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 × 10$^5$ Pa	0.200 × 10$^5$ Pa
Normal melting point of sulphur	1.797× 10$^5$ Pa	0.287 × 10$^5$ Pa

1. What is the absolute temperature of normal melting point of sulphur as read by thermometers A and B?
2. 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 = 273.16 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.

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(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.