1. Tardigrade
  2. Question
  3. Chemistry
  4. A reversible isothermal evaporation of 90 g of water is carried out at 100° C. Heat of evaporation of water is 9.72 kcal mol -1. Assuming water vapour to behave like an ideal gas, what is the change in internal energy of the system?

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Q. A reversible isothermal evaporation of $90 \,g$ of water is carried out at $100^{\circ} C$. Heat of evaporation of water is $9.72\,kcal \,mol ^{-1}$. Assuming water vapour to behave like an ideal gas, what is the change in internal energy of the system?

Thermodynamics

Solution:

Change in enthalpy $(\Delta H)=$ Heat of evaporation $\times$ Number of moles $=9.72 \times 5$
$\Delta H=\Delta E+\Delta n R T$
$\Delta E=48.6-\left(5 \times 2 \times 10^{-3} \times 373\right)$
$=44.87 \,kcal$