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	$1.250 \times {10^5}pa$	$0.200 \times {10^5}pa$
Normal melting point of sulphur	$1.797 \times {10^5}pa$	$0.287 \times {10^5}pa$

(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.

Answer 1

Ideal gas is a perfect gas in which there will be no interaction between the constituent molecules. One molecule will not be affected by the other molecule. These molecules occupy a negligible amount of space and they always obey gas laws. We solve this problem using one of the gas laws.

Formula used:
\eqalign{ & PV = nRT \cr & \dfrac{P}{T} = {\text{ constant}} \cr}

Complete step by step answer:
These pressure thermometers measured the pressures at the triple point of water temperature and at the normal melting point of sulphur.
Usually thermometers are used to measure the temperature. So here also we will measure the unknown temperature with two different thermometers by comparing with the known temperature by using Charles law.
The ideal gas equation states that
$PV = nRT$
Where P is the pressure of the ideal gas and V is the volume and ‘n’ is the number of moles and R is the universal gas constant and T the temperature.
Charles law states that as long as the number of moles and volume are constant for ideal gas, pressure will be proportional to the temperature. We use that law here.
$\dfrac{P}{T} = {\text{ constant}}$
Triple point of water is 273.16 kelvin. So for thermometer A
\eqalign{ & \dfrac{{{P_W}}}{{{T_W}}} = \dfrac{{{P_S}}}{{{T_S}}} \cr & \Rightarrow {T_S} = \dfrac{{{P_S}{T_W}}}{{{P_W}}} \cr & \Rightarrow {T_S} = \dfrac{{\left( {1.797 \times {{10}^5}pa} \right)\left( {273.16K} \right)}}{{1.250 \times {{10}^5}pa}} \cr & \therefore {T_{{S_A}}} = 392.69K \cr}
$\dfrac{P}{T} = {\text{ constant}}$
Triple point of water is 273.16 kelvin. So for thermometer B
\eqalign{ & \dfrac{{{P_W}}}{{{T_W}}} = \dfrac{{{P_S}}}{{{T_S}}} \cr & \Rightarrow {T_S} = \dfrac{{{P_S}{T_W}}}{{{P_W}}} \cr & \Rightarrow {T_S} = \dfrac{{\left( {0.287 \times {{10}^5}pa} \right)\left( {273.16K} \right)}}{{0.200 \times {{10}^5}pa}} \cr & \therefore {T_{{S_B}}} = 391.98K \cr}
Hence the normal melting point of sulphur as read by thermometers A and B are 392.69 and 391.98 kelvin respectively.
The reason for the slightly different answers from A and B are because the oxygen and hydrogen are not perfect ideal gases here. They behave as ideal gases at low pressure and very high temperature. So if we reduce the pressure then we can eliminate the disparities.

Note:
All the ideal gas laws we apply are valid only for the ideal gases. At low temperatures and high pressures there will be interactions between the molecules of the gases. Hence they deviate from the ideal gas behaviour. Usually gases behave as ideal gases at higher temperatures and lower pressures.