- #1
ChrisW
- 8
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Ok so i was doing a problem involving finding the pressure of mercury at its boiling point (630.05K) using Troutons rule and the final answer seems a bit strange to me.
Integration of the Clausius-Clapeyron Equation:
[tex]{ln(\frac{{P}_{2}}{{P}_{1}}) = -\frac{{\Delta}_{vap}\overline{H}}{R}(\frac{1}{T}_2-\frac{1}{T}_1)[/tex]
So... you end up with a number, but it's unitless? What am I missing? Thanks.
Troutons rule states that at standard T and P:
[tex]{\Delta}_{vap}\overline{S} \approx {88} {J}\cdot{K}^{-1}\cdot{mol}^{-1}[/tex]
and the change in entropy is related to the change in enthalpy by:
[tex]{\Delta}_{vap}\overline{H}={\Delta}_{vap}\overline{S}\cdot{T}_{vap}[/tex]
Integration of the Clausius-Clapeyron Equation:
[tex]{ln(\frac{{P}_{2}}{{P}_{1}}) = -\frac{{\Delta}_{vap}\overline{H}}{R}(\frac{1}{T}_2-\frac{1}{T}_1)[/tex]
So... you end up with a number, but it's unitless? What am I missing? Thanks.
Troutons rule states that at standard T and P:
[tex]{\Delta}_{vap}\overline{S} \approx {88} {J}\cdot{K}^{-1}\cdot{mol}^{-1}[/tex]
and the change in entropy is related to the change in enthalpy by:
[tex]{\Delta}_{vap}\overline{H}={\Delta}_{vap}\overline{S}\cdot{T}_{vap}[/tex]