Clausius clapeyron and heat capacity

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SUMMARY

The discussion centers on the relationship between the Clausius-Clapeyron equation and heat capacity, specifically addressing the variable 'C' in the context of thermodynamic equations. Participants clarify that 'C' represents temperature multiplied by the left side of the final equation, emphasizing the importance of understanding this relationship in thermodynamics. The conversation highlights the need for precise definitions and interpretations of variables in thermodynamic equations to avoid confusion.

PREREQUISITES
  • Understanding of the Clausius-Clapeyron equation
  • Basic knowledge of thermodynamics
  • Familiarity with heat capacity concepts
  • Ability to interpret thermodynamic equations
NEXT STEPS
  • Research the derivation of the Clausius-Clapeyron equation
  • Study the relationship between heat capacity and phase changes
  • Explore advanced thermodynamic concepts such as entropy and enthalpy
  • Learn about the application of the Clausius-Clapeyron equation in real-world scenarios
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Students and professionals in physics, chemistry, and engineering fields who are studying thermodynamics and seeking to deepen their understanding of the Clausius-Clapeyron equation and heat capacity.

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Homework Statement
Consider a substance whose vapor is at equilibrium at temperature T with it's liquid phase. Calculate the heat capacity C = dQ / dT as a function of n, c_p(T) and q. Use Maxwell relationships and the clausius clapeyron equation and treat the vapor as an ideal gas.
How can I go on?
Relevant Equations
dp/dT=q/(T(v_g-v_f))
Clausius Clapeyron.JPG
 
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I'm guessing that what they mean by C is T times the left side of your final equation. If it isn't that, then I have no idea what they're getting at.
 

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