Finding latent heat of vaporisation as a function of temperature

[Saitama]

Homework Statement

The vapour pressure of a certain liquid is given by the equation:
$$\log_{10}P=3.54595-\frac{313.7}{T}+1.40655\log_{10}T$$
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!

 

[Chestermiller]

Homework Statement

The vapour pressure of a certain liquid is given by the equation:
$$\log_{10}P=3.54595-\frac{313.7}{T}+1.40655\log_{10}T$$
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.

 

[Saitama]

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.
$$\frac{dP}{dT}=\frac{L}{T\Delta v}$$
But how am I supposed to find $\Delta v$ here?

 

[Chestermiller]

I checked the Clausius-Clapeyron equation on wikipedia.
$$\frac{dP}{dT}=\frac{L}{T\Delta v}$$
But how am I supposed to find $\Delta v$ here?

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:
$$\frac{dP}{dT}=\frac{Lp}{RT^2}$$

or, equivalently,
$$\frac{d\ln{P}}{d(1/T)}=-\frac{L}{R}$$

 

[Saitama]

or, equivalently,
$$\frac{d\ln{P}}{d(1/T)}=-\frac{L}{R}$$

I don't see how would I get the $log_{10} T$ term.
Solving,
$$\ln P=\frac{-L}{RT}+C$$
where C is some constant.

 

[Chestermiller]

I don't see how would I get the $log_{10} T$ term.
Solving,
$$\ln P=\frac{-L}{RT}+C$$
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.

 

[Saitama]

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.

 

[Brijesh Upadhyay]

 

[Chestermiller]

View attachment 225672

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?

 

[Borek]

Not to mention the fact posting full solutions is in general against forum rules.