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| Temperature | Pressure thermometer A | Pressure thermometer B |
| Triple-point of water | 1.250 × 105 Pa | 0.200 × 105 Pa |
| Normal melting point of sulphur | 1.797 × 105 Pa | 0.287 × 105 Pa |
(a) What is the absolute temperature of normal melting point of sulphur as read by thermometers Aand B?
(b) What do you think is the reason behind the slight difference in answers of thermometers Aand B? (The thermometers are not faulty). What further procedure is needed in the experiment to reduce the discrepancy between the two readings?
Asked by Harsha Last Modified
Answer 
Harshit Bhardwaj
(a) Triple point of water, T = 273.16 K.
At this temperature, pressure in thermometer A, PA = 1.250 × 105 Pa
Let T1 be the normal melting point of sulphur.
At this temperature, pressure in thermometer A, P1 = 1.797 × 105 Pa
According to Charles’ law, we have the relation:
= 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, PB = 0.200 × 105 Pa
At temperature T1, the pressure in thermometer B, P2 = 0.287 × 105 Pa
According to Charles’ law, we can write the relation:
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.
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