JAMB Chemistry 2018 Paper

Concept:Free expansion of an ideal gas into a vacuum involves no work and no heat exchange.Explanation:In a free expansion, the external pressure is zero, so the work done is $W = 0$.For an ideal gas, the internal energy depends only on temperature.Since the gas expands into a vacuum without doing work and without absorbing heat, $\Delta U = 0$ and the temperature remains constant.Enthalpy is $H = U + PV$. For an ideal gas, $PV = nRT$, so $\Delta H = \Delta U + nR\Delta T$.Both $\Delta U = 0$ and $\Delta T = 0$, therefore $\Delta H = 0$.However, the gas now occupies a larger volume, so its entropy increases: $\Delta S > 0$.Thus, $T\Delta S$ is positive because $\Delta S$ is positive at constant temperature.Answer:D. $\Delta H$ is zero and $T\Delta S$ is positive.