Calculate the change in internal energy when $1 \,g$ of water is converted into steam at atmospheric pressure $(1.013 \times 10^{5} \,Pa)$. Given the latent heat of vaporisation is $2256 \,J/g$, the volume of $1 \,g$ of water is $1 \,cm^{3}$, and the volume of $1 \,g$ of steam is $1671 \,cm^{3}$.

(N/A) Let us consider $1 \,g$ of water for conversion into steam.
Latent heat of vaporisation of water, $L_{v} = 2256 \,J/g$.
Therefore, the heat absorbed is $\Delta Q = m \times L_{v} = 1 \,g \times 2256 \,J/g = 2256 \,J$.
Volume of $1 \,g$ water at atmospheric pressure is $V_{l} = 1 \,cm^{3} = 1 \times 10^{-6} \,m^{3}$.
Volume of $1 \,g$ steam is $V_{g} = 1671 \,cm^{3} = 1671 \times 10^{-6} \,m^{3}$.
Change in volume, $\Delta V = V_{g} - V_{l} = (1671 - 1) \times 10^{-6} \,m^{3} = 1670 \times 10^{-6} \,m^{3}$.
Work done at constant atmospheric pressure, $\Delta W = P \Delta V = (1.013 \times 10^{5} \,Pa) \times (1670 \times 10^{-6} \,m^{3}) \approx 169.2 \,J$.
From the first law of thermodynamics, $\Delta U = \Delta Q - \Delta W$.
$\Delta U = 2256 \,J - 169.2 \,J = 2086.8 \,J$.
Thus, the change in internal energy is $2086.8 \,J$.