Boiling point of cyclohexane at 620 mm hg?
Standard atmospheric pressure is 760 mm Hg. Boiling point at 760 mm= 80.74˚C
A liquid boils when its vapor pressure is equal to the atmospheric pressure. The vapor pressure of a liquid is proportional the absolute temperature of the liquid.
As the atmospheric pressure decreases, less vapor pressure is needed to cause the liquid to boil. Less vapor pressure means lower temperature.
The boiling point of cyclohexane at 620 mm Hg is less than 80.74˚C
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Answer:
Boiling point of cyclohexane at 620 mm Hg is 440.6 K
Explanation:
According to clausius-clapeyron equation for a liquid-vapour equilibrium-
[tex]ln(\frac{P_{2}}{P_{1}})=\frac{-\Delta H_{vap}}{R}(\frac{1}{T_{2}}-\frac{1}{T_{1}})[/tex]
where [tex]P_{2}[/tex] and [tex]P_{1}[/tex] are vapor pressure of liquid at [tex]T_{2}[/tex] and [tex]T_{1}[/tex] temperature respectively
[tex]\Delta H_{vap}[/tex] is the molar enthalpy of vaporization of liquid
Let's assume [tex]\Delta H_{vap}[/tex] is equal to standard molar enthalpy of vaporization of a liquid
For cyclohexane, standard molar enthalpy of vaporization is 32.83 kJ/mol
Here, [tex]P_{2}[/tex] = 620 mm Hg, [tex]T_{1}[/tex] is 450.8 K (boiling point of cyclohexane) and [tex]P_{1}[/tex] is 760 mm Hg
So, [tex]ln(\frac{620mm Hg}{760mm Hg})=\frac{-32.83\times 10^{3}J/mol}{8.314 J/(mol.K)}\times (\frac{1}{T_{2}}-\frac{1}{450.8K})[/tex]
So, [tex]T_{2}=440.6K[/tex]
