Verified Solution • 2025–26 Session
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Q10.5: NCERT Physics - Part 2 THERMAL PROPERTIES OF MATTER 10.5 Solution: Two ideal gas thermometers A and B use oxygen and hydrogen…
⚠️ Common Exam Traps & Conceptual Pitfalls
Students frequently interpret the pressure values like “1.250 _ 10 Pa” as $1.250\times10\,$Pa (≈12.5 Pa) instead of the intended $1.250\times10^{5}$ Pa, which corrupts the $T=T_{tp}\frac{P}{P_{tp}}$ calculation. Always confirm the exponent and keep all pressures in the same units before applying the linear gas‑thermometer relation.
Two ideal gas thermometers A and B use oxygen and hydrogen respectively. The following observations are made: Temperature: Triple-point of water: Pressure thermometer A = 1.250 _ 10_ Pa, Pressure thermometer B = 0.200 _ 10_ Pa. Normal melting point of sulphur: Pressure thermometer A = 1.797 _ 10_ Pa, Pressure thermometer B = 0.287 _ 10_ 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 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?
Step-by-Step Verified Solution
**Answer:** (a) , . (b) Different gases deviate from ideal behaviour. Reduce pressure to zero.
**Step-by-Step Explanation:**
(a) Using the constant-volume gas thermometer relation:
For thermometer A:
For thermometer B:
(b) The slight difference is because oxygen and hydrogen do not behave perfectly ideally. To reduce the discrepancy, measurements should be taken at lower pressures and extrapolated to zero pressure. [citation:7]
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