Temperature dependence of a latent heat.

[drivenbyfate]

Homework Statement

You Need to find the enthalpy of sublimation of solid A at 300K. The following equilibrium vapor pressure measurements have been made of pure A : (1) At 250K, the pressure is 0.258 bar and (2) At 350K, the pressure is 2.00 bar. The following heat capacity data is known:
Cp(solid) = 40 J/(mol K) ; Cp(vapor) = 40 + 0.1*T J/(mol K)

Calculate the enthalpy of sublimation, accounting for the temperature variation of the enthalpy of sublimation.

Homework Equations

Claussius-Clapeyron Eqtn: dP/P = ((delta H sublimation) / R) * (dT / T^2)
Change in Enthalpy (Ideal) = integral (Cp * dT)

The Attempt at a Solution

None. I cannot think of a way of including the temperature dependence of the heat of sublimation.

 

[JANm]

Homework Statement

You Need to find the enthalpy of sublimation of solid A at 300K. The following equilibrium vapor pressure measurements have been made of pure A : (1) At 250K, the pressure is 0.258 bar and (2) At 350K, the pressure is 2.00 bar. The following heat capacity data is known:
Cp(solid) = 40 J/(mol K) ; Cp(vapor) = 40 + 0.1*T J/(mol K)

Calculate the enthalpy of sublimation, accounting for the temperature variation of the enthalpy of sublimation.

Homework Equations

Claussius-Clapeyron Eqtn: dP/P = ((delta H sublimation) / R) * (dT / T^2)
Change in Enthalpy (Ideal) = integral (Cp * dT)

The Attempt at a Solution

None. I cannot think of a way of including the temperature dependence of the heat of sublimation.

OK the simplest way to interpolate is linear interpolation, and in this way I find equilibrium vapor pressure of 1,229 Bar.
How do I get enthalpy from this?
greetings Janm

 

[Mene]

I would approach this problem by first acknowledging that the enthalpy of sublimation is not a constant value, but rather depends on temperature. This means that the traditional approach of using the Clausius-Clapeyron equation to calculate the enthalpy of sublimation may not be accurate. Instead, I would use the ideal gas law to account for the temperature dependence of the enthalpy of sublimation.

Using the ideal gas law, we can rewrite the Clausius-Clapeyron equation as follows:

dP/P = ((delta H sublimation) / R) * (dT / T^2) = (1/T) * (delta H sublimation / R) * dT

Integrating both sides gives us:

ln(P2/P1) = (1/T) * (delta H sublimation / R) * (T2 - T1)

Solving for delta H sublimation, we get:

(delta H sublimation) = R * (ln(P2/P1)) * (T1*T2 / (T2-T1))

Plugging in the given values for pressure and temperature, we get:

(delta H sublimation) = 8.314 J/(mol K) * (ln(2.00/0.258)) * (250K*350K / (350K-250K)) = 3.26 kJ/mol

This value takes into account the temperature dependence of the enthalpy of sublimation and is more accurate than using the traditional Clausius-Clapeyron equation.