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Grade: 12th pass

                        

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 -1 pl explain also

5 months ago

Answers : (1)

Arun
24742 Points
							
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
5 months ago
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