Solution:

(a) Given:

Triple point of water, T= 273.16K.

Pressure in thermometer A at the  triple point, P_A =1.25×10⁵ Pa

Normal melting point of sulphur = T₁

Pressure in thermometer A at this temperature, P₁=1.797×10⁵ Pa

According to Charles law, the following relationship exists:

P_A/T=P₁/T₁

T₁ = P₁ T/P_A

= (1.797×10⁵×273.16)/1.25×10⁵
​
=392.69 K

This means that while using thermometer A, we can determine the absolute temperature of sulfur’s typical melting point: 392.69 K.

(b) The pressure in thermometer B at the triple point of water, P_B =0.2×10⁵ Pa

The temperature in thermometer B at the normal melting point (T₁) of sulfur is, P₁ =  0.287×10⁵ Pa

According to Charles law, we can write the relation:
P_B/T=P₁/T₁
T₁ = P₁ T/P_B​​

T₁ = (0.287×10⁵ × 273.16)/0.2×10⁵
=391.98 K

As a result, the absolute temperature of sulfur’s usual melting point, as measured by thermometer B, was determined as 391.98 K.

(b)