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Thermodynamics — Chemistry STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceChemistryThermodynamics2 Marks
Question
Standard vaporisation enthalpy of benzene at boiling point is $30.8 \mathrm{~kJ} \mathrm{~mol}^{-1}$. For how long would 100 W electric heater have to operate in order to vaporise a 100 g sample at that temperature (power = energy/ time and $1 \mathrm{~W}=1 \mathrm{~J} \mathrm{~s}^{-1}$ )?
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Answer

$\Delta_{\text {vap }} \mathrm{H}^{\circ}$ (benzene $)=30.8 \mathrm{~kJ} \mathrm{~mol}^{-1}$ Molar mass of benzene, $\mathrm{C}_6 \mathrm{H}_6=(6 \times 12+6 \times 1) \mathrm{g} \mathrm{mol}^{-1}=78 \mathrm{~g} \mathrm{~mol}^{-1}$ Energy needed to vaporise benzene $=30.8 \mathrm{~kJ} \mathrm{~mol}^{-1} \times \frac{100 \mathrm{~g}}{78 \mathrm{~g} \mathrm{~mol}^{-1}}$
$=39.49 \mathrm{~kJ}$
So, Time $=\frac{\text { energy }}{\text { power }}=\frac{39.49 \mathrm{~kJ}, \mathrm{~J}}{100 \mathrm{~W}}$
$=\frac{39.49 \times 10^3 \mathrm{~J}}{100 \mathrm{~J} \mathrm{~s}^{-1}}=394.9 \mathrm{~s}=6.6 \mathrm{~min}$
 

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