One mole of water is converted into steam at $373 \, K$. The heat absorbed at $1 \, atm$ pressure is $40.68 \, kJ$. If the molar volumes of water and steam are $18 \, mL$ and $30600 \, mL$ respectively,find $\Delta U$ for the process in $kJ$.

Enthalpy of sublimation of iodine is $24 \ cal \ g^{-1}$ at $200 \ ^oC$. If specific heat of $I_{2(s)}$ and $I_{2(vap)}$ are $0.055$ and $0.031 \ cal \ g^{-1} K^{-1}$ respectively,then enthalpy of sublimation of iodine at $250 \ ^oC$ in $cal \ g^{-1}$ is

The heat of reaction for $C_6H_{12}O_{6(s)} + 6O_{2(g)} \to 6CO_{2(g)} + 6H_2O_{(l)}$ at constant pressure is $-651 \, kcal$ at $17 \, ^oC$. Calculate the heat of reaction at constant volume at $17 \, ^oC$ in $kcal$.

$2.2 \, g$ of nitrous oxide $(N_{2}O)$ gas is cooled at a constant pressure of $1 \, atm$ from $310 \, K$ to $270 \, K$ causing the compression of the gas from $217.1 \, mL$ to $167.75 \, mL$. The change in internal energy of the process,$\Delta U$ is $-x \, J$. The value of $x$ is $....$ [nearest integer] (Given: atomic mass of $N = 14 \, g \, mol^{-1}$ and of $O = 16 \, g \, mol^{-1}$. Molar heat capacity of $N_{2}O$ is $100 \, J \, K^{-1} \, mol^{-1}$)

For an ideal gas, consider only $P-V$ work in going from an initial state $X$ to the final state $Z$. The final state $Z$ can be reached by either of the two paths shown in the figure. Which of the following choice(s) is (are) correct? [take $\Delta S$ as change in entropy and $w$ as work done].
(A) $\Delta S_{X \to Z} = \Delta S_{X \to Y} + \Delta S_{Y \to Z}$
(B) $w_{X \to Z} = w_{X \to Y} + w_{Y \to Z}$
(C) $w_{X \to Y \to Z} = w_{X \to Y} + w_{Y \to Z}$
(D) $\Delta S_{X \to Y \to Z} = \Delta S_{X \to Y}$

The ratio of heats liberated at $298 \ K$ from the combustion of one $kg$ of coke and by burning water gas obtained from $1 \ kg$ of coke is. (Assume coke to be $100 \%$ carbon.) (Given enthalpies of combustion of $CO_{2}, CO$ and $H_{2}$ as $393.5 \ kJ/mol, 283.5 \ kJ/mol, 285.5 \ kJ/mol$ respectively all at $298 \ K$.) (in $: 1$)