Answer to Question #118601 in Physics for Susan Williams

Question

A thread of mercury of length 16 cm is used to trap some air in a capillary tube of uniform cross – sectional area and closed at one end. When the tube is held vertically, with the closed end at the bottom, the length of the trapped air column is 30 cm. calculate the length of the air column when the tube is held,
1. Horizontally
2. Vertically with the open end underneath. [Atmospheric pressure = 76 cm of mercury

Answer 1

We can measure the pressure in cm because we will express pressures in terms of mercury and replace volumes of the trapped air with heights because the tube has uniform cross-section.

With the closed end down and "volume" of 30 cm, the pressure inside is

$$P_1=h_m+h_a,$$

where $h_m$ is pressure of the mercury and $h_a$ is the pressure of the atmosphere.

1) When the tube is held horizontally, the mercury is just a barrier, trapped air pressure equals the atmospheric ($P_2=P_0$ or $h_2=h_0$), the temperature is constant, according to Boyle's law:

$$P_1V_1=P_2V_2,\\ V_2=\frac{P_1V_1}{P_2}=\frac{h_1(h_m+h_0)}{h_0}=\frac{30(16+76)}{76}=36\text{ cm}.$$

2) When the tube is again turned 90°, the atmosphere pushes the mercury upward, the mercury tries to flow downward because of gravity:

$$P_3=P_0-P_m=h_0-h_m.$$

Therefore, the "volume" or height $V_3$ of the air in the tube is:

$$P_3V_3=P_1V_1,\\ V_3=\frac{P_1V_1}{P_3}=\frac{h(h_0+h_m)}{h_0-h_m}=\frac{30(76+16)}{76-16}=46\text{ cm}.$$