Question

A new flurocarbon of molar mass $102gmo{l}^1$ was placed in an electrically heated vessel. When the pressure was 650 torr, the liquid boiled at ${77}^{\circ }C$. After the boiling point had been reached, it was found that a current of 0.25 A from a 12.0 volt supply passed for 600 sec vaporises 1.8 g of the sample. The molar enthalpy & internal energy of vaporisation of new flourocarbon will be :

• A. $\Delta H=102kJ/mol,\Delta E=99.1kJ/mol$
• B. $\Delta H=95kJ/mol,\Delta E=100.3kJ/mol$
• C. $\Delta H=107kJ/mol,\Delta E=105.1kJ/mol$
• D. $\Delta H=92.7kJ/mol,\Delta E=97.4kJ/mol$

Answer 1

The correct option is A $\Delta H=102kJ/mol,\Delta E=99.1kJ/mol$

Molar mass = 102 gram/mole
P = 650 torr ; T = 77 + 273 = 350 K
Q = i x t = 0.25 x 600 = 150 C
E = Q x V = 150 x 12 = 1800 J
This heat is supplied to the system at constant pressure that swhy this is used for change in enthalpy
$\because$ For vaporisation of 1.8 gram, amount of heat required q =1800 J
$\therefore$ For vaporisation of 102 gram, amount of heat required $q=\frac{1800}{1.8}\times 102$ J
=$102{10}^3$ J = 102 KJ/mole
$\Delta H=\Delta U+P\Delta V$
$\Delta H=\Delta U+\Delta n_{g}RT$
For determination of $\Delta U$ per mol $(\Delta n_g=1)$
$102\left(KJ/mol\right)=\Delta U+(1\times 8.3\times 350)\times {10}^{-3}\Rightarrow \Delta U=102-2.9=99.1KJ/mole$