Solveeit Logo
Login

Question

Question: The enthalpy of vaporization of water at \[{{100}^{\circ }}C\] is \(40.63\,KJ\text{ }mo{{l}^{-1}}\) ...

The enthalpy of vaporization of water at 100∘C{{100}^{\circ }}C is 40.63 KJ mol−140.63\,KJ\text{ }mo{{l}^{-1}} The value ΔU\Delta U for this process:
(a)- 37.35 KJ mol−137.35\text{ }KJ\text{ }mo{{l}^{-1}}
(b)- 39.08KJ mol−139.08KJ\text{ }mo{{l}^{-1}}
(c)- 42.19KJ mol−142.19KJ\text{ }mo{{l}^{-1}}
(d)- 43.73KJ mol−143.73KJ\text{ }mo{{l}^{-1}}

Explanation

Solution

The internal energy of a system is calculated by the formula ΔH=ΔU+ΔngRT\Delta H=\Delta U+\Delta {{n}_{g}}RT where ΔH\Delta H is the change in enthalpy of the system, ΔU\Delta U is the change in internal energy of the system, R is the gas constant, T is the temperature of the system, and Δng\Delta {{n}_{g}} is a difference of the number of moles of product and reactants.

Complete step by step answer:
Internal energy change is the heat absorbed or evolved at constant volume. It is denoted by ΔU\Delta U.
Enthalpy change of a system is equal to the heat absorbed or evolved by the system at constant pressure. It is denoted by ΔH\Delta H.

The internal energy change and change in enthalpy can be related as ΔH=ΔU+ΔngRT\Delta H=\Delta U+\Delta {{n}_{g}}RT
Where, ΔH\Delta H is the change in enthalpy of the system and the question it is given 40.63 KJ mol−140.63\, KJ\text{ }mo{{l}^{-1}}
R is the gas constant and its value is taken 8.314 x 10−3 JK−1mol−18.314\text{ x 1}{{\text{0}}^{-3}}\text{ J}{{\text{K}}^{-1}}mo{{l}^{-1}}
T is the temperature of the system.
It is always taken in Kelvin (K).

The temperature of the water is given 100∘C{{100}^{\circ }}C. It has to be converted into Kelvin.
T=100+273=373KT=100+273=373K
The temperature is 373 K.

Δng\Delta {{n}_{g}} is the difference between the number of moles of products and reactants.
So, the reaction is:
H2O(l)→H2O(g){{H}_{2}}O(l)\to {{H}_{2}}O(g)
Δng=np−nr\Delta {{n}_{g}}={{n}_{p}}-{{n}_{r}}

So, the number of moles in the product is 1, and the number of moles in the reactant is 0 because the moles of liquid and solid are taken 0.
Δng=1−0=1\Delta {{n}_{g}}=1-0=1
So, the internal energy change can be calculated.
ΔH=ΔU+ΔngRT\Delta H=\Delta U+\Delta {{n}_{g}}RT
ΔU=ΔH−ΔngRT\Delta U=\Delta H-\Delta {{n}_{g}}RT
ΔU=40.63−(1 x 8.314 x 10−3 x 373)\Delta U=40.63-(1\ \text{x 8}\text{.314 x 1}{{\text{0}}^{-3}}\text{ x 373)}
ΔU=37.53KJ/mol\Delta U=37.53 KJ/mol
So, the correct answer is “Option A”.

Note: The number of moles must be taken correctly. Only the moles of gaseous elements or compounds are taken for the calculation. The temperature must be converted to Kelvin if present in other forms.