1. Tardigrade
  2. Question
  3. Chemistry
  4. Standard enthalpy of vaporization Δ H ° for water at 100° C is 40.66 kJ / mol -1 . The internal energy change of vaporization of water at 100° C (in kJ / mol -1 ) is:

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Q. Standard enthalpy of vaporization $\Delta H ^{\circ}$ for water at $100^{\circ} C$ is $40.66\, kJ / mol ^{-1} .$ The internal energy change of vaporization of water at $100^{\circ} C$ (in $kJ / mol ^{-1}$ ) is:

Thermodynamics

Solution:

$H _{2} O (\ell) \rightarrow H _{2} O ( g )$

$\Delta H =\Delta E +\Delta n _{ g } RT$

$40.66=\Delta E +1 \times \frac{8.314}{1000} \times 373$

$40.66=\Delta E +3.101$

$ \Delta E =37.56$