Assuming the water vapour to be a perfect gas, calculate the internal energy when 1 mol of water at 100°C and 1 pressure is converted to ice at O°C. Given the enthalpy of fusion of ice is 6.00 kJ mol and heat capacity of water is 4.2J/g°C
Answers
Answer:
Answer:
∆U=-13.56KJmol^-1
Given:
- Number of moles,n=1
- Heat capacity of water=4.2J/g°C
- Enthalpy of fusion of ice=6KJ
Now,
∆H1 ∆H2
H2O(l)→H2O(l)→H2O(s)
100°c 0°c 0°C
To find:
- The change in internal energy, ∆U=?
Solution:
As we know that,
∆H=∆H1+∆H2
∆H1=(-18×4.2×100)Jmol^-1
∆H1=-75.6×100Jmol^-1
∆H1=-7560Jmol^-1
∆H1=-7.56KJmol^-1
∆H2=-6KJmol^-1
Total enthalpy change will be,
∆H=∆H1+∆H2
∆H=-7.56+(-6)KJmol^-1
∆H=-13.56KJmol^-1
So,
Change in enthalpy=-13.56KJmol^-1
Since, Change in volume is negligible during change from the liquid to solid state.
∆V=0
So, P∆V=∆ngRT=0
From this,
∆H=∆U+∆ngRT
∆H=∆U+0
∆U=∆H=-13.56KJmol^-1
So,Change in internal energy=-13.56KJmol^-1
Answer:
∆U=-13.56KJmol^-1
Given:
Number of moles,n=1
Heat capacity of water=4.2J/g°C
Enthalpy of fusion of ice=6KJ
Now,
∆H1 ∆H2
H2O(l)→H2O(l)→H2O(s)
100°c 0°c 0°C
To find:
The change in internal energy, ∆U=?
Solution:
As we know that,
∆H=∆H1+∆H2
∆H1=(-18×4.2×100)Jmol^-1
∆H1=-75.6×100Jmol^-1
∆H1=-7560Jmol^-1
∆H1=-7.56KJmol^-1
∆H2=-6KJmol^-1
Total enthalpy change will be,
∆H=∆H1+∆H2
∆H=-7.56+(-6)KJmol^-1
∆H=-13.56KJmol^-1
So,
Change in enthalpy=-13.56KJmol^-1
Since, Change in volume is negligible during change from the liquid to solid state.
∆V=0
So, P∆V=∆ngRT=0
From this,
∆H=∆U+∆ngRT
∆H=∆U+0
∆U=∆H=-13.56KJmol^-1
So,Change in internal energy=-13.56KJmol^-1