Question

What is $\Delta U$ when $2.0$ mole of liquid water vaporises at ${100}^{o}C$? The heat of vaporisation $\left(\Delta H_{vap}\right)$ of water at ${100}^{o}C$ is $40.66KJ{mol}^{-1}$

Answer 1

Solution:-

As we know that,

$\Delta H_{vap.}=\Delta U+\Delta n_{g}RT$

whereas,

$\Delta H_{vap.}=$ standard enthalpy of vaporization

$\Delta U=$ Internal energy

$T=$ Temperature of liquid $=100℃=(100+273)K=373K$

$R=$ Universal gas constant $=8.314\times {10}^{-3}KJ{K}^{-1}{mol}^{-1}$

$\Delta n_g=$ Difference between no. of moles of gaseous reactant and gaseous product.

Given that the heat of vaporisation of water at $100℃$ is $40.66KJ{mol}^{-1}$.

For vaporization of $2$ moles of water,

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

$\Delta H_{vap.}=2\times 40.66=81.32KJ{mol}^{-1}$

$\Delta n_g=n_P-n_R=2-0=2$

Now fro ${eq}^n\left(1\right)$, we have

$81.32=\Delta U+2\times 8.314\times {10}^{-3}\times 373$

$\Rightarrow \Delta U=81.32-6.202=74.118KJ/mol$

Hence when $2$ mole of liquid water vaporises at $100℃,\Delta U$ will be $74.118KJ/mol$.