Let the normal melting point of sulphur = ${\mathrm{T}}_1$

At this temperature, pressure in thermometer A, ${\mathrm{P}}_1$ = 1.797 × 10${}^5$ Pa

According to Charles’ law,

$\frac{{\mathrm{P}}_{\mathrm{A}}}{\mathrm{T}}=\frac{{\mathrm{P}}_1}{{\mathrm{T}}_1}$

$\therefore {\mathrm{T}}_1=\frac{{\mathrm{P}}_1\mathrm{T}}{{\mathrm{P}}_{\mathrm{A}}}=\frac{1.797\times {10}^5\times 273.16}{1.250\times {10}^5}$= 392.69 K

Therefore, the absolute temperature of the normal melting point of sulphur as read by thermometer A is 392.69 K.

At triple point 273.16 K, the pressure in thermometer B, P${}_B$ = 0.200 × 10${}^5$ Pa

At temperature ${\mathrm{T}}_1$, the pressure in thermometer B, P${}_2$ = 0.287 × 10${}^5$ Pa

According to Charles’ law,

$$\begin{gathered} \frac{{\mathrm{P}}_{\mathrm{B}}}{\mathrm{T}}=\frac{{\mathrm{P}}_1}{{\mathrm{T}}_1} \\ \frac{0.200\times {10}^5}{273.16}=\frac{0.287\times {10}^5}{{\mathrm{T}}_1} \\ \therefore {\mathrm{T}}_1=\frac{0.287\times {10}^5}{0.200\times {10}^5}\times 273.16=391.98\mathrm{K} \end{gathered}$$

Therefore, the absolute temperature of the normal melting point of sulphur as read by thermometer B is 391.98 K.

The oxygen and hydrogen gas present in thermometers A and B respectively are not perfect ideal gases. Hence, there is a slight difference between the readings of thermometers A and B.

To reduce the discrepancy between the two readings, the experiment should be carried under low-pressure conditions. At low pressure, these gases behave as perfect ideal gases.