The enthalpy of vaporization of water at 100 degree Celsius is 40.63KJ...
H2O (liquid)-------->H2O(gas)
∆n =1
∆E=∆H --n×R×T
=40630 -- 1×2×373
=40630 --3101.1
=37528.9J/mol
=37.528kJ/mol
The enthalpy of vaporization of water at 100 degree Celsius is 40.63KJ...
Enthalpy of Vaporization
The enthalpy of vaporization, denoted as ΔHvap, is the amount of heat required to convert one mole of a substance from the liquid phase to the gaseous phase at a given temperature and pressure. In this case, we are given that the enthalpy of vaporization of water at 100 degrees Celsius is 40.63 kJ/mol.
Definition of ∆E
The symbol ΔE represents the change in internal energy of a system. It is defined as the final internal energy minus the initial internal energy. In the context of this problem, we need to determine the value of ΔE for the process of vaporizing water at 100 degrees Celsius.
Relationship between ΔHvap and ΔE
The relationship between ΔHvap and ΔE can be expressed using the equation:
ΔHvap = ΔE + PΔV
where PΔV represents the work done on or by the system. In the case of vaporization, the work done is typically negligible, as the volume change is small. Therefore, we can assume that PΔV is close to zero, and the equation simplifies to:
ΔHvap ≈ ΔE
Calculation
Since ΔHvap is approximately equal to ΔE, we can conclude that the value of ΔE for the vaporization of water at 100 degrees Celsius is also 40.63 kJ/mol.
Therefore, the correct answer is:
C) 40.63 kJ/mol
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