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A swimmer coming out from a pool is covered with a film of water weighing about $18$g. How much heat must be supplied to evaporate this water At $298K$ ? Calculate the internal energy of vaporization at $100^\circ C$ .
$\Delta {{\rm H}_{vap}}$ for water at $373K = $$40.66$$kJ/mo{l^{ - 1}}$

Last updated date: 22nd Feb 2024
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Hint:The enthalpy of vaporization may be written below:
$\Delta {H_{vap}} = \Delta {U_{vap}} + p\Delta V$
Where $\Delta {U_{vap}}$ is the internal energy of vaporisation, $p$ is pressure and $\Delta V$ is the change of volume.
It is the increased internal energy of the vapour phase compared to the liquid phase, plus the work done against the ambient pressure. The rise within the internal energy is seen because the energy required to beat the intermolecular interactions within the liquid (or solid), within the case of the process of sublimation.

Complete step by step answer:
${H_2}O(l) \to {H_2}O(g)$ at $298K$
To calculate the moles of water to be converted into the gaseous state
Moles=given mass/molecular mass
We know that given mass of water$ = $$18$g
The molecular mass of ${H_2}O = 18$g
So the no. of moles of water$ = \dfrac{{18}}{{18}} = 1$mol
Heat is supplied to evaporate $18$g of water
= mol \times \Delta {{\rm H}_{vap}} \\
Given, $\Delta {{\rm H}_{vap}}$ for water at $373K = $$40.66$$kJ/mo{l^{ - 1}}$
Substituting the values in the above equation, we can get the value of heat supplied to a given mass of water and we get:
= 1 \times 40.66 \\
= 40.66kJ \\
Now we can calculate the value of internal energy of vaporization,$\Delta {U_{vap}}$ at $373K$
Using steam as behaving like an ideal gas
$\Delta U = \Delta {\rm H} - p\Delta V$
$\Delta U = \Delta {\rm H} - \Delta {n_{(g)}}RT$
$\Delta {n_g}$ = change in gaseous moles of the products - change in gaseous moles of the reactants
$ = $ $1 - 0 = 1$
We know, the value of $R = 8.314 \times {10^{ - 3}}J/molK$
 $T = 373K$
$ = 40.66kJ - $$\left[ {(1mol)(8.314 \times {{10}^{ - 3}}kJ/molK) \times 373K} \right]$
$\Delta {U_{Vap}} = 37.56kJ$

Thus correct answer is option D.

-The enthalpy of vaporization (symbol $\Delta {{\rm H}_{vap}}$ ), also called the (latent) heat of vaporization or heat of evaporation, can be defined as the amount of energy (enthalpy) that has to be added to a liquid substance to convert a quantity of that substance into a gas.
-The enthalpy of vaporization is actually a function of the pressure at which that transformation takes place.
-The gas law, also mentioned because the general gas equation is the equation of the state of a hypothetical perfect gas.
-In its empirical form, it is written as $PV = nRT$
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