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Calculating Enthalpy of Vaporization using Clapeyron Equation

  1. Feb 28, 2013 #1
    Hi everyone. I am having a problem that hopefully someone here can help me with. For the purposes of flash calculations, I’m trying to find the enthalpy of vaporization of a compound using the Antoine equation and Clapeyron equation. I am using heptane at 15.5597 psia as an example. For the Antoine equation in the form log10(Psat) = A-B/(Tsat+C), where Psat is in units of mmHg and Tsat is in units of °C, I have Antoine coefficients for heptane of A = 6.89385, B = 1264.37, and C = 216.636. At Psat = 15.5597 psia = 804.592 mmHg, this gives me a saturation temperature of Tsat = 100.386 °C. The Clapeyron equation for a vapor-liquid phase change is dPsat/dTsat=Δhvap/(TsatΔνvap). Rearranging yields Δhvap = (dPsat/dTsat)(Tsat•Δνvap). Using the Rackett equation and the Peng-Robinson equation of state, I have found within reasonable accuracy that Δνvap = 4.385 ft3/lb. Taking the derivative of the Antoine equation with respect to temperature yields dPsat/dTsat = BPsatln(10)/(C+Tsat)2. Thus, Δhvap = BPsatTsatΔνvapln(10)/(C+Tsat)2. Plugging everything in results in a value of Δhvap = 198.348 ft3•psi/lb, which is equivalent to 36.703 BTU/lb. The actual value of Δhvap for heptane at 15.5597 psia should be somewhere around 135 BTU/lb. I’ve looked over my method and calculations multiple times but I can’t find an error. Perhaps the Antoine equation cannot be differentiated because it uses empirical parameters? Thanks. Any help would be much appreciated.
  2. jcsd
  3. Feb 28, 2013 #2
    In the Clapyron equation, you did remember that Tsat is absolute temperature, correct?
  4. Feb 28, 2013 #3
    Perfect, that's exactly what I did wrong. I left the Tsat in the denominator in °C because the Antoine coefficients are for temperatures in °C, but I converted the other Tsat to Kelvin and now I'm getting 136.6 BTU/lb. Thanks a lot. It didn't make sense to me at first that I should convert to Kelvin because then the temperature units wouldn't seem to cancel out (K/ °C). But then I remembered that "per degree celsius" is the same as "per degree Kelvin" because a degree Kelvin is equal to a degree Celsius. Thanks again.
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