Question

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

Answer 1

Benzene (l) $\underrightarrow{Vapourization}$Benzene (g)

$△H=△V+△n_{g}RT,$ here ${△}_{ng}=1-0=1$

$\therefore △H=△V+RT$

$\therefore V=30.84-8.314\times 353\times {10}^3=27.9KJ/mol$

$7.8g=0.1mol$ of benzene.

$\therefore VIt=△H\times$ No.of moles)

$\therefore t=\frac{△H\times n}{VI}=\frac{30.84\times {10}^3\times 0.1}{12\times 00.5}=514Seconds$

$=8min34Sec$