The enthalpy of vaporization of water at $298 \ K$ is $386 \ kJ \ mol^{-1}$. What is the entropy of vaporization of water in $kJ \ K^{-1} \ mol^{-1}$?

$A$ reversible adiabatic process is one in which:

Melting point of a solid is $x \ K$ and its latent heat of fusion is $600 \ cal \ mol^{-1}$. The entropy change for fusion of $1 \ mol$ solid is $2 \ cal \ mol^{-1} \ K^{-1}$. The value of $x$ will be......$K$

The entropy change,in the conversion of one mole of liquid water at $373 \, K$ to vapour at the same temperature is (Latent heat of vaporization of water,$\Delta H_{vap} = 2.257 \, kJ/g$) ........ $J \, K^{-1}$ (in $.9$)

Statement-$1$: For a process at equilibrium,the total entropy is maximum and the change in total entropy is zero.
Statement-$2$: For a reversible process,if the same amount of heat is supplied to a system at two different temperatures,then the entropy change is less at a lower temperature.

An irreversible process occurring isothermally in an isolated system leads to