#### Question

Two ideal gas thermometers Aand Buse 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 |

(a) What is the absolute temperature of the 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?

#### Solution 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:

`P_A/T = P_1/T_1`

`:. T_1 =( P_1T)/P_A = (1.797 xx 20^(5) xx 273.16)/(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 × 10^{5} Pa

At temperature *T*_{1}, the pressure in thermometer B, *P*_{2} = 0.287 × 10^{5} Pa

According to Charles’ law, we can write the relation:

`P_B/T = P_1/T_1`

`(0.200xx10^5)/273.16 = (0.287 xx 10^5)/T_1`

`:. T_1 = (0.287xx10^5)/(0.200xx10^5) xx 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.

#### Solution 2

Let T be the melting point of sulphur for thermometer A

`P_"tr" = 1.250 xx 10^5 Pa`; `P = 1.797 xx 10^5 Pa`

Now `T_A = T_(tr) xx P/P_"tr"`

`T_A = (273.16 xx 1.797 xx 10^5)/(.250xx10^5)k = 392.69 K`

For thermometer b

`P_"tr" = 0.200 xx 10^5 Pa`; `P = 0.287 xx 10^5 `Pa

`T_B = = T_"tr" xx P/P_"tr" = (273.16 xx 0.287 xx 10^5)/(0.200xx10^5) K`

b) The value of the melting point of sulphur found from the two thermometers differ slightly due to the reason that in practice, the gases do not behave strictly as perfect gases i.e., gases are not perfectly ideal.

To reduce the discrepency, readings should be taken for lower and lower pressures and the plot between temperature measured versus absolute pressure of the gas at triple point should be extrapolated to obtain the temperature in the limit pressure tends to zero (if P —> 0), when the gases approach ideal gas behaviour.