Question

A glass tube in the shape of a letter J has the shorter limb sealed and the longer limb open. Mercury is poured into the tube until the levels in either limb is the same when the tube is vertical.In this position, the length of the air column in the sealed limb is 6.3cm.More mercury is then poured into the tube until the length of the trapped air column is 42cm.Calculate the difference in the levels of mercury in the limbs if a nearby mercury barometer reads 75.0cm and the reading of a nearby thermometer has not changed?​

Answer 1

Answer:

35.4 cm

Explanation:

We have that when the level of mercury on either limb is the same, the pressure of the trapped air, P₁ = Atmospheric pressure

Also the initial height of the mercury in the tube = The reading of the barometer = 75.0 cm

The initial length of the air column, l₁ = 6.3 cm

The final length of the air column, l₂ = 4.2 cm (The length is expected to decrease due to compression)

The volume, V = l × A

Where;

A = The cross sectional area of the tube

Therefore, the volume of the air column is directly proportional to the length of the air column

∴ V ∝ l

According to Boyles law, we have;

P₁·V₁ = P₂·V₂

Where;

P₁ = The initial pressure in the air column before more mercury is added

V₁ = The initial volume occupied by the air in the air column

P₂, and V₂ are the final pressure and volume of the air column respectively

Given that V = l·A, we can write;

P₁·l₁·A = P₂·l₂·A

P₂ = P₁·l₁·A/(l₂·A) = P₁·l₁/(l₂) = P₁ × 6.3/4.2 = 1.5·P₁

The pressure in the air column after more mercury is added, P₂ = 1.5 × P₁

P₁ = Atmospheric pressure, therefore;

The pressure in the air column after more mercury is added, P₂ = 1.5 × Atmospheric pressure

Pressure = h·ρ·g

Where;

ρ = The density of the substance

g = The acceleration due to gravity

h = The height of the column of the fluid

Given that the density and the gravitational force, can be taken as constant, we have that the pressure of the fluid is directly proportional to the height of the fluid column

Therefore, when the pressure doubles, the height of the fluid column doubles, and when the factor of increase is 1.5, we have;

The final level of the mercury, h₂ = 1.5·h₁ = 1.5×75 cm = 112.5 cm

The initial length of the closed end of the J tube, $h_{closed1}$ = 6.3 cm + 75 cm = 81.3 cm

The final length of the mercury in the closed end, $h_{closed2}$ = 81.3 cm - 4.2 cm = 77.1 cm

The difference in the level of mercury, Δh = h₂ - $h_{closed2}$

∴ Δh = 112.5 cm - 77.1 cm = 35.4 cm

The difference in the levels of mercury in the limbs, Δh = 35.4 cm