Is my proof valid for the formula of elevation in boiling point?


By Clausius clayperon equation, $$\ln(P)=-\frac{∆H_v}{RT}+c$$ Differentiating, $$\frac{dP}{P}=\frac{∆H_v dT}{RT^2}$$ Approximating $dP\approx ∆P$ , $dT\approx ∆T$, $$\frac{∆P}{P}=\frac{∆H_v∆T}{RT^2}$$ Let $∆P=P_{solution}-∆P_{solvent}$ and $T^2\approx T_{boiling}^2$ also $∆T=$ Change in boiling point,

By the formula for relative lowering of vapour pressure, $\frac{∆P}{P}=mf_{solute}$ $$mf_{solute}=\frac{∆H_v∆T_{boiling}}{RT_{boiling,solvent}^2}$$ $$∆T=\Big(\frac{RT_b^2}{∆H_v}\Big)mf_{solute}$$ But i don't know how to proceed further, can anyone help?

The actual formula differs from mine as, $$∆T=\Big(\frac{RT_b^2}{∆H_v}\Big)molality_{solute}$$


closed as unclear what you're asking by Mithoron, jerepierre, airhuff, Tyberius, Jon Custer Nov 13 '17 at 20:52

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  • $\begingroup$ I'm not sure that you can go further; is there a reason you expect to be able to do so? $\endgroup$ – a-cyclohexane-molecule Nov 13 '17 at 16:22
  • $\begingroup$ I updated the question @a-cyclohexane-molecule $\endgroup$ – neonpokharkar Nov 14 '17 at 7:04