A thin tube of uniform cross section has trapped air columns of lengths 46 and 44.5 cm above and below a middle pellet of mercury 5 cm in length when held at an angle of 60° to the vertical. When plac

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A thin tube of uniform cross section has trapped air columns of lengths 46 and 44.5 cm above and below a middle pellet of mercury 5 cm in length when held at an angle of 60° to the vertical. When placed horizontally, the columns are equal in length. The temperature is 27°C. The pressure of air columns, in cm Hg, when the tube is horizontal, is

TestHub · Accepted Answer

Tilted, the pellet's vertical height is 5cos 60° = 2.5 cm, so P lower = P upper + 2.5 cm Hg. Horizontal, each column is (46+44.5)/2 = 45.25 cm at common pressure P. Boyle's law on each column: P u · 46 = 45.25P and P l · 44.5 = 45.25P. Subtracting: 45.25P(1/44.5 - 1/46) = 2.5 ⇒ 0.03317P = 2.5 ⇒ P ≈ 75.4 cm Hg. Matches (d).

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