Respuesta :
Answer:
Option E is correct. 320K
Explanations:
In order to get the normal boiling point of the liquid, we will use the Clausius-Clapeyron equation expressed according to the equation:
[tex]\ln (\frac{P_2}{P_1}_{})=-\frac{\triangle H_{vap}}{R}\cdot(\frac{1}{T_2}-\frac{1}{T_1})[/tex]where:
P1 is the vapor pressure of that substance at T1
P2 is the vapor pressure of that substance at T2
ΔHvap is the enthalpy of vaporization.
R is the gas constant - usually expressed as 8.314J/Kmol
Given the following parameters:
[tex]\begin{gathered} P_1=102\operatorname{mm}Hg \\ P_2=760\operatorname{mm}Hg(cons\tan t) \\ T_1=273K \\ \triangle H_{\text{vap}}=30.8\text{kJ/mol} \\ R=8.31\text{J/Kmol} \end{gathered}[/tex]Substitute the given parameters into the formula to get T2
[tex]\ln (\frac{760_{}}{102_{}}_{})=-\frac{30.8\times10^3\frac{J}{mol}_{}}{\frac{8.31J}{\operatorname{km}ol}}\cdot(\frac{1}{T_2_{}}-\frac{1}{273_{}})[/tex]Simplify the result to get the value of T2
[tex]\begin{gathered} \ln (7.4509)=-3,706.3778(\frac{1}{T_2}-\frac{1}{273}) \\ 2.0083=\frac{-3,706.3778}{T_2}+13.5765 \\ \frac{-3,706.3778}{T_2}=2.0083-13.5765 \\ \frac{-3,706.3778}{T_2}=-11.5682 \\ -11.5682T_2=-3,706.3778 \\ T_2=\frac{-3,706.3778}{-11.5682} \\ T_2=320.39K \\ T_2\approx320K \end{gathered}[/tex]Hence the normal boiling point of this liquid is approximately 320K
