Question

Two ideal gas thermometers A and B use oxygen and hydrogen respectively. The following observations are made

Temperature	Pressure thermometer A	Pressure thermometer B
Triple-point of water	$1.250\times {10}^{5}Pa$	$0.200\times {10}^{5}Pa$
Normal melting point of sulphur	$1.797\times {10}^{5}Pa$	$0.287\times {10}^{5}Pa$

(a) What is the absolute temperature of normal melting point of sulphur as read by thermometers A and B ?

(b) What do you think is the reason behind the slight difference in answers of thermometers A and B ? (The thermometers are not faulty) What further procedure is needed in the experiment to reduce the discrepancy between the two readings ?

Answer 1

(a)

Triple point of water, $T=273.16K$.

At this temperature, pressure in thermometer $A,P_A=1.25\times {10}^{5}Pa$

Let $T_1$ be the normal melting point of sulphur.

At this temperature, pressure in thermometer $A,P_1=1.797\times {10}^{5}Pa$
According to Charles law, we have the relation:

$P_A/T=P_1/T_1$

$T_1=\frac{1.797\times {10}^5\times 273.16}{1.25\times {10}^5}=392.69K$

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

(b)

At triple point 273.16 K, the pressure in thermometer B, $P_B=0.2\times {10}^{5}Pa$
At temperature $T_1$, the pressure in thermometer B, $P_2=0.287\times {10}^{5}Pa$
According to Charles law, we can write the relation:
$P_B/T=P_1/T_1$

$0.2\times {10}^5/273.16=0.287\times {10}^5/T_1$
$T_1=\left(\frac{0.287\times {10}^5}{0.2\times {10}^5}\right)\times 237.16=391.98K$

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