Clausius clapeyrion equation Hfg

[johnsmith456]

Hi guys

I'm trying to figure out a way of calculating the enthalpy of vaporisation using the clapeyron equation.

So far I have two different forms of the clapeyron equation. However, I'm not sure how to go any further.

Ln(P2/P1) = (ΔHfg / R) ( 1/T1 - 1/T2) eqn(1)

Ln(P) = -(ΔHfg / R) (1/T) + C eqn(2)

Both of these will give the value of delta Hfg between two T's and P's. Whereas I want a particular value of Hfg at a given T and P. Is this possible?

Also from eqn (2), which is from the eqn of a straight line, give the gradient of the line to be (-Hfg/R). but surely Hfg isn't constant ?

i would have though Hfg decreases as T increases.

Anyway, the point of my question is to see how I can calculate Hfg NOT ΔHfg. I really hope this is possible?

Thanks very much in advance

kind Regards
JS

 

[Chestermiller]

$\Delta Hfg$ IS the heat of vaporization. If you want the heat of vaporization at temperature T, evaluate the slope over the region from $T+\delta T$ to $T-\delta T$ where $\delta T$ is say about 10 C. This will give you a fairly accurate numerical approximation to the derivative.