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* 10^3K` is 465 mm Hg

Step-by-step explanation:

The vapor pressure is determined by Clausius Clapeyron equation:

`ln((P_2)/(P_1))=(\Delta H_(vap))/(R)((1)/(T_1)-(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 ((P_2)/(400))=(351000)/(2.303* 8.314J/mole.K)[[(1)/(2890K)-(1)/(2920K)]`

`\log ((P_2)/(400))=0.06517`

`((P_2)/(400))=1.162`

`P_2=465mmHg`

Thus the vapor pressure of Fe at
`2.92* 10^3K` is 465 mm Hg