Question

at constant volume Gas thermometer using sulphur records pressure of 2*10 ^4Pascal at triple point of water and 2.87 *ten^ 4 pascals at temperature of melting sulphur. calculate the melting point of sulphur.

Answer 1

$$\begin{gathered} DearStudent, \\ Triplepointofwater=T_{tr}=273.16K \\ Pressureatthetriplepointofwater{P}_{tr}=2\times {10}^{4}Pa \\ Pressureatthetempratureofmeltingsulphor=2.87\times {10}^{4}Pa \\ weknowthat \\ \frac{P_{tr}}{P}=\frac{T_{tr}}{T} \\ ThereforetempratureofMeltingsulphorT=\frac{P_{tr}}{P}\times T_{tr} \\ T=\frac{2.87\times {10}^{4}Pa}{2\times {10}^{4}Pa}\times 273.16K \\ T=1.434\times 273.16K=391.98K \\ Regards \end{gathered}$$