Question

A swimmer coming out from a pool is covered with a film of water weighing about 18 g. Calculate the internal energy of vaporisation at ${100}^{0}C$.

[ ∆ vap H for water at 373 K = 40.66 kJ mol -1 ]

• A. $37.55kJmo{l}^{-1}$
• B. $3.67kJmo{l}^{-1}$
  
• C. $30.67kJmo{l}^{-1}$
• D. $3.567kJmo{l}^{-1}$
  35.67kJmol−

Answer 1

Answer: (A)

$37.55kJmo{l}^{-1}$

$H_{2}O\left(l\right)\stackrel{\text{vaporization}}{\to }{H}_{2}O\left(g\right)$

$m=18g$ $m=18g$

Number of moles $=\frac{18g}{18gmo{l}^{-1}}=1mol$

${\Delta }_{vap}U={\Delta }_{vap}{H}^{⊝}-P\Delta V$

$={\Delta }_{vap}{H}^{⊝}-\Delta ngRT$

$=40.66-(1\times 8.314\times {10}^{-3}\times 373)$

$=40.66-3.10$

${\Delta }_{vap}U=37.56kJmo{l}^{-1}$

Hence, the correct option is $\text{A}$