One mole of an ideal gas at $900 \ K$ undergoes two reversible processes,$I$ followed by $II$,as shown in the graph. If the work done by the gas in the two processes is the same,the value of $\ln \frac{V_3}{V_2}$ is. . . . . . . . ($U$: internal energy,$S$: entropy,$p$: pressure,$V$: volume,$R$: gas constant). (Given: molar heat capacity at constant volume,$C_{V, m}$ of the gas is $\frac{5}{2} R$)

The enthalpy of combustion of methane at $25\,^{\circ}C$ is $890\,kJ$. The heat liberated when $3.2\,g$ of methane is burnt in air is.....$kJ$

Match the List-$I$ with List-$II$

List-$I$ Thermodynamic Process	List-$II$ Magnitude in $kJ$
$A$. Work done in reversible,isothermal expansion of $2 \ mol$ of ideal gas from $2 \ dm^3$ to $20 \ dm^3$ at $300 \ K$.	$I$. $4$
$B$. Work done in irreversible isothermal expansion of $1 \ mol$ ideal gas from $1 \ m^3$ to $3 \ m^3$ at $300 \ K$ against a constant pressure of $3 \ kPa$.	$II$. $11.5$
$C$. Change in internal energy for adiabatic expansion of a $1 \ mol$ ideal gas with change of temperature $= 320 \ K$ and $\overline{C}_V = \frac{3}{2} R$.	$III$. $6$
$D$. Change in enthalpy at constant pressure of $1 \ mole$ ideal gas with change of temperature $= 337 \ K$ and $\overline{C}_P = \frac{5}{2} R$.	$IV$. $7$

Choose the correct answer from the option given below:

The heat liberated when $1.89 \ g$ of benzoic acid is burnt in a bomb calorimeter at $25^{\circ} C$ increases the temperature of $18.94 \ kg$ of water by $0.632^{\circ} C$. If the specific heat of water at $25^{\circ} C$ is $0.998 \ cal / (g^{\circ} C)$,then find the heat of combustion of benzoic acid.

The reaction of cyanamide,$NH_{2}CN_{(s)}$ with oxygen was run in a bomb calorimeter and $\Delta U$ was found to be $-742.24 \ kJ \ mol^{-1}$. The magnitude of $\Delta H_{298}$ for the reaction
$NH_{2}CN_{(s)} + \frac{3}{2} O_{2(g)} \rightarrow N_{2(g)} + CO_{2(g)} + H_{2}O_{(l)}$
is $............ \ kJ$. (Rounded off to the nearest integer)
[Assume ideal gases and $R = 8.314 \ J \ mol^{-1} K^{-1}$]

Given
$C_{(graphite)} + O_{2(g)} \rightarrow CO_{2(g)};$
$\Delta_rH^o = -393.5 \, kJ \, mol^{-1}$
$H_{2(g)} + \frac{1}{2} O_{2(g)} \rightarrow H_2O_{(l)};$
$\Delta_rH^o = -285.8 \, kJ \, mol^{-1}$
$CO_{2(g)} + 2H_2O_{(l)} \rightarrow CH_{4(g)} + 2O_{2(g)};$
$\Delta_rH^o = + 890.3 \, kJ \, mol^{-1}$
Based on the above thermochemical equations,the value of $\Delta_rH^o$ at $298 \, K$ for the reaction
$C_{(graphite)} + 2H_{2(g)} \rightarrow CH_{4(g)}$ will be ........... $kJ \, mol^{-1}$.