Question

The molar enthalpy of vaporization of benzene at its boiling point $\left(353K\right)$ is $30.84kJmo{l}^{-1}$. What is the molar internal energy change? For how long would a $12$ volt source need to supply a $0.5A$ current in order to vaporise $7.8g$ of the sample at its boiling point?

Answer 1

${C_6{H}_6}_{\left(l\right)}⟶{C_6{H}_6}_{\left(g\right)}$

$\therefore \Delta n=n_P-n_R=1-0=1$

Given that:-

$\Delta H=30.84KJ/mol=30840J/mol$

$T=353K$

As we know that,

$\Delta H=\Delta E+\Delta n_{g}RT$

$30840=\Delta E+1\times 8.314\times 353$

$\Delta E=30840-2934.842=27905.158=27.9KJ$

Now

Molecular weight of benzene $=78g/mol$

Weight of benzene $=7.8g$

No. of moles of benzene $=\frac{7.8}{78}=0.1\text{mole}$

$\therefore$ Energy provided to evaporate $0.1$ mole of benzene $\left(E\right)=30.84\times 0.1=3084J$

As we know that,

$E=V\times I\times t$

Given that,

$V=12\text{volt}$

$t=?$

$I=0.5A$

$3084=12\times 0.5\times t$

$\Rightarrow t=\frac{3084}{6}=514s$

Hence the molar internal energy change will be $27.9KJ$ and for $t=514s$, a $12$ volt source need to supply a $0.5A$ current in order to vaporise $7.8g$ of the sample at its boiling point.