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)?

Answer 1

Given:
Pressure measured by a constant volume gas thermometer at the triple point of water, P_tr = 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:
$$\begin{gathered} T=\frac{P}{P_{tr}}\times 273.16\mathrm{K} \\ \Rightarrow P=\frac{T\times P_{tr}}{273.16} \\ \Rightarrow P=\frac{373.16\times 40\times {10}^3}{273.16} \\ \Rightarrow P=54643\mathrm{Pa} \\ \Rightarrow P=54.6\times {10}^3\mathrm{Pa} \\ \Rightarrow P\simeq 55\mathrm{kPa} \end{gathered}$$
Therefore, the pressure measured at the boiling point of water is 55 kPa.