Answer: The vapor pressure of mercury at 322°C is 0.521 atm
Explanation:
To calculate the final pressure, we use the Clausius-Clayperon 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] = initial pressure which is the pressure at normal boiling point = 1 atm
[tex]P_2[/tex] = final pressure which is vapor pressure of mercury = ?
[tex]\Delta H_{vap}[/tex] = Enthalpy of vaporization = 58.51 kJ/mol = 58510 J/mol (Conversion factor: 1 kJ = 1000 J)
R = Gas constant = 8.314 J/mol K
[tex]T_1[/tex] = initial temperature which is normal boiling point = [tex]356.7^oC=[356.7+273]K=629.7K[/tex]
[tex]T_2[/tex] = final temperature = [tex]322^oC=[322+273]K=595K[/tex]
Putting values in above equation, we get:
[tex]\ln(\frac{P_2}{1})=\frac{58510J/mol}{8.314J/mol.K}[\frac{1}{629.7}-\frac{1}{595}]\\\\\ln P_2=-0.6517atm\\\\P_2=e^{-0.6517}=0.521atm[/tex]
Hence, the vapor pressure of mercury at 322°C is 0.521 atm
The normal boiling point of Hg is 356.7 °C. The vapor pressure of Hg at 322 °C is 0.519 atm.
The normal boiling point of mercury (Hg) is 356.7 °C (T₁). The word "normal" refers to a vapor pressure of 1 atm (P₁).
Given the enthalpy of vaporization (∆Hvap = 58.51 kJ/mol), we can calculate the vapor pressure (P₂) at 322 °C (T₂) using the Clausius-Clapeyron equation.
[tex]ln(\frac{P_2}{P_1} ) = \frac{-\Delta H_{vap}}{R} (\frac{1}{T_2}-\frac{1}{T_1} )[/tex]
where,
To apply this equation, we need to convert the temperatures to Kelvin using the following expression.
[tex]K = \° C + 273.15\\\\T_1: K = \° C + 273.15 = 356.7\° C + 273.15 = 629.9 K\\\\T_2: K = \° C + 273.15 = 322\° C + 273.15 = 595 K[/tex]
The vapor pressure of mercury at 322 °C is:
[tex]ln(\frac{P_2}{1atm} ) = \frac{-(58.51 \times 10^{3} J/mol.K)}{(8.314 J/mol.K)} (\frac{1}{595K}-\frac{1}{629.9K} )\\\\P_2 = 0.519 atm[/tex]
The normal boiling point of Hg is 356.7 °C. The vapor pressure of Hg at 322 °C is 0.519 atm.
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