Question

An open glass tube is immersed in mercury in such a way that a length of 8 cm extends above the mercury level. The open end of the tube is then closed and sealed and the tube is raised vertically up by additional 46 cm. What will be length of the air column above mercury in the tube now? (Atmospheric pressure = 76 cm of Hg)

• A. 16 cm
• B. 22 cm
• C. 38 cm
• D. 6 cm

Answer 1

Answer: (A)

16 cm

Answer 2

Let the initial state have trapped air column of length 8 cm at atmospheric pressure 76 cm Hg. Thus,

$$P_1V_1 = 76 \times 8.$$

After sealing, the tube is raised by 46 cm so that the total height of the sealed tube above the external mercury level becomes

$$8 + 46 = 54 \text{ cm}.$$

Let the new air column (inside) have length $x$ cm. Then the mercury column inside the tube is

$$54 - x \text{ cm}.$$

At the mercury–air interface, the pressure in the trapped air equals the external pressure reduced by the mercury column height:

$$P_2 = 76 - (54 - x) = x + 22.$$

Since the process is isothermal,

$$P_1V_1 = P_2V_2 \quad \Longrightarrow \quad 76\times 8 = (x+22)\times x.$$

This gives:

$$608 = x^2 + 22x \quad \Longrightarrow \quad x^2 + 22x -608 = 0.$$

The discriminant is:

$$\Delta = 22^2 + 4\times608 = 484 + 2432 = 2916,\quad \sqrt{2916}=54.$$

Thus,

$$x = \frac{-22\pm54}{2}.$$

Taking the positive value:

$$x = \frac{32}{2} = 16\,\text{cm}.$$