Help Internal Energy of a Liquid

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SUMMARY

The discussion centers on the calculation of internal energy during the vaporization of Neon, a monoatomic gas. The formula for change in internal energy is established as ΔU = Q - W, where Q is the heat (Q = mL) and W is the work (W = PΔV). The user calculates the internal energy of Neon gas at its boiling point and finds Uliquid to be negative, leading to confusion regarding the physical implications of negative internal energy. The conversation highlights the importance of understanding that during boiling, the internal energy per unit mass of both the liquid and gas remains constant, with energy input contributing to the enthalpy of vaporization.

PREREQUISITES
  • Understanding of thermodynamics principles, specifically the first law of thermodynamics.
  • Familiarity with the concepts of latent heat and enthalpy of vaporization.
  • Knowledge of the ideal gas law and its application to monoatomic gases.
  • Ability to perform calculations involving internal energy and state changes in substances.
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  • Research the concept of enthalpy of vaporization and its role in phase transitions.
  • Learn about the thermodynamic properties of Neon and other monoatomic gases.
  • Explore the implications of negative internal energy in thermodynamic systems.
  • Study the derivation and application of the ideal gas law in real-world scenarios.
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Students of thermodynamics, physicists, and engineers interested in the properties of gases and liquids during phase transitions, particularly those studying internal energy and vaporization processes.

tamiller07
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In thermodynamics
Change in Internal Energy = Heat - Work (\DeltaU=Q-W)

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

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.

\DeltaU = Ugas - Uliquid therefore Uliquid=Ugas - \DeltaU

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,

Uliquid=Ugas - \DeltaU
Uliquid=16740 Joules - 77993 Joules
Uliquid= -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!
 
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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.
 
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.
 

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