Question

The internal energy change (in J) when $90g$ of water undergoes complete evaporation at $100ºC$ is

( Given :$∆H_{vap}$for water at$373k=41kJ/mol,R=8.314J{K}^{-1}mo{l}^{-1})$,

Answer 1

Step 1: Given data:

Enthalpy of vaporization at $373K$ is $∆H_{vap}$= $41\mathrm{kJ}/\mathrm{mol}$

Universal gas constant value $R=8.314\mathrm{J}/\mathrm{Kmol}$

Step 2: Formula used:

The internal energy can be calculated as

$∆E_{vap}=∆H_{vap}-∆n_{g}RT$ where,$∆E_{vap}$ is internal energy of vaporization, $n_g$ no of moles, $∆H_{vap}$ is enthalpy of vaporization. and $T$ temperature.

Step 3: Calculating the values of internal energy:

No of moles in 90 grams of water, $n_g=\frac{90g}{18g}=5$

Enthalpy of vaporization in Joule $∆H_{vap}=5\times 41000\mathrm{J}$

$$\begin{gathered} ∆E_{vap}=41000\times 5-5\times 8.314\times 373 \\ ∆E_{vap}=189494.39 \end{gathered}$$

Therefore the internal energy change (in $J$) when $90g$ of water undergoes complete evaporation at $100ºC$ is$∆E_{vap}=189494.39$.