If 2.5 moles of an ideal gas at a certain temperature are allowed to expand isothermally and reversibly from an initial volume of $2 \mathrm{dm}^3$ to $20 \mathrm{dm}^3$, the work done by the gas is -16.5 kJ . The temperature (in K ) of the gas is (Round off to the nearest value) ( $R=8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$ )

The bond enthalpies of heavy hydrogen, $\mathrm{O}-\mathrm{O}$ and $\mathrm{D}-\mathrm{O}$ are $+400,+498$ and $+490 \mathrm{kJmol}^{-1}$, respectively. The $\Delta_r H^{\circ}$ of the reaction to produce $\mathrm{D}_2 \mathrm{O}$ is

In the reaction, $\mathrm{H}_2 \mathrm{O}(l) E \longrightarrow \mathrm{H}_2 \mathrm{O}(s)$ at $0^{\circ} \mathrm{C}$ and 1 atom, the internal energy change is $-41 \mathrm{~kJ} / \mathrm{mol}$. What will be the value of molar enthalpy change?

The correct order of " $\Delta H_f^{\circ}$ " values of diamond (I), graphite (II) and fullerene (III) is