question_answer

The entropy change associated with the conversion of 1 kg of ice at 273 K to water vapours at 383 K is: (Specific heat of water liquid and water  vapours are 4.2 \[{{\text{K}}^{-1}}\]\[\text{k}{{\text{g}}^{-1}}\] and 2.0 kJ \[{{\text{K}}^{-1}}\]\[{{\text{k}}^{-1}}\] heat of liquid fusion and vapourisation of water are 334 kJ \[\text{k}{{\text{g}}^{-1}}\]and 2491 kJ \[\text{k}{{\text{g}}^{-1}}\]respectively) (log 273 = 2.436, log 373 = 2.572, log 383 = 2.583)

A) 7.90 kJ \[{{\text{K}}^{-1}}\]\[\text{k}{{\text{g}}^{-1}}\]

B) 2.64 kJ \[{{\text{K}}^{-1}}\]\[\text{k}{{\text{g}}^{-1}}\]

C) 8.49 kJ \[{{\text{K}}^{-1}}\]  \[\text{k}{{\text{g}}^{-1}}\]

D) 9.26 kJ \[{{\text{K}}^{-1}}\]\[\text{k}{{\text{g}}^{-1}}\]

Correct Answer: D

Solution :

\[\Delta S=\Delta {{S}_{1}}+\Delta {{S}_{2}}+\Delta {{S}_{3}}+\Delta {{S}_{4}}\] \[=\frac{334}{273}+4.2\ell n\frac{373}{273}+\frac{2491}{373}+2\ell n\frac{383}{373}\] \[=9.267kJ\,k{{g}^{-\,1}}{{K}^{-\,1}}.\]