Question

Assuming that, water vapour is an ideal gas the internal energy change (ΔU)(ΔU) when 11 mol of water is vaporized at 11 bar pressure and 100o100o C will be:

[given: molar enthalpy of vaporisation of water at $1$ bar and $373$ K$=41$ kJ $mo{l}^{-1}$ and $R=8.3$ J $K^{-1}$ $mo{l}^{-1}$]

• A. $41.00$ kJ $mo{l}^{-1}$
• B. $4.100$ kJ $mo{l}^{-1}$
• C. $3,7904$ J $mo{l}^{-1}$
• D. None of these

Answer 1

The correct option is C $3,7904$ J $mo{l}^{-1}$

Solution:- (C) $37904J/mol$

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:-

$\Delta H=41kJ/mol=41000J/mol$

$T=373K$

$R=8.3J/mol-K$

$\therefore \Delta U=41000-1\times 8.3\times 373$

$\Rightarrow \Delta U=41000-3095.1=37904KJ/mol$

Hence the internal energy change is $37904kJ/mol$.