrm{lg}$ Volume of $1 \mathrm{ g}$ of water $=1 \mathrm{ cm}^3=10^{-6} \mathrm{ m}^3$ Volume of $1 \mathrm{ g}$ of steam $=1671 \mathrm{ cm}^3=1671 \times 10^{-6} \mathrm{ m}^3$ Pressure, $\mathrm{p}=1 \times 10^5 \mathrm{ Pa}$ Latent heat of vaporization of water, $\mathrm{L}=2256 \mathrm{ J} / \mathrm{g}$ Change in volume, $\Delta V=(1671-1) \times 10^{-6} \mathrm{ m}^3$ =1670 $\times$10^{-6} \mathrm{ m}^3 Heat supplied, $\Delta$\mathrm{Q}=\mathrm{mL}$=1 \times 2256=2256 \mathrm{ J}$ As the steam expands, so the work done in expansion is $$\begin{array}{rlr}\Delta \mathrm{W}& =\mathrm{p} \Delta \mathrm{V}=1 \times10^5 \times1670 \times10^{-6} \quad \text{ [from Eq. (i)] }\\ & =167 \mathrm{ J}& \ldots \text{.(iii) } \end{array}$$According to first law of thermodynamics, $$\begin{array}{rlrl}\Delta \mathrm{Q}& =\Delta \mathrm{U}+\Delta \mathrm{W} \Rightarrow \Delta \mathrm{U}=\Delta \mathrm{Q}-\Delta \mathrm{W}\\ & =2256-167 & \text{ [from Eq. (ii) and (iii)] }\\ & =2089 \mathrm{ J}& \end{array}$$