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Naphthalene, a substance present in some mothballs, has an enthalpy of vaporization of 49.4kl/mol. If the vapor pressure of naphthalene is .300mmhg at 298K, what is the temp at which the vapor pressure is 7.6x10^2mmhg?

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Revise phase equilibria and Raoults law.


P is the vapour pressure of the solution, [math]P^y[/math]=vapour pressure of the pure solvent [math]m[/math]=mole fraction of solvent in the solution.

Use [math]pV=nRT[/math] to find kelvin.

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honestly, the ideal gas law might or might not help, but the clausius clapeyron equation is designed for exactly this situation:


[math]\ln(\frac{P_{2}}{P_{1}}) = \frac{\Delta H_{vap}}{R}(\frac{1}{T_{1}}-\frac{1}{T_{2}})[/math]


apologies for the non-greek spelling for delta, but latex wouldnt accept the symbol. Note that you're given the deltaHvap and T1 and P1 and P2. just solve for T2

Edited by Cap'n Refsmmat
fixed the greek - you just needed to put a space after it and a \ before it. - ecoli (also fixed the ln - Cap'n)
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