Question

Vapour pressure and temperature relationship in shown as, $logP=\frac{-500\left(K\right)}{T}+15$. Heat of vaporization will be:

• A. $2.303$ Kcal
• B. $4.606$ Kcal
• C. $500$ Kcal
• D. $500$ Cal

Answer 1

Answer: (A)

$2.303$ Kcal

$\log P=\frac{-\Delta H_{vap}}{2.303TR}+C$ ($R=$ Gas constant)

$\log P=\frac{-500}{T}+15$

Comparing with equation:

$\frac{\Delta H_{vap}}{8.314\times 2.303}=500$

$⟹\Delta H_{vap}=9573.57$ $J$

In Calories $=\frac{9573.57}{4.184}=2303$ $cal=2.303$ $kcal$