Chemistry - Thermodynamics - II Question with Solution | TestHub

${\mathrm{H}}_2\mathrm{O}\left(\mathrm{g},100^{\circ }\mathrm{C}\right)⟶{\mathrm{H}}_2\mathrm{O}\left(\ell ,100^{\circ }\mathrm{C}\right)\to {\mathrm{H}}_2\mathrm{O}\left(\ell ,0^{\circ }\mathrm{C}\right)$ (A) (B)

(D) $\Delta {\mathrm{S}}_{\mathrm{AB}}=\frac{-\mathrm{m}\times \Delta {\mathrm{H}}_{\text{vap }}}{{\mathrm{T}}_{\mathrm{BP}}}=\frac{-18\times 540}{373}=-26.06\mathrm{Cal}.{}^{\circ }\mathrm{C}$ $\Delta {\mathrm{S}}_{\mathrm{BC}}=\mathrm{ms}\ell \frac{{\mathrm{T}}_2}{{\mathrm{T}}_1}=18\times 1\times \ell n\left(\frac{273}{373}\right)=-5.62\mathrm{Cal}/{}^{\circ }\mathrm{C}$ $\Delta {\mathrm{S}}_{\mathrm{CD}}=\frac{-\mathrm{m}\times \Delta {\mathrm{H}}_{\text{fusion }}}{{\mathrm{T}}_{\mathrm{mp}}}=\frac{-18\times 80}{273}=-5.27\mathrm{Cal}/{}^{\circ }\mathrm{C}$ $\Delta {\mathrm{S}}_{\mathrm{A}\to \mathrm{D}}=\Delta {\mathrm{S}}_{\mathrm{AB}}+\Delta {\mathrm{S}}_{\mathrm{BC}}+\Delta {\mathrm{S}}_{\mathrm{CD}}=-36.95\mathrm{Cal}/{}^{\circ }\mathrm{C}$

Correct options are ABCD .