Help Internal Energy of a Liquid

[tamiller07]

In thermodynamics
Change in Internal Energy = Heat - Work ($$\Delta$$U=Q-W)

During vaporization my Heat is mass*Latent Heat (Q=mL) and Work is Pressure*change in volume (W=P$$\Delta$$V), so $$\Delta$$U=mL-P$$\Delta$$V.

Let's take any of the natural monoatomic gases, Neon in this case.

$$\Delta$$ = Volume of gas state - Volume of liquid state, you can find this using a standard 1kg of mass and the density at the boiling point of in each state. I used [noparse]http://encyclopedia.airliquide.com/[/noparse] as my source.

$$\Delta$$U = U_gas - U_liquid therefore U_liquid=U_gas - $$\Delta$$U

I can find the internal energy(U) of Neon gas at the boiling point. U=$$\frac{3}{2}$$nRT.(n=mols of Neon, R=8.31, T=boiling temp in K or 27.1K

When it works out,

U_liquid=U_gas - $$\Delta$$U
U_liquid=16740 Joules - 77993 Joules
U_liquid= -61252 Joules

my question is, how can I have negative internal energy for the liquid? My professors can't give me an answer and they've been researching for a few days now. Please help!

 

[Curl]

The amount of liquid also changes, so what you did there is wrong. When you are boiling a liquid, the internal energy per unit mass of the liquid does not change. Neither is the internal energy per unit mass of the gas. 100% of the energy you put in goes into the "hidden energy" aka "enthlapy of vaporization" for the system.

I bet your professors haven't even thought about this because they don't care, and they say "i don't know" just to get you off their back.

 

[Drakkith]

U = Ugas - Uliquid therefore Uliquid=Ugas - U

Just throwing out a guess, but wouldn't this be -Uliquid = u/ugas? If my memory on equations is correct...

Thats U divided by Ugas btw.