Question
Standard enthalpy of vaporisation $${\Delta _{vap}}{H^ \circ }$$ for water at $${100^ \circ }C$$ is $$40.66\,kJ\,mo{l^{ - 1}}.$$ The internal energy of vaporisation of water at $${100^ \circ }C\left( {{\text{in}}\,kJ\,mo{l^{ - 1}}} \right)$$ is ( assume water vapour to behave like an ideal gas ).
A.
+ 37.56
B.
- 43.76
C.
+ 43.76
D.
+ 40.66
Answer :
+ 37.56
Solution :
\[\begin{align}
& {{H}_{2}}O\left( l \right)\xrightarrow{{{100}^{\circ }}C}{{H}_{2}}O\left( g \right) \\
& {{\Delta }_{vap}}{{H}^{\circ }}={{\Delta }_{vap}}{{E}^{\circ }}+\Delta {{n}_{g}}RT \\
& {{\Delta }_{vap}}{{H}^{\circ }}=\text{enthalpy of vaporisation} \\
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=40.66\,kJ\,mo{{l}^{-1}} \\
& \text{For the above reaction,} \\
& \Delta {{n}_{g}}={{n}_{p}}-{{n}_{r}} \\
& \,\,\,\,\,\,\,\,\,\,=1-0 \\
& \,\,\,\,\,\,\,\,\,\,=1 \\
& R=8.314 \\
& T={{100}^{\circ }}C \\
& \,\,\,\,=273+100 \\
& \,\,\,\,=373K \\
\end{align}\]
$$\therefore 40.66\,kJ\,mo{l^{ - 1}} = $$ $${\Delta _{vap}}{E^ \circ } + 1 \times 8.314 \times {10^{ - 3}} \times 373$$
$$\eqalign{
& {\Delta _{vap}}{E^ \circ } = 40.66\,kJ\,mo{l^{ - 1}} - 3.1\,kJ\,mo{l^{ - 1}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = + 37.56\,kJ\,mo{l^{ - 1}} \cr} $$