Clausius-Clapeyron equation and absorbed heat

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SUMMARY

The discussion focuses on the Clausius-Clapeyron equation, specifically its application in calculating the efficiency of a Carnot cycle involving vaporization. The equation is derived from the relationship between the work done during isothermobaric processes and the heat absorbed during vaporization, expressed as dp/dT = n λ / T(V_g - V_ℓ). The participants highlight the importance of including expansion work in the heat absorbed calculation, noting that the latent heat of water is approximately 2 MJ/kg while the expansion work at atmospheric pressure contributes around 0.1 MJ. Clarifications regarding the definitions of heat of vaporization and the role of pressure in vaporization processes are also discussed.

PREREQUISITES
  • Understanding of the Clausius-Clapeyron equation
  • Familiarity with Carnot cycle thermodynamics
  • Knowledge of latent heat and its significance in phase changes
  • Basic principles of calorimetry and vaporization
NEXT STEPS
  • Study the derivation of the Clausius-Clapeyron equation in detail
  • Explore the implications of expansion work in thermodynamic processes
  • Investigate the definitions and calculations of latent heat in various substances
  • Learn about the applications of the Carnot cycle in real-world thermodynamic systems
USEFUL FOR

Students and professionals in thermodynamics, physicists, and engineers interested in heat transfer and phase change phenomena will benefit from this discussion.

MaxLinus
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Homework Statement
Work out the Clausius-Clapeyron equation using the efficiency of a Carnot cycle.
Relevant Equations
Clausius-Clapeyron equation; Carnot efficiency
In working out the Clausius-Clapeyron equation in an elementary method, we usually consider the work done in a Carnot cycle, built with two isothermobarics at ##p## , ##T## and ##p- dp##,##T-dT## pressures and temperatures and two adiabatics, as ##\mathcal{L} = dp(V_g - V_\ell)##, where ##V_g## is the volume of ##n## vaporized moles and ##V_\ell## is their volume in the liquid phase. The heat absorbed in the isothermobaric vaporization is usually set to ##Q_a = n \lambda##, where ##\lambda## is the molar latent heat of vaporization. So the efficiency is ##\eta =\frac{\mathcal{L}}{Q_a} = \frac{dp(V_g - V_\ell)} {n \lambda}##. Then the solution is easy: the efficiency must be equal to ##dT/T## (Carnot cycle), and you get
$$ \frac{dp}{dT} = \frac{n \lambda}{T(V_g - V_\ell)} $$But why the work ##\mathcal{L}_1 = p (V_g - V_\ell)## done in the isothermobaric expansion is not considered in the balance? The heat absorbed should account for vaporization and for expansion, whose effect appears not so easily neglibile: latent heat of water is about ##2 \, MJ/kg##, while an expansion of ##V_g - V_\ell = 1 \, m^3## (not an exaggerated value) at atmospheric pressure (##\sim 100 \, kPa ##) should contribute with approximately ##100 \, kJ \sim 0.1 \, MJ##.
 
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MaxLinus said:
The heat absorbed should account for vaporization and for expansion
If I'm understanding it correctly, the definition of heat of vaporization does include the expansion work ##p(V_g-V_l)##. For example, see the definitions here and here.
 
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Ok. The first link you posted states it very clearly.
Many thanks.
 
Back again, just a bit.
In first problems about calorimetry, we usually do not deal with volumes, but only with moles and masses. Yet, vaporization takes place at a definite pressure. So, if I understand correctly your hint, we can still define a work in the expansion even if it occurs (for example) in an open vessel, but in an environment with pressure. Something related to partial volumes ...
 

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