Clausius-Clapeyron equation and absorbed heat

In summary, the Clausius-Clapeyron equation states that the efficiency of a Carnot cycle must be equal to the heat absorbed in the vaporization and expansion of the heated substance.
  • #1
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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  • #2
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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Likes MaxLinus
  • #3
Ok. The first link you posted states it very clearly.
Many thanks.
 
  • #4
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 ...
 

FAQ: Clausius-Clapeyron equation and absorbed heat

What is the Clausius-Clapeyron equation?

The Clausius-Clapeyron equation is a fundamental thermodynamic equation that relates the change in vapor pressure of a substance to its temperature and enthalpy of vaporization. It is often used to model phase transitions, such as the transition from liquid to gas.

How is the Clausius-Clapeyron equation derived?

The Clausius-Clapeyron equation is derived from the combined gas law and the definition of enthalpy. It can also be derived from the first and second laws of thermodynamics.

What is the significance of the Clausius-Clapeyron equation?

The Clausius-Clapeyron equation is significant because it allows us to make predictions about phase transitions and understand the behavior of substances at different temperatures and pressures. It is also a key tool in studying thermodynamic processes and calculating the enthalpy of vaporization.

How is the Clausius-Clapeyron equation used in real-world applications?

The Clausius-Clapeyron equation has various real-world applications, such as in weather forecasting, climate modeling, and industrial processes. It is also used in the design and operation of refrigeration systems and in the production of pharmaceuticals and other chemical products.

What is the relationship between the Clausius-Clapeyron equation and absorbed heat?

The Clausius-Clapeyron equation includes the enthalpy of vaporization, which is the amount of heat absorbed when a substance changes from liquid to gas. This means that the equation is directly related to the amount of heat absorbed or released during a phase transition. The greater the enthalpy of vaporization, the more heat is required for the substance to change phase.

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