For the process $H_2O_{(l)} (1 \ bar, 373 \ K) \rightarrow H_2O_{(g)} (1 \ bar, 373 \ K)$,the correct set of thermodynamic parameters is:

Calculate the work done in $kJ$ when $3$ moles of an ideal gas at $27^{\circ} C$ expand isothermally and reversibly from $10 \ atm$ to $1 \ atm$ $[R=8.314 \ J \ K^{-1} \ mol^{-1}]$

Observe the following reactions:
$AB_{(g)} + 25 H_2O_{(l)} \rightarrow AB_{(25 H_2O)} ; \Delta H = x \ kJ \ mol^{-1}$
$AB_{(g)} + 50 H_2O_{(l)} \rightarrow AB_{(50 H_2O)} ; \Delta H = y \ kJ \ mol^{-1}$
The enthalpy of dilution $(\Delta H_{dil})$ in $kJ \ mol^{-1}$ is

$A$ solution of $500 \ mL$ of $2 \ M \ KOH$ is added to $500 \ mL$ of $2 \ M \ HCl$ and the mixture is well shaken. The rise in temperature $T_1$ is noted. The experiment is then repeated using $250 \ mL$ of each solution and the rise in temperature $T_2$ is again noted. Assume all heat is absorbed by the solution.

Given below are two statements:
Statement-$I$: For isothermal irreversible change of an ideal gas,$q = -w = P_{\text{ext}} (V_{\text{final}} - V_{\text{initial}})$
Statement-$II$: For adiabatic change,$\Delta U = W_{\text{adiabatic}}$
The correct answer is

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 $R$)