Clausius-Clapeyron equation and absorbed heat

AI Thread Summary
The discussion focuses on the Clausius-Clapeyron equation, emphasizing the relationship between absorbed heat during vaporization and the work done in a Carnot cycle. It highlights that the efficiency of the cycle can be expressed as η = dp(Vg - Vℓ) / nλ, leading to the derivation of dp/dT = nλ / T(Vg - Vℓ). A key point raised is the omission of the work done during isothermobaric expansion in the heat balance, despite its significant contribution. The conversation clarifies that the heat of vaporization inherently includes expansion work, even in scenarios like open vessels under pressure. Overall, the discussion underscores the importance of considering both vaporization and expansion in thermodynamic calculations.
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.
 
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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