$1 \mathrm{~kg}$ of water at $100^{\circ} \mathrm{C}$ is converted into steam at $100^{\circ} \mathrm{C}$ by boiling at atmospheric pressure. The volume of water changes from $1.00 \times 10^{-3} \mathrm{~m}^{3}$ as a liquid to $1.671 \mathrm{~m}^{3}$ as steam. The change in internal energy of the system during the process will be

(Given latent heat of vaporisaiton $=2257 \mathrm{~kJ} / \mathrm{kg}$, Atmospheric pressure = $\left.1 \times 10^{5} \mathrm{~Pa}\right)$

<p>To find the change in internal energy, we need to consider both the heat added during the process and the work done during the process.</p> <p>First, let&#39;s calculate the heat added ($Q$) to convert 1 kg of water at 100°C into steam at 100°C using the latent heat of vaporization:</p> <p>$Q = m \times L$</p> <p>where $m = 1 \mathrm{~kg}$ and $L = 2257 \mathrm{~kJ/kg}$:</p> <p>$Q = 1 \mathrm{~kg} \times 2257 \mathrm{~kJ/kg} = 2257 \mathrm{~kJ}$</p> <p>Next, let&#39;s calculate the work done ($W$) on the system during the process. The work done is given by:</p> <p>$W = -P \Delta V$</p> <p>where $P$ is the atmospheric pressure and $\Delta V$ is the change in volume. We are given that the atmospheric pressure is $P = 1 \times 10^5 \mathrm{~Pa}$, and the change in volume is<br/><br/> $$\Delta V = 1.671 \mathrm{~m}^3 - 1.00 \times 10^{-3} \mathrm{~m}^3 = 1.670 \mathrm{~m}^3$$.</p> <p>Now, we can calculate the work done:</p> <p>$$W = -(1 \times 10^5 \mathrm{~Pa})(1.670 \mathrm{~m}^3) = -167000 \mathrm{~J} = -167 \mathrm{~kJ}$$</p> <p>Finally, we can find the change in internal energy ($\Delta U$) using the first law of thermodynamics:</p> <p>$\Delta U = Q + W$</p> <p>Substitute the values of $Q$ and $W$:</p> <p>$\Delta U = 2257 \mathrm{~kJ} - 167 \mathrm{~kJ} = 2090 \mathrm{~kJ}$</p> <p>The change in internal energy of the system during the process is +2090 kJ. </p>