Two ideal gas thermometer `A` and `B` use oxygen and hydrogen respectively . The following observations are made:
Temperature, Pressure therometer A, Pressure therometer B
Triple point of water, `1.250xx10^(5)Pa`, `0.200xx10^(5)Pa`
Normal melting point of sulphur, ` 1.797xx10^(5)Pa`, `0.287xx10^(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 for the slightly different answers from `A` and `B` ? (The thermometers are not faulty). what further procedure is needed in the experiment to reduce the discrepancy between the two readings.

Explanation

(a) Triple point of water, T = 273.16 K.
At this temperature, pressure in thermometer `A, P_(A) = 1.250 xx 10^(5) `Pa
Let `T_(1)` be the normal melting point of sulphur.
At this temperature, pressure in thermometer` A, P_(1) = 1.797x× 10^(5)` Pa
According to Charles’ law, we have the relation:
`(P_(A))/T=(P_(1))/(T_(1))`
`:.T_(1)=(P_(1)T)/(P_(A))=(1.797xx10^(5)xx273.16)/(1.250xx10^(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 ×x 10^(5)` Pa
At temperature T1, the pressure in thermometer `B, P_(2) = 0.287 x× 10^(5)` Pa
According to Charles’ law, we can write the relation:
`(P_(B))/T=(P_(1))/(T_(1))`
`(0.200xx10^(5))/(273.16)=(0.287xx10^(5))/(T_(1))`
`:. T_(1)=(0.287xx10^(5))/(0.200xx10^(5))xx273.16=391.98 K`
Therefore, the absolute temperature of the normal melting point of sulphur as read by thermometer B is 391.98 K.
(b) 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.