- #1
MaxLinus
- 5
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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##.
$$ \frac{dp}{dT} = \frac{n \lambda}{T(V_g - V_\ell)} $$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##.