Question

The vapor pressure of liquid iron is 400 mm Hg at 2.89Ã103 K. Assuming that its molar heat of vaporization is constant at 351 kJ/mol, the vapor pressure of liquid Fe is ________ mm Hg at a temperature of 2.92Ã103 K.

Answer 1

Answer: The vapor pressure of Fe at $2.92\times 10^3K$ is 465 mm Hg

Explanation:

The vapor pressure is determined by Clausius Clapeyron equation:

$ln(\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}(\frac{1}{T_1}-\frac{1}{T_2})$

where,

$P_1$= initial pressure at $2890K$ = 400 mm Hg

$P_2$ = final pressure at $2920K$ = ?

$\Delta H_{vap}$ = enthalpy of vaporisation = 351 kJ/mol = 351000 J/mol

R = gas constant = 8.314 J/mole.K

$T_1$= initial temperature = 2890 K

$T_2$ = final temperature = 2920 K

Now put all the given values in this formula, we get

$\log (\frac{P_2}{400})=\frac{351000}{2.303\times 8.314J/mole.K}[[\frac{1}{2890K}-\frac{1}{2920K}]$

$\log (\frac{P_2}{400})=0.06517$

$(\frac{P_2}{400})=1.162$

$P_2=465mmHg$

Thus the vapor pressure of Fe at $2.92\times 10^3K$ is 465 mm Hg