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
- What is the absolute temperature of normal melting point of sulphur as read by thermometers A and B?
- 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?
| 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 |
- What is the absolute temperature of normal melting point of sulphur as read by thermometers A and B?
- 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?
Solution and Explanation
(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.
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Concepts Used:
Ideal Gas Equation
An ideal gas is a theoretical gas composed of a set of randomly-moving point particles that interact only through elastic collisions.
What is Ideal Gas Law?
The ideal gas law states that the product of the pressure and the volume of one gram molecule of an ideal gas is equal to the product of the absolute temperature of the gas and the universal gas constant.
PV=nRT
where,
P is the pressure
V is the volume
n is the amount of substance
R is the ideal gas constant
Ideal Gas Law Units
When we use the gas constant R = 8.31 J/K.mol, then we have to plug in the pressure P in the units of pascals Pa, volume in the units of m3 and the temperature T in the units of kelvin K.