Question

For $A\left(l\right)\rightleftharpoons A\left(g\right),{\Delta }_{vap}H=460.6calmo{l}^{-1}$ and boiling point = $50K$. Then the boiling point at 10 atm is

• A. $150$ K
• B. $75$ K
• C. $100$ K
• D. $0$ K

Answer 1

The correct option is C $100$ K

$\Delta H=2.303R[\frac{1}{T_1}-\frac{1}{T_2}]log\frac{P_2}{P_1}$
$460.6=2.303\times 2[\frac{1}{50}-\frac{1}{T_2}]log10$
$\therefore$ $T_2=100K$