Calculate the work done in $kJ$ when $3$ moles of an ideal gas at $27^{\circ} C$ expand isothermally and reversibly from $10 \ atm$ to $1 \ atm$ $[R=8.314 \ J \ K^{-1} \ mol^{-1}]$

Match the transformations in Column-$I$ with the appropriate options in Column-$II$.

Column-$I$	Column-$II$
$(A) \; CO_{2(s)} \to CO_{2(g)}$	$(p) \; \text{Transition state}$
$(B) \; CaCO_{3(s)} \to CaO_{(s)} + CO_{2(g)}$	$(q) \; \text{Allotropic change}$
$(C) \; 2H^{\cdot} \to H_{2(g)}$	$(r) \; \Delta H > 0$
$(D) \; P_{\text{(white solid)}} \to P_{\text{(red solid)}}$	$(s) \; \Delta S > 0$
	$(t) \; \Delta S < 0$

The $\Delta H$ for vaporisation of a liquid is $20 \, kJ/mol$. Assuming ideal behaviour,the change in internal energy for the vaporisation of $1 \, mole$ of the liquid at $60^{\circ} C$ and $1 \, bar$ is close to $.... \, kJ/mol$

$1 \, \text{mole}$ of $H_2SO_4$ is mixed with $2 \, \text{moles}$ of $NaOH$. The heat evolved will be

The enthalpy of sublimation of $I_{2(s)}$ is $57.3 \, kJ \, mol^{-1}$ and the enthalpy of fusion is $15.5 \, kJ \, mol^{-1}$. The enthalpy of vaporization of $I_2$ is ..................... $kJ \, mol^{-1}$. (in $.8$)

The difference between the reaction enthalpy change $(\Delta _r H)$ and reaction internal energy change $(\Delta _r U)$ for the reaction $2C_6H_{6(l)} + 15O_{2(g)} \longrightarrow 12CO_{2(g)} + 6H_2O_{(l)}$ at $300 \ K$ is $....$ $J \ mol^{-1}$ $(R = 8.314 \ J \ mol^{-1} \ K^{-1})$