Question

The pressure measured by a constant volume gas thermometer is $40kPa$ at the triple point of water. What will be the pressure measured at the boiling point of water (${100}^{0}C$)?

Answer 1

Given,

Pressure measured by a constant volume gas thermometer at the triple point of water,

$P_{tr}=40kPa=40\times {10}^{3}Pa$

The boiling point of water,$T=100{}^{\circ }C=373.16K$

Let the pressure measured at the boiling point of water be P.

For a constant volume gas thermometer, temperature pressure relation is given below

$T=\frac{P}{P_{tr}}\times 273.16K$

$\Rightarrow P=\frac{T\times P_{tr}}{273.16}$

$=\frac{373.16\times 40\times {10}^3}{273.16}=54620Pa$

$=5.42\times {10}^{3}Pa\approx 55KPa$