Question

A new flurocarbon of molar mass $102\text{g/mol}$ was placed in a electrically heated vessel. When the pressure was $650\text{torr}$, the liquid boiled at ${57}^o\text{C}$. After the boiling point had been reached it was found that a current of $0.25\text{A}$ from a $12\text{V}$ supply passed for $600\text{s}$ vapourises $1.8\text{g}$ of the sample.
$(\text{Take}R=\frac{25}{3}\text{J/molK}$)

• A. $\Delta H=102\text{kJ/mol}$
• B. $\Delta H=99.75\text{kJ/mol}$
• C. $\Delta U=102\text{kJ/mol}$
• D. $\Delta U=99.75\text{kJ/mol}$

Answer 1

Answer: (A), (D)

The correct options are
A $\Delta H=102\text{kJ/mol}$
D $\Delta U=99.75\text{kJ/mol}$

Molar mass $=102\text{g/mol}$
Pressure $=650\text{torr}$
Temperature $={57}^o\text{C}=330\text{K}$
$Q=i\times t=0.25\times 600=150\text{C}$
$\Delta H=Q\times V=150\times 12=1800\text{J}$
This heat is supplied to the system at constant pressure that is why this is used for change in enthalpy.
$\therefore$
For vaporisation of $1.8\text{g}$, amount of heat required $\text{q}=1800\text{J}$
For vaporisation of $102\text{g}$, amount of heat required $\text{q}=\frac{1800}{1.8}\times 102=102\times {10}^3\text{J}=102\text{kJ/mole}$
Also,
$\Delta H=\Delta U+\Delta n_{g}RT$
$102=\Delta U+(1\times \frac{25}{3}\times {10}^{-3}\times 330)$
$\Delta U=102-2.75=99.25\text{kJ/mole}$