Answer:
36.43 atm is the pressure in the container at 142.0°C.
Explanation:
The Clausius Clapeyron equation:
[tex]\ln \frac{P_1}{P_2}=\frac{\Delta H_{vap}}{R}\times (\frac{1}{T_2}-\frac{1}{T_1})[/tex]
[tex]P_1[/tex] = Pressure at temperature [tex]T_1[/tex]
[tex]P_2[/tex] = Pressure at temperature [tex]T_2[/tex]
[tex]\Delta H_{vap][/tex] = Enthalpy of vaporization of substance
R = Universal gas constant = 8.314 J/ mol K
[tex]P_1=2.30 atm,T_1=25^oC=298.15 K[/tex]
[tex]P_2=?,T_2=142^oC=415.15 K[/tex]
[tex]\Delta H_{vap]= 24.3 kJ/mol=24300 J/mol[/tex]
[tex]\ln \frac{2.30 atm}{P_2}=\frac{24300 J/mol}{8.314 J/mol K}\times (\frac{1}{415.15 K}-\frac{1}{298.15 K})[/tex]
[tex]P_2=36.43 atm[/tex]
36.43 atm is the pressure in the container at 142.0°C.
