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Question: The enthalpy of vaporization of water at \({{100}^{\circ }}C\) is 40.63\(kJmo{{l}^{-1}}\). The value...

The enthalpy of vaporization of water at 100∘C{{100}^{\circ }}C is 40.63kJmol−1kJmo{{l}^{-1}}. The value of ΔU\Delta U for this process would be:-
(A) 37.53kJmol−1kJmo{{l}^{-1}}
(B) 39.08kJmol−1kJmo{{l}^{-1}}
(C) 42.19kJmol−1kJmo{{l}^{-1}}
(D) 43.73kJmol−1kJmo{{l}^{-1}}

Explanation

Solution

As we know that enthalpy of vaporization is also known as heat of vaporization or heat of evaporation. It is the amount of energy or enthalpy which must be added to a liquid substance in order to transform a quantity of that substance into a gaseous form. So here we have to calculate internal energy (ΔU\Delta U) using the data of enthalpy of vaporization.
Formula used:
We will require the following formula in this solution:-
ΔHvap=ΔU+ΔngRT\Delta {{H}_{vap}}=\Delta U+\Delta {{n}_{g}}RT
where,
ΔHvap\Delta {{H}_{vap}}= enthalpy of vaporization
ΔU\Delta U= internal energy
Δng\Delta {{n}_{g}}= difference between number of gaseous moles of reactants and products

Complete answer:
Let us first discuss about enthalpy of vaporization and internal energy followed by the relation between them as follows:-
Enthalpy of vaporization: It is also known as heat of vaporization which is the amount of energy that must be added to a liquid substance in order to transform a quantity of that substance into a gaseous form.
Internal energy: It is the energy or enthalpy required to create or prepare the system in any given internal state. It is an extensive property and can’t be measured directly.
-The relation between enthalpy of vaporization and internal energy is shown below:-
ΔHvap=ΔU+ΔngRT\Delta {{H}_{vap}}=\Delta U+\Delta {{n}_{g}}RT
where,
ΔHvap\Delta {{H}_{vap}}= enthalpy of vaporization
ΔU\Delta U= internal energy
Δng\Delta {{n}_{g}}= difference between number of gaseous moles of reactants and products
R = universal gas constant
T = absolute temperature
-The vaporization of water is shown below:-
H2O(l)→H2O(g){{H}_{2}}O\left( l \right)\to {{H}_{2}}O\left( g \right)
So Δng\Delta {{n}_{g}} = number of moles on product side – number of moles on reactant side
Δng\Delta {{n}_{g}}= 1-0 = 1
The values given are:-
ΔHvap\Delta {{H}_{vap}}= 40.63kJmol−1kJmo{{l}^{-1}}= 40630 Jmol−1Jmo{{l}^{-1}}
T = 100∘C{{100}^{\circ }}C= (100+273) =373K
The value of R used will be 8.314JK−1mol−1J{{K}^{-1}}mo{{l}^{-1}}
-On substituting all the given values inΔHvap=ΔU+ΔngRT\Delta {{H}_{vap}}=\Delta U+\Delta {{n}_{g}}RT, we get:-
⇒ΔHvap=ΔU+ΔngRT Rearrange the formula:- ⇒ΔU=ΔHvap−ΔngRT ⇒ΔU=40630Jmol−1−(1×8.314JK−1mol−1×373K) ⇒ΔU=40630Jmol−1−3093Jmol−1 ⇒ΔU=37537Jmol−1 ⇒ΔU=37.53kJmol−1 \begin{aligned} & \Rightarrow \Delta {{H}_{vap}}=\Delta U+\Delta {{n}_{g}}RT \\\ & \text{Rearrange the formula:-} \\\ & \Rightarrow \Delta U=\Delta {{H}_{vap}}-\Delta {{n}_{g}}RT \\\ & \Rightarrow \Delta U=40630Jmo{{l}^{-1}}-(1\times 8.314J{{K}^{-1}}mo{{l}^{-1}}\times 373K) \\\ & \Rightarrow \Delta U=40630Jmo{{l}^{-1}}-3093Jmo{{l}^{-1}} \\\ & \Rightarrow \Delta U=37537Jmo{{l}^{-1}} \\\ & \Rightarrow \Delta U=37.53kJmo{{l}^{-1}} \\\ \end{aligned}

Hence the value of ΔU\Delta U for vaporization of water at 100∘C{{100}^{\circ }}C is (A) 37.53kJmol−1kJmo{{l}^{-1}}

Note:
Remember to substitute the values in formula accordingly with the similar units of universal gas constant and perform the conversion as well. Other gas constant values that can be used in these types of questions are as follows:-
-8.314J⋅K−1⋅mol−1J\cdot {{K}^{-1}}\cdot mo{{l}^{-1}}
-8.314m3⋅Pa⋅K−1⋅mol−1{{m}^{3}}\cdot Pa\cdot {{K}^{-1}}\cdot mo{{l}^{-1}}
-2 cal⋅K−1⋅mol−1cal\cdot {{K}^{-1}}\cdot mo{{l}^{-1}}
-8.205m3⋅atm⋅K−1⋅mol−1{{m}^{3}}\cdot atm\cdot {{K}^{-1}}\cdot mo{{l}^{-1}}