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The enthalpy of vaporization of water at 100 degree Celsius is 40.63KJ/mol. The value of ∆E for this process would be A) 37.53 kJ/mol B) 39 kJ/mol C) 39 kJ/mol D) 43.73 kJ/mol?
Most Upvoted Answer
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
∆n =1
∆E=∆H --n×R×T
=40630 -- 1×2×373
=40630 --3101.1
=37528.9J/mol
=37.528kJ/mol
Community Answer
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
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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The enthalpy of vaporization of water at 100 degree Celsius is 40.63KJ/mol. The value of ∆E for this process would be A) 37.53 kJ/mol B) 39 kJ/mol C) 39 kJ/mol D) 43.73 kJ/mol?
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The enthalpy of vaporization of water at 100 degree Celsius is 40.63KJ/mol. The value of ∆E for this process would be A) 37.53 kJ/mol B) 39 kJ/mol C) 39 kJ/mol D) 43.73 kJ/mol? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about The enthalpy of vaporization of water at 100 degree Celsius is 40.63KJ/mol. The value of ∆E for this process would be A) 37.53 kJ/mol B) 39 kJ/mol C) 39 kJ/mol D) 43.73 kJ/mol? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The enthalpy of vaporization of water at 100 degree Celsius is 40.63KJ/mol. The value of ∆E for this process would be A) 37.53 kJ/mol B) 39 kJ/mol C) 39 kJ/mol D) 43.73 kJ/mol?.
The enthalpy of vaporization of water at 100 degree Celsius is 40.63KJ/mol. The value of ∆E for this process would be A) 37.53 kJ/mol B) 39 kJ/mol C) 39 kJ/mol D) 43.73 kJ/mol? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about The enthalpy of vaporization of water at 100 degree Celsius is 40.63KJ/mol. The value of ∆E for this process would be A) 37.53 kJ/mol B) 39 kJ/mol C) 39 kJ/mol D) 43.73 kJ/mol? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The enthalpy of vaporization of water at 100 degree Celsius is 40.63KJ/mol. The value of ∆E for this process would be A) 37.53 kJ/mol B) 39 kJ/mol C) 39 kJ/mol D) 43.73 kJ/mol?.
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