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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) TA=392.69 KT_A = 392.69 \text{ K}, TB=391.98 KT_B = 391.98 \text{ K}. (b) Different gases deviate from ideal behaviour. Reduce pressure to zero. **Step-by-Step Explanation:** (a) Using the constant-volume gas thermometer relation:
T=Ttp×PPtpT = T_{tp} \times \frac{P}{P_{tp}}
For thermometer A:
TA=273.16×1.797×1051.250×105=392.69 KT_A = 273.16 \times \frac{1.797 \times 10^5}{1.250 \times 10^5} = 392.69 \text{ K}
For thermometer B:
TB=273.16×0.287×1050.200×105=391.98 KT_B = 273.16 \times \frac{0.287 \times 10^5}{0.200 \times 10^5} = 391.98 \text{ K}
(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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