[Nanothree]

   I'm new to vaporization. I've been reading about it and am trying to work some problems.

Here's the one I can't figure out. Like, I know Delta Hvap = Hvap - Hliquid, and I see that some sort of PV=nRT calculation is required here, but where do I start?

The enthalpy of vaporization of benzene, C6H6(l), is 33.9 kJ mol-1 at 298K. How many liters of C6H6(g), measured at 298 K and 95.1 mmHg, are formed when 1.54 kJ of heat is absorbed by C6H6(l) at a constant temperature of 298K?


I'm having difficulty figuring out what's happening conceptually there. C6H6(l) is absorbing heat to form a vapor, but volume and moles are unknown. How do I go about approaching these kinds of problems?

[Nanothree]

Nevermind!! I solved it!!!!!   


For those curious:

(1.54 kJ) * (1 mol / 33.9 kJ) = 0.0454 mol C6H6 (g)

v=(nrt)/p
   = (.0454) (0.0821) (298) / (95.1/760)
   = 8.88 L

[generalsky]

Yeah, I agree with the former result. How to say ,in this kind of calculation, we should be careful with the units of every quantity.