[MitchTwitchita]

Calculate the enthalpy of vaporization for a compound if its vapor pressure is 56 mmHg at -16 degrees C and 292 mmHG at 6 degrees C.
The final answer is in kJ.

P1/P2 = (delta Hvap/R)(T1 - T2/T1*T2)
(55 mmHg/292 mmHg) = (delta Hvap/8.314 J/K * mol)[(257 K - 279 K)/257 K * 279 K)]
(0.188356164 mmHg) = (delta Hvap/8.314 J/K * mol)(-0.000306821 K)
delta Hvap = [(0.188356164 mmHg)(8.314 J/mol)]/(-0.000306821)

This is as far as I can get it.  When calculating for an unknown pressure, the antilog is taken from both sides.  However, I'm not sure what to do in this situation, plus the units look really messed up.  Any advice?

[Yggdrasil]

P1/P2 is a dimensionless number, 0.188356164, not 0.188356164mmHg.

[MitchTwitchita]

Oh, that makes sense.  The units would cancel initially.  Now how would I solve for delta Hvap?  Would I have to employ some antilog trickery or simply solve for delta Hvap normally?

[Yggdrasil]

I didn't notice, but in your first equation, you ommitted the natural log:

ln(P1/P2) = (delta Hvap/R)(T1 - T2/T1*T2)

Since you can calculate ln(P1/P2), you just need simple algebra to solve for ΔH.  Antilogs are needed only when solving for one of the values of pressure.

[MitchTwitchita]

Thanks Yggdrasil, I didn't know you could use ln.  You've been a great *delete me* 

[MitchTwitchita]

Why does *delete me* come up when I type "help?"  It makes it look like I've typed something vulgar and insulting.

[Yggdrasil]

Its a result of the twisted humor of our mods and admins.