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Question
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)?
Solution
Given:
Pressure measured by a constant volume gas thermometer at the triple point of water, Ptr = 40 kPa = 40 × 103 Pa
Boiling point of water, T = 100°C = 373.16 K
Let the pressure measured at the boiling point of water be P.
For a constant volume gas thermometer, temperature-pressure relation is given below:
\[T = \frac{P}{P_{tr}} \times 273 . 16 K\]
\[ \Rightarrow P = \frac{T \times P_{tr}}{273 . 16}\]
\[ \Rightarrow P = \frac{373 . 16 \times 40 \times {10}^3}{273 . 16}\]
\[ \Rightarrow P = 54643 Pa\]
\[ \Rightarrow P = 54 . 6 \times {10}^3 Pa \]
\[ \Rightarrow P \simeq 55 \ kPa\]
Therefore, the pressure measured at the boiling point of water is 55 kPa.
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