Answer:
Information shadow:
Amount of water to
evaporate = 18 g
= 1 mole water
\[{{\Delta
}_{vap}}{{H}^{{}^\circ }}=40.66kJmo{{l}^{-1}}\]
Problem solving
strategy:
Required
heat for evaporation \[(q)=n{{\Delta }_{vap}}H\]
Internal
energy of vaporisation can be calculated as,
\[{{\Delta }_{vap}}H={{\Delta }_{vap}}U+\Delta {{n}_{g}}RT\]
Working
it out:
\[q=n{{\Delta
}_{vap}}H\]
\[=1\times
40.66kJ\]
\[=40.66kJ\]
\[{{\Delta
}_{vap}}U={{\Delta }_{vap}}H-\Delta {{n}_{g}}RT\]
\[{{H}_{2}}O(l)\to
{{H}_{2}}O(g)\]
\[\Delta
{{n}_{g}}=1-0=1\]
\[{{\Delta
}_{vap}}U=40.66-1\times 8.314\times {{10}^{-3}}\times 373\]
\[=37.56\,\,kJ\,mo{{l}^{-1}}\]
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