During the isothermal expansion of an ideal gas:

In an isothermal expansion of a gaseous sample, the correct relation is: (consider $\mathrm{w}$ (work) with sign according to new IUPAC convention)

[The reversible and irreversible processes are carried out between same initial and final states.]

Calculate magnitude of $\mathrm{\Delta U}$ of the reaction in $\mathrm{kcal} {\mathrm{mol}}^{-1}$ for the hydrogenation of acetylene to ethene at constant volume and at $77°\mathrm{C}.$ Given that ${\mathrm{\Delta H}}_{\mathrm{f}}\left({\mathrm{H}}_2\mathrm{O}\right)=-67.8 \mathrm{kcal} {\mathrm{mol}}^{-1}$; ${\mathrm{\Delta H}}_{\text{comb }}\left({\mathrm{C}}_2{\mathrm{H}}_2\right)=-310.1 \mathrm{kcal} {\mathrm{mol}}^{-1}, {\mathrm{\Delta H}}_{\text{comb }}\left({\mathrm{C}}_2{\mathrm{H}}_4\right)=-337.2 \mathrm{kcal} {\mathrm{mol}}^{-1}$.

Round off your answer to the nearest integer.

Assuming that $50\%$ of the heat is useful, how many $\mathrm{kgs}$ of water, at $15°\mathrm{C}$, can be heated to $95°\mathrm{C}$ by burning $200 \mathrm{litres}$ of methane at $\mathrm{NTP}? {\mathrm{\Delta H}}_{\text{combustion }}\left({\mathrm{CH}}_4\right)$ $=211 \mathrm{kcal}/\mathrm{mole},\text{ specific heat of water }=1 \mathrm{kcal}/\mathrm{kg} \mathrm{K}.$

Answer after rounding off to the nearest integer value.

Calculate the $\mathrm{C}-\mathrm{C}$ bond enthalpy from the following data:

$\left(\mathrm{a}\right) \mathrm{C}\left(\mathrm{s}\right)⟶\mathrm{C}\left(\mathrm{g}\right); \mathrm{\Delta H}=170 \mathrm{kcal}$

$\left(\mathrm{b}\right) \frac{1}{2}{\mathrm{H}}_2\left(\mathrm{g}\right)⟶\mathrm{H}\left(\mathrm{g}\right); \mathrm{\Delta H}=52 \mathrm{kcal}$

$\left(\mathrm{c}\right)$ Heat of formation of ethane $=-20 \mathrm{kcal}$

$\left(\mathrm{d}\right) \mathrm{C}-\mathrm{H}$ bond enthalpy$=99 \mathrm{kcal}$

The heat of combustion of graphite and diamond at $298 \mathrm{K}$ are $-393$ and $-395 \mathrm{kJ}/\mathrm{mole}$, respectively. The specific heat of these substances are $720$ and $505 \mathrm{J} {\mathrm{kg}}^{-1}{\mathrm{K}}^{-1}$, respectively. Calculate the heat of transformation of graphite into diamond at $273 \mathrm{K}$.

[Hint: $\mathrm{\Delta H}=(-393-(395\left)\right)+(0.720-0.505)\times \frac{12}{1000}\times (298-273)$]

Assume that the only change in volume is due to the production of hydrogen and calculate the work done in joules(in multiple of ${10}^3$) when $2.0$ moles of $\mathrm{Zn}$ dissolve in hydrochloric acid, giving ${\mathrm{H}}_2$ at $35°\mathrm{C}$ and $1 \mathrm{atm}$.

$\mathrm{Zn} \left(\mathrm{s}\right)+2\mathrm{HCl} \left(\mathrm{aq}\right)\to {\mathrm{ZnCl}}_2 \left(\mathrm{aq}\right)+{\mathrm{H}}_2 \left(\mathrm{g}\right)$

[Hint: $\mathrm{W}=-\mathrm{p\Delta V}=-{\mathrm{\Delta n}}_{{\mathrm{H}}_2}\mathrm{RT}$]