Internal Energy & 1st Law of Thermodynamics: P vs V Graph?

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SUMMARY

The discussion centers on the relationship between internal energy and the First Law of Thermodynamics, specifically in the context of a P vs V graph. It is established that if the graph forms an enclosed shape, such as a square or triangle, the total internal energy change over one complete cycle is zero. This conclusion is based on the principle that internal energy is a state function, dependent solely on temperature, as expressed by the equation u = c_vT for ideal gases and perfect liquids. Therefore, if the initial and final states are identical, the change in internal energy is zero, regardless of the path taken.

PREREQUISITES
  • Understanding of the First Law of Thermodynamics
  • Familiarity with state functions in thermodynamics
  • Knowledge of P vs V diagrams
  • Basic principles of ideal gases and perfect liquids
NEXT STEPS
  • Study the implications of the First Law of Thermodynamics in closed systems
  • Explore the concept of state functions in thermodynamics
  • Learn about the derivation and applications of the equation u = c_vT
  • Investigate different thermodynamic cycles and their effects on internal energy
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Students of physics, thermodynamics enthusiasts, and professionals in engineering fields who seek to deepen their understanding of energy transformations and thermodynamic principles.

jaidon
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I have a question concerning internal energy and the First Law of Thermodynamics. If you plotted P vs V and the shape of the graph is an enclosed shape ie) square or triangle, would the total internal energy in one cycle be zero? (one cycle is for example-start at top left of square and finish at top left of square). I think this is right, but am a little uncertain being a Physics moron and all. I would appreciate anyone's input. Thanks
 
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If anyone has any idea on this I would love to hear from you. This thought is what I have based several questions on, and if I am incorrect I need to figure out something else. Thanks.
 
Internal Energy is a variable of state. Given some thermodynamic trajectory, for instance from 1 to 2, the change of internal energy is given by:

[tex]\int_1^2 du=u_2-u_1[/tex]

Due to the fact in ideal gases and perfect liquids the internal energy is only a function of the temperature:

[tex]u=c_vT[/tex]

then if 1 is the same thermodynamic state than 2 the change of internal energy is zero, no matter which trajectory it was.
 

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