Question

The boiling point of a liquid is 50 K at 1 atm and $\Delta H_{vap}=460.6$ cal/mol. What will be its boiling point at 10 atm pressure ?

• A. 150 K
• B. 75 K
• C. 100 K
• D. 200 K

Answer 1

The correct option is C 100 K

$2.303log\frac{P_2}{P_1}=\frac{\Delta H}{R}\frac{(T_2-T_1)}{T_1{T}_2}$
On solving we get $T_2=100K$