Finding latent heat of vaporisation as a function of temperature

In summary, the author is asking for help with a homework problem that they are not familiar with, and is asking for help that is not provided in a clear or concise manner.
  • #1
Saitama
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Homework Statement


The vapour pressure of a certain liquid is given by the equation:
[tex]\log_{10}P=3.54595-\frac{313.7}{T}+1.40655\log_{10}T[/tex]
where ##P## is the vapour pressure in mm and T is temperature in kelvin. Determine the molar latent heat of vapourisation as a function of temperature. Calculate its value at 80 K.


Homework Equations





The Attempt at a Solution


I am really clueless about where to start. The liquid vapourises at a temperature when its vapour pressure is equal to the atmospheric pressure. I can calculate the boiling point from here but I don't think that would help here as I need molar latent heat of vapourisation as a function of temperature.

Any help is appreciated. Thanks!
 
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  • #2
Pranav-Arora said:

Homework Statement


The vapour pressure of a certain liquid is given by the equation:
[tex]\log_{10}P=3.54595-\frac{313.7}{T}+1.40655\log_{10}T[/tex]
where ##P## is the vapour pressure in mm and T is temperature in kelvin. Determine the molar latent heat of vapourisation as a function of temperature. Calculate its value at 80 K.


Homework Equations





The Attempt at a Solution


I am really clueless about where to start. The liquid vapourises at a temperature when its vapour pressure is equal to the atmospheric pressure. I can calculate the boiling point from here but I don't think that would help here as I need molar latent heat of vapourisation as a function of temperature.

Any help is appreciated. Thanks!

A liquid boils at the temperature at which its equilibrium vapor pressure is equal to atmospheric pressure. It is capable of vaporizing at temperatures below the boiling point, in which case its partial pressure in the gas phase is less than or equal to the equilibrium vapor pressure. The equilibrium vapor pressure is a function of temperature. We know that liquids evaporate at temperatures below the boiling point because, if we leave a small bowl of water out, it will eventually evaporate.

Now for the equation you presented. Look up the Clausius-Clapeyron equation in your physical chemistry or thermo book, or google it on the internet. You can use this equation to calculate the heat of vaporization from the relationship between the equilibrium vapor pressure and the temperature.
 
  • #3
Chestermiller said:
A liquid boils at the temperature at which its equilibrium vapor pressure is equal to atmospheric pressure. It is capable of vaporizing at temperatures below the boiling point, in which case its partial pressure in the gas phase is less than or equal to the equilibrium vapor pressure. The equilibrium vapor pressure is a function of temperature. We know that liquids evaporate at temperatures below the boiling point because, if we leave a small bowl of water out, it will eventually evaporate.

Now for the equation you presented. Look up the Clausius-Clapeyron equation in your physical chemistry or thermo book, or google it on the internet. You can use this equation to calculate the heat of vaporization from the relationship between the equilibrium vapor pressure and the temperature.

I checked the Clausius-Clapeyron equation on wikipedia.
[tex]\frac{dP}{dT}=\frac{L}{T\Delta v}[/tex]
But how am I supposed to find ##\Delta v## here? :confused:
 
  • #4
Pranav-Arora said:
I checked the Clausius-Clapeyron equation on wikipedia.
[tex]\frac{dP}{dT}=\frac{L}{T\Delta v}[/tex]
But how am I supposed to find ##\Delta v## here? :confused:

This is a good start, but, with all due respect to Wikipedia, this is not the Clausius-Clapeyron equation. This is the Clapeyron equation. The Clausius-Clapeyron makes two additional assumptions:
1. The molar volume of the liquid is much less than the molar volume of the vapor
2. The vapor can be treated as an ideal gas.

When these additional assumptions are made, one obtains:
[tex]\frac{dP}{dT}=\frac{Lp}{RT^2}[/tex]

or, equivalently,
[tex]\frac{d\ln{P}}{d(1/T)}=-\frac{L}{R}[/tex]
 
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  • #5
Chestermiller said:
or, equivalently,
[tex]\frac{d\ln{P}}{d(1/T)}=-\frac{L}{R}[/tex]

I don't see how would I get the ##log_{10} T## term.
Solving,
[tex]\ln P=\frac{-L}{RT}+C[/tex]
where C is some constant.
 
  • #6
Pranav-Arora said:
I don't see how would I get the ##log_{10} T## term.
Solving,
[tex]\ln P=\frac{-L}{RT}+C[/tex]
where C is some constant.

You can't integrate it this way because L is a function of T. You have to get L directly from the differential equation.
 
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  • #7
Chestermiller said:
You can't integrate it this way because L is a function of T. You have to get L directly from the differential equation.

Thank you! That worked. :smile:
 
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  • #9
Brijesh Upadhyay said:
You do realize that the last response to this thread was 5 years ago, and the OP has not been seen since last October, correct?
 
  • #10
Not to mention the fact posting full solutions is in general against forum rules.
 
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1. What is latent heat of vaporisation?

Latent heat of vaporisation is the amount of energy required to convert a substance from liquid to gas at a constant temperature.

2. How is latent heat of vaporisation measured?

To measure latent heat of vaporisation, the substance is heated at a constant rate until it reaches its boiling point. The amount of energy required to vaporise the substance is then calculated using the equation Q = mL, where Q is the energy, m is the mass of the substance, and L is the latent heat of vaporisation.

3. Why is latent heat of vaporisation important?

Latent heat of vaporisation is important because it plays a crucial role in many natural and industrial processes, such as evaporation, condensation, and distillation. It also affects the amount of energy required for a substance to change states, which can have significant impacts on the environment and energy consumption.

4. How does temperature affect latent heat of vaporisation?

As temperature increases, the latent heat of vaporisation decreases. This is because at higher temperatures, the molecules in a substance have more energy and are able to overcome the intermolecular forces that hold them together in the liquid state. As a result, less energy is required to convert the substance from liquid to gas.

5. What factors can affect the accuracy of measuring latent heat of vaporisation as a function of temperature?

The accuracy of measuring latent heat of vaporisation as a function of temperature can be affected by factors such as experimental errors, impurities in the substance, and changes in atmospheric pressure. It is important to carefully control these variables in order to obtain accurate results.

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