For the transition, rhombic sulfur → monoclinic sulfur, the value of $ΔS$ is positive. The transition temperature increases with increase in pressure. Which is denser, the rhombic or the monoclinic form? Prove your answer mathematically.

I tried by using Clapeyron equation but still failed. How should I approach and solve it? This is so tricky.


$$\frac{\mathrm dP}{\mathrm dT} = \frac{\Delta_\mathrm{trs}S}{\Delta_\mathrm{trs}V}$$

$\Delta_\mathrm{trs}S$ is positive. $\displaystyle\frac{\mathrm dP}{\mathrm dT}$ is said to be positive as well. Thus, $\Delta_\mathrm{trs}V$ has to be positive, i.e. monoclinic sulfur has a greater molar volume, and thus rhombic sulfur is denser.

  • 3
    $\begingroup$ Note that IUPAC recommends to denote "transition" with upright "trs" (subscript "tr" refers to either "transmitted" or "triple point") as subscript and there is no analogy to superscripted "transition state" notation (e.g. "≠" or "‡"). $\endgroup$ – andselisk Aug 3 '19 at 15:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.