Question

The enthalpy of vaporization of water at ${100}^{\circ }C$ is $40.63KJmo{l}^{-1}$. The value $\Delta$U for this process would be:

• A. $37.53KJmo{l}^{-1}$
• B. $39.08KJmo{l}^{-1}$
• C. $42.19KJmo{l}^{-1}$
• D. $43.73KJmo{l}^{-1}$

Answer 1

The correct option is A $37.53KJmo{l}^{-1}$

Vapourisation of water-

${H_{2}O}_{\left(l\right)}⟶{H_{2}O}_{\left(g\right)}$

$\Delta n_g=n_P-n_R=1-0=1$

As we know that,

$\Delta H=\Delta U+\Delta n_{g}RT$

$\Rightarrow \Delta U=\Delta H-\Delta n_{g}RT$

Given:-

$\Delta H=40.63kJ/mol$

$T=100℃=(100+273)K=373K$

$\therefore \Delta U=40.63-(1\times 8.314\times {10}^{-3}\times 373)$

$\Rightarrow \Delta U=40.63-3.101=37.53kJ/mol$

Hence the value of $\Delta U$ is $37.53kJ/mol$.

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