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.(adsbygoogle = window.adsbygoogle || []).push({});

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]

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# Natural log of a pressure is ?

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