The entropy change associated with the conversion

Question

The entropy change associated with the conversion of 1kg of ice at 273K to water vapours at 383K is:(Specific heat of water liquid and water vapours are 4.2 kJ K-1 kg-1 and 2.0 kJ K-1 kg-1, heat of liquid fusion and vapourisation of water are 334 kJ kg-1 and 2491 kJ kg-1 respectively) (log 273 = 2.436, log 373 = 2.572, log 383 = 2.583)	7.90 kJ K-1 kg-1	2.64 kJ K-1 kg-1	8.49 kJ K-1 kg-1	9.26 kJ K-1 kg-1pl explain also

Arun · Accepted Answer

Entropy for phase transition at constant pressure - \Delta S= \frac{\Delta H_{Transition}}{T} - wherein Transition \RightarrowFusion, Vaporisition, Sublimation \Delta H\RightarrowEnthalpy \Delta E\RightarrowInternal Energy T\RightarrowTransitional temperature Entropy for solid and liquid - \Delta S= nC\, \ln \frac{T_{f}}{T_{i}} or \Delta S= ms\, \ln \frac{T_{f}}{T_{i}} - wherein n= no. of moles C= molar heat capacity m= mass s= specific heat capacity Phase change path H_{2}O(S)\underset{\Delta S_{1}}{\rightarrow}H_{2}O(l)\overset{\Delta S_{2}}{\rightarrow}H_{2}O(l)\overset{\Delta S_{3}}{\rightarrow}H_{2}O(g)\overset{\Delta S_{4}}{\rightarrow}H_{2}O(g) 273 K 273 K 373 K 373 K 383 K \Delta S_{1}=\frac{\Delta H_{fusion}}{273}=\frac{334}{273}=1.22 \Delta S_{2}=4.2 ln (\frac{373}{273})=1.31 \Delta S_{3}=\frac{\Delta H_{vap}}{373}=\frac{2491}{373}=6.67 \Delta S_{4}=2.0 ln (\frac{383}{373})=0.05 \Delta S \: Total = \Delta S_{1}+\Delta S_{2}+\Delta S_{3}+\Delta S_{4}= 9.26 KJ \: Kg^{-1}K^{-1} Hence option D is correct

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AD , 5 Years ago

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