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asdf1
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When proving the equation,
dU=TdS-PdV
why is dw= PdV not dw= PdV +vdP?
dU=TdS-PdV
why is dw= PdV not dw= PdV +vdP?
Proof of dU=TdS-PdV: dw=PdV Not dw=PdV+vdP
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SUMMARY
The discussion centers on the thermodynamic equation dU=TdS-PdV and clarifies why the differential work done, dw, is expressed as PdV rather than PdV + VdP. It is established that work is only performed when there is a change in volume; thus, if volume remains constant, the pressure does not contribute to work done. The term VdP is identified as part of the internal energy change rather than work, reinforcing the distinction between work and internal energy in thermodynamic processes.
PREREQUISITES- Understanding of thermodynamic principles, specifically the first law of thermodynamics.
- Familiarity with differential calculus as applied to physical equations.
- Knowledge of the concepts of internal energy and work in thermodynamics.
- Basic grasp of pressure-volume relationships in gases.
- Study the derivation of the first law of thermodynamics in detail.
- Explore the implications of constant volume processes in thermodynamics.
- Learn about the Maxwell relations and their applications in thermodynamic equations.
- Investigate the concept of internal energy and its relation to work and heat transfer.
Students of thermodynamics, physicists, and engineers seeking a deeper understanding of energy transfer and work in thermodynamic systems.
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