Question

A glass tube, sealed at both ends, is 100 cm long. It lies horizontally with the middle 10 cm containing mercury. The two ends of the tube contain air at 27°C and at a pressure 76 cm of mercury. The air column on one side is maintained at 0°C and the other side is maintained at 127°C. Calculate the length of the air column on the cooler side. Neglect the changes in the volume of mercury and of the glass.

Answer 1

$\begin{array}{l}\text{Let CSA of the tube be A.}\\ \text{On the colder side:}\\ P_1=0.76\text{m Hg}\\ T_1=300K\\ V_1=V\\ T_2=273K\\ V_2=Ax\\ \frac{P_{1}V}{T_1}=\frac{P_{2}Ax}{T_2}\\ \Rightarrow P_2=\frac{P_{1}V{T}_2}{T_{1}Ax}\\ \text{On the hotter side:}\\ P_1=0.76\text{m Hg}\\ T_1=300K\\ {V_1}^{\text{'}}=V\\ {T_2}^{\text{'}}=400K\\ {V_2}^{\text{'}}=Ay\\ \frac{{P_1}^{\text{'}}V}{T_1}=\frac{{P_2}^{\text{'}}Ay}{{T_2}^{\text{'}}}\\ \Rightarrow {P_2}^{\text{'}}=\frac{P_{1}V{T_2}^{\text{'}}}{T_{1}Ay}\\ \text{In equilibrium, the pressures on both side will balance each other.}\\ \Rightarrow {P_2}^{\text{'}}=P_2\\ \Rightarrow \frac{P_{1}V{T_2}^{\text{'}}}{T_{1}Ay}=\frac{P_{1}V{T}_2}{T_{1}Ax}\\ \Rightarrow \frac{{T_2}^{\text{'}}}{y}=\frac{T_2}{x}\\ \text{From the length of the tube, we get}\\ x+y+0.1=1\\ \Rightarrow y=0.9-x\\ \frac{400}{(0.9-x)}=\frac{273}{x}\\ \Rightarrow x=0.365\text{m}\\ \Rightarrow x=36.5\text{cm}\end{array}$