91Ó°ÊÓ

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

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:

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

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

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

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

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

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:

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

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

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

Solve further as:

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

Hence, the unknown temperature is 348 K.