Integrating a Differential: Understanding the Steps in a Proof

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SUMMARY

The discussion focuses on the integration steps involved in a proof related to thermodynamics, specifically the equation d(μ/T) = - (3R du)/(2u) - (R dv)/v. The integration leads to the expression (μ/T) - (μ/T)_0 = - (3R/2) ln(u/u_0) - R ln(v/v_0). Participants clarify that the integration process follows standard calculus principles, where integrating dx from x_0 to x_1 results in x_1 - x_0, applied here with the variables μ/T, u, and v. This understanding allows the original poster to proceed with their work confidently.

PREREQUISITES
  • Understanding of basic calculus, specifically integration techniques.
  • Familiarity with thermodynamic variables such as μ (chemical potential) and T (temperature).
  • Knowledge of the ideal gas constant R and its application in thermodynamic equations.
  • Experience with logarithmic functions and their properties in mathematical proofs.
NEXT STEPS
  • Review integration techniques in calculus, focusing on definite and indefinite integrals.
  • Study thermodynamic principles related to chemical potential and temperature relationships.
  • Explore the application of the ideal gas law and its constants in thermodynamic calculations.
  • Learn about logarithmic identities and their use in simplifying expressions in proofs.
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Students and professionals in physics, chemistry, and engineering, particularly those studying thermodynamics and mathematical proofs in these fields.

fluidistic
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Homework Statement


There's a step I don't really understand in some "proof".
[itex]d \left ( \frac{\mu }{T} \right )=-\frac{3R du}{2u}-\frac{Rdv}{v}[/itex]. Now he integrates both sides to get [itex]\frac{\mu}{T}- \left ( \frac{\mu}{T} \right ) _0=-\frac{3R}{2} \ln \frac{u}{u_0}-R \ln \frac{v}{v_0}[/itex].
I don't understand how an integration leads to such an expression.

Homework Equations





The Attempt at a Solution

Clueless.
 
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hi fluidistic! :smile:

this is standard …

if we integrate dx from xo to x1, we get x1 - xo

we then put x = x1, and it becomes ∫ dx = x - xo

they've done it for x = µ/T, dx = du/u, and dx = dv/v :wink:
 
Thanks a lot. Crystal clear to me now. I can proceed further. :smile:
 

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