Clausius-Clapeyron equation and absorbed heat

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Homework Help Overview

The discussion revolves around the Clausius-Clapeyron equation, specifically focusing on the relationship between heat absorbed during vaporization and the work done during isothermobaric expansion. Participants explore the implications of including expansion work in the heat balance and its significance in the context of calorimetry.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants examine the role of expansion work in the Clausius-Clapeyron equation and question whether it is adequately accounted for in the heat absorbed during vaporization. There is a discussion about the definitions of heat of vaporization and the conditions under which vaporization occurs.

Discussion Status

Some participants express understanding of the definitions related to heat of vaporization and its relation to expansion work. Others continue to explore the implications of these definitions, particularly in the context of calorimetry and vaporization in different environments.

Contextual Notes

There is a mention of typical assumptions in calorimetry problems, such as focusing on moles and masses rather than volumes, which may influence the discussion on expansion work and its relevance in various scenarios.

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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