Question

A new flurocarbon of molar mass 102 g mol-1 was placed in an electrically heated vessel. When the pressure was 650 torr, the liquid boiled at 770C. 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 1.8g of the sample. The molar enthalpy & internal energy of vaporisation of new flourocarbon will be :

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

Answer 1

Answer: (A)

$\Delta$H = 102 kJ/mol, $\Delta$E = 99.1 kJ/mol

Answer 2

Molar mass = 102 gram/mole

P = 650 torr ; T = 77 + 273 = 350 K

Q = i × t = 0.25 × 600 = 150

E = Q × V = 150 × 12 = 1800 J

This heat is supplied to the system at constant pressure that’s why 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 = $\frac{1800}{1.8}$× 102 J

= 102 × 103 J = 102 KJ/mole

$\Delta$H = $\Delta$U + P$\Delta$V

$\Delta$H = $\Delta$U + $\Delta$ngRT

For determination of $\Delta$U per mol ($\Delta$ng = 1)

(KJ/mol) = $\Delta$U + (1 × 8.3 × 350) × 10–3

$\Rightarrow$ $\Delta$U = 102 – 2.9 = 99.1 KJ/mole