Taking the derivative of the van der waals equation

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SUMMARY

The discussion focuses on calculating the derivative dT/dVm for the van der Waals equation. The user correctly identifies the derivatives of the terms RT(Vm-b) and -a/Vm^2 but struggles with the application of the product rule in the context of the equation. The correct application of the product rule leads to the term [R/(Vm-b)](dT/dVm), clarifying the relationship between temperature and molar volume in the van der Waals context.

PREREQUISITES
  • Understanding of calculus, specifically differentiation and the product rule.
  • Familiarity with the van der Waals equation of state.
  • Knowledge of thermodynamic concepts such as temperature, pressure, and molar volume.
  • Basic understanding of physical chemistry principles.
NEXT STEPS
  • Review the product rule in calculus to solidify understanding of its application.
  • Study the van der Waals equation and its implications in real gas behavior.
  • Explore thermodynamic derivatives and their significance in physical chemistry.
  • Practice additional problems involving differentiation of thermodynamic equations.
USEFUL FOR

Students of physical chemistry, particularly those studying thermodynamics and the behavior of real gases, as well as educators looking for examples of calculus applications in chemistry.

Razael
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Homework Statement



I'm trying to find dT/dVm for the van der waals equation. I was looking at http://courses.washington.edu/bhrchem/c456/vdw_jtc.pdf" page, but I'm confused about how they reached their result:

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The Attempt at a Solution



I honestly have no idea what they did. dT/dVm of RT(Vm-b) is obviously -RT/(Vm-b)^2, and dT/dVm of -a/Vm^2 is 2a/Vm^3, and dT/dVm of P is 0, but I have no idea where [R/(Vm-b)](dT/dVm) came from. Am I missing something obvious here? It's been a while since calc.
 
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The first term is the product of RT and 1/(V_m-b), so you need to use the product rule: (fg)'(x)=f'(x)g(x)+f(x)g'(x).
 

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