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.
Top Questions on Thermal Expansion
- A uniform metal solid sphere is rotating with angular speed \( \omega_0 \) about its diameter. If the temperature is raised by 50°C, the angular speed will be: Given \( \alpha_{\text{metal}} = 20 \times 10^{-5} \, \text{°C}^{-1} \)
- AP EAMCET - 2024
- Physics
- Thermal Expansion
- A metal rod of length 1 m at 20°C is made up of a material of coefficient of linear expansion \( 2 \times 10^{-5} \, \text{°C}^{-1} \). The temperature at which its length is increased by 1 mm is:
- KEAM - 2024
- Physics
- Thermal Expansion
- When water is heated from $0^\circ$C to $8^\circ$C, its volume
- KEAM - 2024
- Physics
- Thermal Expansion
- On a temperature scale X, the boiling point of water is 65° X and the freezing point is 15° X. Assume that the X scale is linear. The equivalent temperature corresponding to 95° X on the Fahrenheit scale would be:
- JEE Main - 2023
- Physics
- Thermal Expansion
A thin rod having a length of \(1 m\) and area of cross-section \(3 \times 10^{-6} m ^2\) is suspended vertically from one end. The rod is cooled from \(210^{\circ} C\) to \(160^{\circ} C\) After cooling, a mass \(M\) is attached at the lower end of the rod such that the length of rod again becomes \(1m\). Young's modulus and coefficient of linear expansion of the rod are \(2 \times 10^{11} N m ^{-2}\) and \(2 \times 10^{-5} K ^{-1}\), respectively. The value of \(M\) is ____ \(kg\). (Take \(g =10 \ m s ^{-2}\))
- JEE Main - 2023
- Physics
- Thermal Expansion
Questions Asked in CBSE Class XI exam
- If U = {a, b, c, d, e, f, g, h}, find the complements of the following sets:
(i) A = {a, b, c}
(ii) B = {d, e, f, g}
(iii) C = {a, c, e, g}
(iv) D = {f, g, h, a}- CBSE Class XI
- Complement of a Set
- Consider the following species: N3- , O2-, F-, Na+, Mg2+ and Al3+
- What is common in them?
- Arrange them in the order of increasing ionic radii.
- CBSE Class XI
- Modern Periodic Law And The Present Form Of The Periodic Table
- Find and write down structures of 10 interesting small molecular weight biomolecules. Find if there is any industry which manufactures the compounds by isolation. Find out who are the buyers.
- CBSE Class XI
- How To Analyse Chemical Composition?
- How many litres of water will have to be added to 1125 litres of the 45% solution of acid so that the resulting mixture will contain more than 25% but less than 30% acid content?
- CBSE Class XI
- inequalities
Find the mean and variance for the following frequency distribution.
Classes 0-10 10-20 20-30 30-40 40-50 Frequencies 5 8 15 16 6 - CBSE Class XI
- Variance and Standard Deviation
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.