| Temperature | Pressure thermometer A | Pressure thermometer B |
| Triple-point of water | 1.250 × 10\(^5\) Pa | 0.200 × 10\(^5\) Pa |
| Normal melting point of sulphur | 1.797× 10\(^5\) Pa | 0.287 × 10\(^5\) Pa |
(a) Triple point of water, T = 273.16 K.
At this temperature, pressure in thermometer A, \(P_A\)= 1.250 × 10\(^5\) Pa
Let \(T_1\) be the normal melting point of sulphur.
At this temperature, pressure in thermometer A, \(P_1\)= 1.797 × 10\(^5\) Pa
According to Charles’ law, we have the relation:
\(\frac{PA}{T}= \frac{P1}{T1}\)
\(T_1 = \frac{P_1T}{P_A}\) = \(\frac{1.797\times 10^5\times 273.16}{1.250\times 10^5}\)
= 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 T1, the pressure in thermometer B, \(P_2\) = 0.287 × 10\(^5\) Pa
According to Charles’ law, we can write the relation:
\(\frac{PB}{T}=\frac{P_1}{T_1}\)
\(\frac{0.200\times10^5}{273.16}\)= \(\frac{0.287\times 10^5}{T_1}\)
\(T_1\) = \(\frac{0.287\times10^5}{0.200\times10^5}\) × 273.16 = 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.