The value of $\Delta H$ for cooling $2 \ mol$ of an ideal monoatomic gas from $225^{\circ} C$ to $125^{\circ} C$ at constant pressure will be [given $C_{p} = \frac{5}{2} R$]. (in $R$)

The reversible expansion of an ideal gas under adiabatic and isothermal conditions is shown in the figure. Which of the following statement$(s)$ is (are) correct?
$(A)$ $T_1 = T_2$
$(B)$ $T_3 > T_1$
$(C)$ $W_{\text{isothermal}} > W_{\text{adiabatic}}$
$(D)$ $\Delta U_{\text{isothermal}} > \Delta U_{\text{adiabatic}}$

The standard entropies of $X_2, Y_2$ and $XY_3$ are $60, 40$ and $50 \ J \ K^{-1} \ mol^{-1}$ respectively. For the reaction $X_2 + 3Y_2 \rightleftharpoons 2XY_3; \Delta H = -60 \ kJ$ to be at equilibrium,the temperature should be.....$K$

For the reaction $X_2Y_{4(l)} \rightarrow 2XY_{2(g)}$ at $300 \ K$, the values of $\Delta U$ and $\Delta S$ are $2 \ kcal$ and $20 \ cal \ K^{-1}$ respectively. The value of $\Delta G$ for the reaction is (in $cal$)

The reaction of cyanamide,$NH_2CN_{(s)}$,with dioxygen was carried out in a bomb calorimeter,and $\Delta U$ was found to be $-742.7 \ kJ \ mol^{-1}$ at $298 \ K$. Calculate the enthalpy change for the reaction at $298 \ K$.
$NH_2CN_{(s)} + \frac{3}{2}O_{2_{(g)}} \to N_{2_{(g)}} + CO_{2_{(g)}} + H_2O_{(l)}$

$A$ liquid confined inside an adiabatic container is taken from state $1$ to state $2$ by a single-stage process as shown in the $P-V$ diagram. Then,$\Delta H$ is: