Latent heat and Clapeyron equation

In summary, the problem involves finding the approximate height of a hill given the boiling point of a certain liquid at the top and bottom of the hill, as well as its latent heat. The solution involves using the ideal gas equation and taking into account the change in pressure and density with altitude, with the pressure at sea level being used as an approximation for the pressure at the bottom of the hill.
  • #1
Telemachus
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Homework Statement


Here there is another problem from callen which I couldn't finish.

It says: It is found that a certain liquid boils at a temperature of 95ºC at the top of a hill, whereas it boils at a temperature of 105º at the bottom. The latent heat is 1000cal/mole. What is the approximate height of the hill?

So this is what I did:
[tex]\frac{\Delta P}{\Delta T}=\frac{l}{T_b\Delta V_{\alpha,\beta}}\rightarrow \frac{\Delta P}{\Delta T}=\frac{l P_1}{T_b^2 R}\rightarrow \Delta P=P_1-P_2=\frac{1000\frac{cal}{mol}P_1\Delta T}{T_b^2 R}\rightarrow P_2=0.15P_1[/tex]
Where [tex]T_b[/tex] is the boiling point temperature. I've used the ideal gas equation [tex]V=\frac{RT}{P}[/tex].
l is the latent heat, which is known.
Then:
[tex]\Delta T=10K[/tex]
[tex]\Delta P= g h \rho[/tex], rho is the air density
[tex]\rho=1.22\frac{kg}{m^3}[/tex]

Then [tex]\Delta P=0.85P_b=11.956\frac{kg}{m^2s^2}h[/tex]

I need the information of the pressure at the bottom of the hill for finishing. Anyway, I'm not sure all the steps are right, and I wanted you to see this. I think I could use the pressure at sea level as an approximation.
 
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  • #2
Telemachus said:

Homework Statement


Here there is another problem from callen which I couldn't finish.

It says: It is found that a certain liquid boils at a temperature of 95ºC at the top of a hill, whereas it boils at a temperature of 105º at the bottom. The latent heat is 1000cal/mole. What is the approximate height of the hill?

So this is what I did:
[tex]\frac{\Delta P}{\Delta T}=\frac{l}{T_b\Delta V_{\alpha,\beta}}\rightarrow \frac{\Delta P}{\Delta T}=\frac{l P_1}{T_b^2 R}\rightarrow \Delta P=P_1-P_2=\frac{1000\frac{cal}{mol}P_1\Delta T}{T_b^2 R}\rightarrow P_2=0.15P_1[/tex]
Where [tex]T_b[/tex] is the boiling point temperature. I've used the ideal gas equation [tex]V=\frac{RT}{P}[/tex].
l is the latent heat, which is known.
Then:
[tex]\Delta T=10K[/tex]
[tex]\Delta P= g h \rho[/tex], rho is the air density
[tex]\rho=1.22\frac{kg}{m^3}[/tex]

Then [tex]\Delta P=0.85P_b=11.956\frac{kg}{m^2s^2}h[/tex]

I need the information of the pressure at the bottom of the hill for finishing. Anyway, I'm not sure all the steps are right, and I wanted you to see this. I think I could use the pressure at sea level as an approximation.
is there anyone who could solve this problem?
 
  • #3
I haven't gone over the rest of your calculations in detail, but you should be using the pressure at sea level (1Bar) as the pressure at the bottom of the hill. There is a more accurate way of doing this problem by taking into account the change in density with altitude, but this probably isn't needed here. In any event, after you complete the calculation, you can recalculate the density at the top and bottom of the hill to see if you judge that it really matters.
 

FAQ: Latent heat and Clapeyron equation

What is latent heat?

Latent heat is the energy required to change the state of a substance without changing its temperature. It is the heat absorbed or released during a phase change, such as melting, vaporization, or sublimation.

What is the Clapeyron equation?

The Clapeyron equation is a thermodynamic equation that describes the relationship between the temperature and pressure at which a substance changes state. It is used to calculate the latent heat of a substance and the slope of its phase boundary on a phase diagram.

How is the Clapeyron equation derived?

The Clapeyron equation is derived from the first and second laws of thermodynamics, along with the ideal gas law. It assumes that the substance is in equilibrium and that the heat of vaporization or fusion is constant over the range of temperatures and pressures being considered.

What is the significance of the Clapeyron equation?

The Clapeyron equation is important because it allows us to predict the conditions at which a substance will change from one state to another. It also helps us understand the behavior of substances under different temperatures and pressures, and is useful in various engineering and scientific applications.

Can the Clapeyron equation be applied to all substances?

No, the Clapeyron equation cannot be applied to all substances. It is most accurate for ideal gases and some simple substances, but becomes less accurate for more complex substances with strong intermolecular forces. Additionally, it assumes that the phase transition is reversible, which may not always be the case in real-world scenarios.

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