Clausius clapeyrion equation Hfg

  • #1

Main Question or Discussion Point

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
 

Answers and Replies

  • #2
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##\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.
 

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