Question

The enthalpy of vaporisation of water at ${100}^{o}C$ is $40.63KJmo{l}^{-1}$. What is the value of $\Delta U$ for this process?

Answer 1

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

For the above reaction-

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

$\Delta H=40.63kJ/mol$

$T=100℃=(100+273)K=373K$

$R=8.314\times {10}^{-3}kJ/mol-K$

As we know that,

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

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

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

$\Rightarrow \Delta U=40.63-3.1=37.53kJ/mol$

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