$A$ gas expands from $3\, dm^3$ to $5.8\, dm^3$ against a constant external pressure of $3\, bar$. The work done during expansion is used to heat $2\, moles$ of water from $290\, K$ to a final temperature of $T\, K$. If the specific heat of water is $4.2\, J\, g^{-1}\, K^{-1}$,then $T = ......\, K$. (in $.6$)

Which of the following is/are true about the reversible isothermal expansion of an ideal gas?
$(a) \ \Delta U = 0$
$(b) \ q = 0$
$(c) \ \Delta T = 0$
$(d) \ q = 2.303 \ nRT \ \log_{10} \left( \frac{V_2}{V_1} \right)$

$0.3 \ g$ of ethane undergoes combustion at $27^{\circ} C$ in a bomb calorimeter. The temperature of the calorimeter system (including the water) is found to rise by $0.5^{\circ} C$. The heat evolved during combustion of ethane at constant pressure is $....... kJ \ mol^{-1}$. (Nearest integer) [Given: The heat capacity of the calorimeter system is $20 \ kJ \ K^{-1}$,$R = 8.3 \ J \ K^{-1} \ mol^{-1}$. Assume ideal gas behaviour. Atomic mass of $C$ and $H$ are $12$ and $1 \ g \ mol^{-1}$ respectively]

$C_{(graphite)} + O_{2(g)} \to CO_{2(g)}; \Delta H = - 94.05 \ k \ cal \ mol^{-1}$
$C_{(diamond)} + O_{2(g)} \to CO_{2(g)}; \Delta H = - 94.50 \ k \ cal \ mol^{-1}$
Therefore:

$1 \ g$ of graphite is burnt in a bomb calorimeter in excess of oxygen at $298 \ K$ and $1 \ atm$ atmospheric pressure according to the equation:
$C \ (graphite) + O_{2(g)} \rightarrow CO_{2(g)}$
During the reaction,the temperature rises from $298 \ K$ to $299 \ K$. If the heat capacity of the bomb calorimeter is $20.7 \ kJ \ K^{-1}$,what is the enthalpy change for the above reaction at $298 \ K$ and $1 \ atm$?

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 $: 1$)