Question

Water is brought to boil under the pressure of $1.0atm$. When an electric current of $0.50A$ from a $12V$ supply is passed for $300s$ through resistance in thermal contact with it is found that $0.798g$ of water is vapourised. Calculate the molar internal energy change at boiling point $\left(373.15K\right)$.

• A. $37.5kJ{mol}^{-1}$
• B. $3.75kJ{mol}^{-1}$
• C. $42.6kJ{mol}^{-1}$
• D. $4.26kJ{mol}^{-1}$

Answer 1

Answer: (A)

$37.5kJ{mol}^{-1}$

$\Delta H=$ work done$=i\times V\times t$
$=0.5\times 12\times 300=1.8kJ$

Molar enthalpy of vaporisation
$\Delta H_m=\frac{\Delta H}{molesof{H}_{2}O}=\frac{\Delta H}{n_{H_{2}O}}$

$=\frac{1.8kJ}{\frac{0.798}{18}}=40.6kJ{mol}^{-1}$

$\Delta H_m=\Delta E_m+p\Delta V$

$\Delta H_m=\Delta E_m+\Delta n_{g}RT$

$\Delta H_m=\Delta E_m+RT$

[$\therefore \Delta n_g=1for{H}_{2}O\left(l\right)\rightleftharpoons H_{2}O\left(g\right)$]

$\therefore$ Molar internal energy change,

$\Delta E_m=\Delta H_m-RT$

$\Delta E_m=40.6-8.314\times {10}^{-3}\times 373.15=37.5kJ{mol}^{-1}$