Can I reverse the PdV integral for work?

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SUMMARY

The discussion centers on the possibility of reversing the work integral W = ∫PdV into the form W = ∫V dP, given the pressure equation P = 40V^2 + 20. The user has known variables including P1, P2, V1, and molecular mass but lacks total mass, final volume, and temperatures. The conclusion drawn is that while the user can express V in terms of P, the integration by parts method indicates that direct reversal of the integral is not feasible without additional considerations.

PREREQUISITES
  • Understanding of thermodynamic work equations
  • Familiarity with integration techniques, specifically integration by parts
  • Knowledge of pressure-volume relationships in thermodynamics
  • Basic concepts of molecular mass and its implications in gas laws
NEXT STEPS
  • Study integration by parts in calculus for thermodynamic applications
  • Explore the implications of the ideal gas law on work calculations
  • Learn about the relationship between pressure and volume in compressible fluids
  • Investigate the derivation and applications of the work integral in thermodynamic systems
USEFUL FOR

Students and professionals in thermodynamics, mechanical engineers, and anyone involved in fluid mechanics or energy systems analysis will benefit from this discussion.

Hercuflea
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Homework Statement


I have a problem where I know P1, P2, V1, and molecular mass. I do not know total mass, final volume, temperature 1 or temperature 2, or the work. I also know that P = 40V^2 + 20

Can I reverse the W = ∫PdV equation into the form W = ∫V dP? Because I could just solve the equation for V = sqrt(P/40 - 1/2) and integrate that. Can I do this?


Homework Equations


W = ∫PdV


The Attempt at a Solution



Qdot - Wdot = mdot ((u2 + v^2/2 + gz2) - (u1 + v^2/2 +gz1))
no heat transfer, constant mass, no change in altitude, no velocity. The system is being compressed.
 
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Hercuflea said:
Can I reverse the W = ∫PdV equation into the form W = ∫V dP? .
Not quite. Integration by parts: ##\int_a^by.dx + \int_{y(a)}^{y(b)}x.dy = [xy]_{x=a}^{x=b}##
 

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