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nadya68 [22]
4 years ago
14

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.
Chemistry
1 answer:
STatiana [176]4 years ago
7 0

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

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