Question: Suppose the temperature of a gas iswhen it is at the boiling point of water. What then is the limiting value of the ratio of the pressure of the gas at that boiling point to its pressure at the triple point of water? (Assume the volume of the gas is the same at both temperatures).

The temperature of the gas is 373.15 K at the boiling point of the water.

By using Equation 18-6 for the temperature as measured with a gas thermometer, it is observed that the limiting value of the pressure ratio is the same as the absolute temperature ratio. Using this, find the limiting value of the ratio of the pressure of the gas at boiling point to its pressure at the triple point of the water.

Formula:

From Equation 18-6, the temperature T as measured with a gas thermometer is given by:

$T=(273.16K)\left(\frac{p}{p_3}\right)$
$T=(273.16K)\left(\frac{p}{p_3}\right)$
$T=(273.16K)\left(\frac{p}{p_3}\right)$

Using formula from equation (i), the ratio of the pressure of the gas at that boiling point to its pressure at the triple point of water is given as:
$$\begin{gathered} \left(\frac{p}{p_3}\right)=\frac{T}{(273.16K)} \\ =\frac{373.15K}{(273.16K)} \\ =1.366 \end{gathered}$$

Therefore, the limiting value of the ratio of the pressure of the gas at boiling point to its pressure at the triple point of the water is 1.366.