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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» (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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