Answer :
Answer: 26.5 mm Hg
Explanation:
The vapor pressure is determined by Clausius Clapeyron equation:
[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]= initial pressure at [tex]26.5^oC[/tex] = ?
[tex]P_2[/tex] = final pressure at [tex]-5.1^oC[/tex] = 100 mm Hg
= enthalpy of vaporisation = 28.0 kJ/mol =28000 J/mol
R = gas constant = 8.314 J/mole.K
[tex]T_1[/tex]= initial temperature = [tex]26.5^oC=273+26.5=299.5K[/tex]
[tex]T_2[/tex] = final temperature =[tex]-5.1^oC=273+(-5.1)=267.9K[/tex]
Now put all the given values in this formula, we get
[tex]\log (\frac{P_1}{100})=\frac{28000}{2.303\times 8.314J/mole.K}[\frac{1}{299.5}-\frac{1}{267.9}][/tex]
[tex]\log (\frac{P_1}{100})=-0.576[/tex]
[tex]\frac{P_1}{100}=0.265[/tex]
[tex]P_1=26.5mmHg[/tex]
Thus the vapor pressure of [tex]CS_2CS_2[/tex] in mmHg at 26.5 ∘C is 26.5