First slide

Gaseous state

Question

A 10.0 cm column of air is trapped by a column of Hg, 4.00 cm long in capillary tube of uniform bore when the tube is held horizontally in a room at 1 atm. Length of the air column when the tube is held vertically with the open end up is

Moderate

A

9.50 cm
B

10.52 cm
C

3.53 cm
D

4.61 cm

Solution

When open end up, the weight of mercury represents an increase in the pressure on the gas of 4.00 cm Hg = 4omm Hg
$∴ Final pressure = 760 + 40 = 800 mmHg$
$p_{1} V_{1} = p_{2} V_{2} \Rightarrow p_{1} \times A \times I_{1} = p_{2} \times A \times l_{2}$
where, A is the area of cross-section (which is constant) and l₁, l₂ length
$∴ p_{1} \times 1_{1} = p_{2} \times I_{2}$
$760 \times 10 = 800 \times l_{2} \Rightarrow l_{2} = 9 .50 cm$

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