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QPages 12-14: Q4: HC Verma: Concepts of Physics - Volume 2 Heat and Temperature Pages 12-14: Q4 Solution: The pressure measured by a constant volume gas thermometer is…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently substitute 0 °C and 100 °C directly into the $P/T$ ratio, forgetting that the ideal‑gas relationship requires absolute temperature in kelvins; always convert $T(°C)$ to $T(K)=T(°C)+273.15$ before applying $P\propto T$.

The pressure measured by a constant volume gas thermometer is 40 kPa at the triple point of water. What will be the pressure measured at the boiling point of water 100�C)?

Step-by-Step Verified Solution

**Answer:**

P = 54.6 \text{ kPa}

**Step-by-Step Explanation:**

T_{tp} = 273.16

K,

P_{tp} = 40

kPa. Boiling point:

T = 373.15

K. Using

P/T = \text{constant}

:

P = P_{tp} \times \frac{T}{T_{tp}} = 40 \times \frac{373.15}{273.16} = 54.6 \text{ kPa}

All Questions in this Chapter

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