Answer: The vapor pressure of Fe at [tex]2.92\times 10^3K[/tex] is 465 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]2890K[/tex] = 400 mm Hg
[tex]P_2[/tex] = final pressure at [tex]2920K[/tex] = ?
[tex]\Delta H_{vap}[/tex] = enthalpy of vaporisation = 351 kJ/mol = 351000 J/mol
R = gas constant = 8.314 J/mole.K
[tex]T_1[/tex]= initial temperature = 2890 K
[tex]T_2[/tex] = final temperature = 2920 K
Now put all the given values in this formula, we get
[tex]\log (\frac{P_2}{400})=\frac{351000}{2.303\times 8.314J/mole.K}[[\frac{1}{2890K}-\frac{1}{2920K}][/tex]
[tex]\log (\frac{P_2}{400})=0.06517[/tex]
[tex](\frac{P_2}{400})=1.162[/tex]
[tex]P_2=465mmHg[/tex]
Thus the vapor pressure of Fe at [tex]2.92\times 10^3K[/tex] is 465 mm Hg
