Clausius Clapeyron Equation of lead

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SUMMARY

The discussion focuses on estimating the melting point of lead at a pressure of 100 atm using the Clausius-Clapeyron equation. The melting point at atmospheric pressure is 327.0 degrees C, with a density change from 1.101 x 104 to 1.065 x 104 kg/m3 and a latent heat of 24.5 kJ/kg. The calculated melting point at 100 atm is approximately 600.754 K. The validity of assuming L/T(ΔV) as constant during this calculation is discussed, emphasizing the need for verification to ensure accuracy.

PREREQUISITES
  • Understanding of the Clausius-Clapeyron equation
  • Knowledge of phase transitions and latent heat
  • Familiarity with thermodynamic properties of materials
  • Basic calculus for differential equations
NEXT STEPS
  • Study the derivation and applications of the Clausius-Clapeyron equation
  • Explore the thermodynamic properties of lead under varying pressures
  • Investigate the impact of latent heat on phase transitions
  • Learn about the relationship between pressure and melting points in other materials
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Students in thermodynamics, materials scientists, and engineers involved in phase transition studies will benefit from this discussion.

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Homework Statement



When lead is melted at atmospheric pressure the melting point is 327.0 degrees C, the density decreases from 1.101 x 10^4 to 1.065 x 10^4 kg/m^3 and the latent heat is 24.5 kJ/kg. Estimate the melting point of lead at a pressure of 100 atm.


Homework Equations





The Attempt at a Solution



Ok So i know how to get the right answer - which i think is 600.754 K as follows:

Clausius-C equation: dp/dT = L/T delta (V)

so i just say that delta p = L/T delta V delta T

Substituting in the values, I get the right answer..

My question is - why is this a valid move? surely it assumes L/T(deltaV) to be constant? Must it be?
 
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Yes, it's assumed that this factor is constant. This is something you'd have to verify to make sure the calculation is accurate.
 

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