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?

Revise phase equilibria and Raoults law.

[math]P=P^ym[/math]

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.

what you need here is nor raoult's law or the ideal gas law, what you need is the clausius clapeyron equation. Look it up in your textbook.

Ideal gas equation should help! I agree with DeanK2

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 June 26, 200817 yr 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)

For future reference,

 

[math]\Delta[/math] is obtained by [math]\Delta[/math]

 

 

Cheers.

 

 

 

EDIT: Never mind. Looks like you figured it out all on your own.

 

EDIT: Never mind. Looks like you figured it out all on your own.

 

that was actually my doing... "last edited by ecoli..."

how'd you do the ln?

You use \ln. You can click on the latex image to display the code, just so you know.