Question: A gas thermometer is constructed of two gas-containing bulbs, each in a water bath, as shown in Figure. The pressure difference between the two bulbs is measured by a mercury manometer as shown. Appropriate reservoirs, not shown in the diagram, maintain constant gas volume in the two bulbs. There is no difference in pressure when both baths are at the triple point of water. The pressure difference is 120 torr when one bath is at the triple point and the other is at the boiling point of water. It is 90.0 torrwhen one bath is at the triple point and the other is at an unknown temperature to be measured. What is the unknown temperature?

1. When both baths are at triple point, $\Delta \text{P}=0$
2. When one bath is at triple point and other is at the boiling point of water, $\Delta \text{P}=120\text{torr}$
3. When one bath is at triple point and other is at unknown temperature T$\Delta \text{P}=90\text{torr}$,

By using the concept of a constant volume gas thermometer and applying temperature-pressure relation to both gases, find the unknown temperature.

Formula:

For constant volume of thermometer of a monatomic gas, the pressure relation to the temperature is given as:

$\frac{T}{P}=\frac{T_3}{P_3}$ …… (i)

From the diagram, let thermometer on the left side be L and that on the right side beR.

Then temperature and pressure of the left side gas is $T_L,P_L$.

Temperature and pressure of the right-side gas is$T_R,P_R$.

For both gases are at triple point, the left and right values are given as:

$$\begin{gathered} T_L\text{(or}{T}_R)=T_3 \\ {P}_L=P_R \\ \Delta P=0 \end{gathered}$$

For the left side gas is at triple point and right side gas is at boiling point of water,

$$\begin{gathered} T_L=T_3,P_L=P_3 \\ {T}_R=373°\text{K} \\ \Delta \text{P}=120\text{torr} \end{gathered}$$

Using equation (i), the pressure at both the left and the right ends is given as:

$$\begin{gathered} \text{Pressure at the left end,}{P}_L=P_3\times \frac{T_L}{T_3} \\ \text{Pressure at the right end,}{P}_R=P_3\times \frac{T_R}{T_3} \end{gathered}$$

So, the pressure difference at both ends is given as:

$$\begin{gathered} P=P_R-P_L \\ 120=P_3\left(\frac{T_R}{T_3}-\frac{T_L}{T_3}\right) \\ 120=P_3\left(\frac{T_R}{T_3}-1\right) \\ 120=P_3\left(\frac{T_R}{T_3}-1\right) \end{gathered}$$ …… (ii)

As the left side gas is at triple point and right side gas is at unknown temperature is as follows:

$$\begin{gathered} T_L=T_3,P_L=P_3 \\ {T}_R=T \end{gathered}$$

Using equation (i), the pressure at both the left and the right ends is given as:

$$\begin{gathered} \text{Pressure at left end,}{P}_L=P_3\times \frac{T_L}{T_3} \\ \text{Pressure at right end,}{P}_R=P_3\times \frac{T}{T_3} \end{gathered}$$

So, the pressure difference at both ends is given as:

$$\begin{gathered} P=P_R-P_L \\ =P_3\left(\frac{T}{T_3}-\frac{T_L}{T_3}\right) \\ =P_3\left(\frac{T}{T_3}-1\right) \\ 90=P_3\left(\frac{T}{T_3}-1\right) \end{gathered}$$ …… (iii)

Taking ratio of equation (ii) and (iii), we get

$$\begin{gathered} \frac{120}{90}=\frac{\frac{T_R}{T_3}-1}{\frac{T}{T_3}-1} \\ \frac{4}{3}=\frac{T_R-T_3}{T-T_3} \\ \frac{4}{3}=\frac{373-273.16}{\text{T}-273.16}\text{(}\because {\text{T}}_R=373°{\text{K andT}}_3=273.16°\text{K)} \\ 4\left(T-273.16\right)=3\left(373-273.16\right) \end{gathered}$$

Solve further as:

$$\begin{gathered} 4T=273.16+3\left(373\right) \\ T=\frac{\left(273.16+1119\right)}{4} \\ T=348.04\text{K} \\ T\approx 348\text{K}\backslash \end{gathered}$$

Hence, the unknown temperature is 348 K.