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In the process : `H_(2)O(s, -10^(@)C, 1atm)rarrH_(2)O(1, 10^(@)C, 1 atm)` `C_(p)` for ice = 9 cal `deg^(-1) mol^(-1), C_(p)` for `H_(2)O=18` cal `deg^(-1)mol^(-1)`. Latent heat of fusion of ice = 1440 cal `mol^(-1)` at `0^(@)C`. The entropy change for the above process is 6.258 cal. `deg^(-1)mol^(-1)` Give the total number of steps in which the third law of thermodynamics is used |
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Answer» Correct Answer - B Step 1. (using the third law of thermodynamics) : (For changing) `H_(2)O(s)(-10^(@)C, 1 atm)rarrH_(2)O(s, 0^(@)C1 atm)` `DeltaS_(1)=underset(-10)overset(0)intn(C_(p))/(T)dT=1xx9xx2.3xxlog.(273)/(263)=0.336" cal deg"^(-1) mol^(-1)` Step 2 (using the second law of thermodynamics) : `H_(2)O(s)(0^(@)C, 1 atm)rarrH_(2)O(l)(0^(@)C, 1 atm)` `DeltaS_(2)=(q_(rev))/(T)=(1440)/(273)=5.258" cal deg"^(-1)mol^(-1)` Step 3 (using the third law of thermodynamics) : `H_(2)O(l)(0^(@)C, 1 atm)rarrH_(2)O(l)(10^(@)C, 1 atm)` `DeltaS_(3)=underset(0)overset(10)intn(C_(p))/(T)dT=1xx18xx2.3xxlog.(283)/(273)=0.647 " cal deg"^(-1)mol^(-1)` `DeltaS=DeltaS_(1)+DeltaS_(2)+DeltaS_(3)=0.336+5.258+0.647 = 6.258 " cal deg"^(-1) mol^(-1)`. |
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