Answer:
T_2=338.9026K
Boiling point of CH3COOCH3 at external pressure is 338.9026K
Explanation:
We are going to use Clausius-Clapeyron Equation:
[tex]ln\frac{P_2}{P_1} =-\frac{\Delta H}{R}(\frac{1}{T_2}-\frac{1}{T_1})[/tex]
Where:
P_2 is the external pressure
P_1 is the atmospheric Pressure=1 atm
ΔH is the heat of vaporization
T_2 boiling point of CH3COOCH3 at external pressure
T_1 normal boiling point of liquid methyl acetate
Now:
[tex]\frac{1}{T_2}=-ln\frac{P_2}{P_1}*\frac{R}{\Delta H}+\frac{1}{T_1} \\\frac{1}{T_2}=-ln\frac{1.29}{1}*\frac{8.314}{30.6*10^3}+\frac{1}{331} \\\frac{1}{T_2}=2.9507*10^-^3\\T_2=\frac{1}{2.9507*10^-^3} \\T_2=338.9026K[/tex]
Boiling point of CH3COOCH3 at external pressure is 338.9026K
