- #1
johnsmith456
- 5
- 0
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
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