JEE Main 2023 — Thermal Properties of Matter Question with Solution

The work to be done in the process is given by

$\begin{array}{rcl}dW& =& PdV\\ & =& 1\times {10}^5 \mathrm{Pa}\times (1.671-0.001) {\mathrm{m}}^3\\ & =& 1.670\times {10}^5 \mathrm{J}\end{array}$

The change in heat energy during the vaporisation process can be calculated as follows-

$\begin{array}{rcl}\Delta Q_{\text{supplied }}& =& 2257\times 1\times {10}^3 \mathrm{J}\\ & =& 22.57\times {10}^5 \mathrm{J}\end{array}$

Hence, the change in internal energy in the process is given by

$\begin{array}{rcl}\Delta U& =& \Delta Q_{\mathrm{supplied}}-\Delta W\\ & =& (22.57-1.67)\times {10}^5 \mathrm{J}\\ & =& 20.9\times {10}^5 \mathrm{J}\\ & =& 2090 \mathrm{kJ}\end{array}$