For a hypothetical reaction $A_{2(g)} + B_{2(g)} \rightleftharpoons 2AB_{(g)}$,at $200 \ K$,$\Delta_r G$ and $\Delta_r S$ are $20 \ kJ \ mol^{-1}$ and $-20 \ J \ K^{-1} \ mol^{-1}$ respectively. If $\Delta_r C_p = 20 \ J \ K^{-1} \ mol^{-1}$,find $\Delta_r H$ at $400 \ K$ in $kJ \ mol^{-1}$.

Consider the following statements:
Statement-$I$: During isothermal expansion of an ideal gas,its enthalpy decreases.
Statement-$II$: When $2.0 \ L$ of an ideal gas expands isothermally into vacuum,$\Delta U = 0$.

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 $R = 8.314 \ J \ mol^{-1} K^{-1}$]

$A$ sample of gas is compressed from an initial volume of $2V_0$ to $V_0$ using three different processes.
First: Using reversible isothermal.
Second: Using reversible adiabatic.
Third: Using irreversible adiabatic under a constant external pressure.
Then

The molar enthalpy change for $H_2O_{(l)} \rightleftharpoons H_2O_{(g)}$ at $373 \ K$ and $1 \ atm$ is $41 \ kJ \ mol^{-1}$. Assuming ideal behavior,the internal energy change for vaporization of $1 \ mol$ of water at $373 \ K$ and $1 \ atm$ in $kJ \ mol^{-1}$ is:

$1 \ g$ 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 $1 \ atm$?