Answer:
The autoclave must operate at 1.94 atm pressure.
Explanation:
To calculate [tex]P_2[/tex] of the reaction, we use Clausius Clapeyron equation, which is:
[tex]\ln(\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}[\frac{1}{T_1}-\frac{1}{T_2}][/tex]
where,
[tex]P_1[/tex] = Pressure of autoclave at temperature [tex]T_1[/tex] = 1 atm
[tex]P_2[/tex] = Pressure of autoclave at temperature [tex]T_2[/tex] = ?
[tex]\Delta H_{vap}[/tex] = Enthalpy of vaporization = 40.656 kJ/mol = 40,656 J/mol
R = Gas constant = 8.314 J/mol K
[tex]T_1[/tex] = initial temperature = [tex]373.12 K[/tex]
[tex]T_2[/tex] = final temperature = [tex]120^oC=[120+273.15]K=393.15 K[/tex]
Putting values in above equation, we get:
[tex]\ln(\frac{P_2}{1 atm})=\frac{40,656 J/mol}{8.314J/mol.K}[\frac{1}{373.12 K}-\frac{1}{393.15 K}]\\\\ P_2=1.94 atm[/tex]
The autoclave must operate at 1.94 atm pressure.
