Question

Assuming that water vapour is an ideal gas, the internal energy change $\left(\Delta U\right)$ when $1$ mol of water is vaporized at $1$ bar pressure and ${100}^o$C (given: molar enthalpy of vaporization of water $41$ $kJ$ $mo{l}^{-1}$ at $1$ bar and $373$ K and $R=8.3$ J $mo{l}^{-1}$ K $mo{l}^{-1}$) will be:

• A. $4.100$ $kJ$ $mo{l}^{-1}$
• B. $3.7904$ $kJ$ $mo{l}^{-1}$
• C. $37.904$ $kJ$ $mo{l}^{-1}$
• D. $41.00$ $kJ$ $mo{l}^{-1}$

Answer 1

The correct option is C $37.904$ $kJ$ $mo{l}^{-1}$

Solution:- (C) $37.904kJ/mol$

Vapourization of water-

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

For the above reaction-

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

As we know that,

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

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

Given:-

$\Delta H=41kJ/mol$

$T=373K$

$R=8.3J/mol-K$

$\therefore \Delta U=41-1\times 8.3\times {10}^{-3}\times 373$

$\Rightarrow \Delta U=37.9041kJ/mol$

Hence the internal energy change is $37.9041kJ/mol$.