1
Question
Assuming the water vapour to be a perfect gas, calculate the internal energy change when 1 mol of water at 100∘C and 1 bar pressure is converted to ice at 0∘C. Given the enthalpy of fusion of ice is 6.00 kJmol−1and the heat capacity of the water is4.2 J/g∘C.
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Solution
Step 1: Individual enthalpy change
The change takes place as follows:
Part 1:1 molH2O(l,100∘)⟶1 mol(l,0∘) enthalpy change △H1
Part 2:1 mol H2O(l,0∘)⟶1 mol(s,0∘C) enthalpy change △H2
Step 2: Total enthalpy change
The total enthalpy change will be:
△H=△H1+△H2
△H1=nCp1△T
Given Cp1=4.2 J/g∘C)
Heat capacity of water =4.2 J/g∘C×18 gmol−1
{mole=Massmol.wt}
=18×4.2 J/mol ∘C
△H1=−(18×4.2 J/mol ∘C)×100 ∘C
=−7560 Jmol−1=−7.56 kJmol−1
Given enthalpy of fusion of ice (△H2)=−6.00 kJmol−1
Here, negative sign represents that energy is released during the conversion.
Therefore, △H=(−7.56 kJmol−1)+(−6.00 kJ mol−1
△H=−13.56 kJ mol−1
Step 3: Internal energy change
There is negligible change in the volume during the change from liquid to solid state.
Therefore, P△V=△ngRT=0
△H=△U+△ngRT or △H=△U+P△V
So, △U=△H=−13.56 kJ mol−1
Final Answer: The internal energy change is −13.56 kJmol−1.
The change takes place as follows:
Part 1:1 molH2O(l,100∘)⟶1 mol(l,0∘) enthalpy change △H1
Part 2:1 mol H2O(l,0∘)⟶1 mol(s,0∘C) enthalpy change △H2
Step 2: Total enthalpy change
The total enthalpy change will be:
△H=△H1+△H2
△H1=nCp1△T
Given Cp1=4.2 J/g∘C)
Heat capacity of water =4.2 J/g∘C×18 gmol−1
{mole=Massmol.wt}
=18×4.2 J/mol ∘C
△H1=−(18×4.2 J/mol ∘C)×100 ∘C
=−7560 Jmol−1=−7.56 kJmol−1
Given enthalpy of fusion of ice (△H2)=−6.00 kJmol−1
Here, negative sign represents that energy is released during the conversion.
Therefore, △H=(−7.56 kJmol−1)+(−6.00 kJ mol−1
△H=−13.56 kJ mol−1
Step 3: Internal energy change
There is negligible change in the volume during the change from liquid to solid state.
Therefore, P△V=△ngRT=0
△H=△U+△ngRT or △H=△U+P△V
So, △U=△H=−13.56 kJ mol−1
Final Answer: The internal energy change is −13.56 kJmol−1.
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