Clapeyron Equation and the triple point (Thermo)

[Telemachus]

Homework Statement

Hi there. I was trying to solve this problem from Callen. It says: In the vecinity of the triple point the vapor pressure of liquid ammonia (in Pascals) is rerpesented by:

[tex]\ln P=24.38-\frac{3063}{T}[/tex]

This is the equation of the liquid-vapor boundary curve in a P-T diagram. Similarly, the vapor pressure of solid ammonia is

[tex]\ln P=27.92-\frac{3754}{T}[/tex]

a) What are the temperature and pressure at the triple point?
b) what is the latent heat of fusion at the triple point?

Homework Equations

I think a relevant equation is
[tex]d(ln P)=-\frac{\Delta h}{R}d\left ( \frac{1}{T} \right ) [/tex]

The Attempt at a Solution

Well, I think I've solved a)
At the triple point we have:
[tex]\ln P=27.92-\frac{3754}{T}=24.38-\frac{3063}{T}[/tex]
Then [tex]T=195.19[/tex]
And [tex]P=5928.9[/tex]

I don't know how to solve b), I think I should work with the equation I've posted in Relevant equations.

 

[anathema]

Hello,

it is important to approach problems with a critical and analytical mindset. First, let's analyze the given information and equations.

The equations for the vapor pressures of liquid and solid ammonia at the triple point are logarithmic functions of temperature. This means that as temperature increases, the vapor pressure decreases. Additionally, the equation for the solid ammonia has a higher coefficient than that of liquid ammonia, indicating that solid ammonia has a lower vapor pressure than liquid ammonia at the same temperature.

a) To find the temperature and pressure at the triple point, we can use the given equations and set them equal to each other, as you have done in your attempt. However, we also need to take into account the fact that at the triple point, all three phases (solid, liquid, and vapor) coexist in equilibrium. This means that the vapor pressure of solid and liquid ammonia must be equal to the vapor pressure of the triple point. So, we can set the two equations equal to each other and solve for T and P:

\ln P_{triple}=27.92-\frac{3754}{T}=24.38-\frac{3063}{T}

Solving for T, we get T = 195.19 K. Substituting this value back into either equation, we can find the vapor pressure at the triple point to be P_{triple}=5928.9 Pa.

b) To find the latent heat of fusion at the triple point, we can use the Clausius-Clapeyron equation, which relates the latent heat of fusion to the slope of the vapor pressure curve at the triple point. This equation is given by:

\frac{\Delta H_{fusion}}{R}=-\frac{d\ln P}{d(1/T)}

We can use the given equations for the vapor pressures of liquid and solid ammonia to calculate the slope at the triple point. Taking the derivative of both equations with respect to (1/T), we get:

\frac{d\ln P_{liquid}}{d(1/T)}=\frac{3063}{T^2}

\frac{d\ln P_{solid}}{d(1/T)}=\frac{3754}{T^2}

Substituting T=195.19 K, we get \frac{d\ln P_{liquid}}{d(1/T)}=0.0083 and \frac{d\ln P_{solid