(O Level Phy) Air bubble rising up the water

You have an air bubble rising up from the bottom of a sea to its surface. The air bubble enlarges as it moves up. By assuming temperature to be constant, you're asked to find the volume or pressure of the air bubble at the bottom or the surface of the sea. This is one of the typical questions on Pressure in O Level.

The equations that you'll need are P₁V₁ = P₂V₂ (Boyle's Law) and P = hρg.

(Nov 2012) A diver works below the surface of a lake. The density of the water in the lake is 1000 kg/m3, the atmospheric pressure at the surface is 1.0 x 105 Pa and the gravitational field strength is 10 N/kg. The diver inflates a balloon with air at a depth of 15 m and attaches the balloon to a tray of objects. (a) Calculate the pressure due to 15 m of water, [2] the total pressure at 15 m below the surface of the lake. [1] (b) The air in the balloon occupies a volume of 0.048 m3 at the pressure calculated in (a)(ii). The diver releases the tray and the balloon and they begin to rise. The temperature of the air in the balloon does not change. Calculate the volume occupied by the air in the balloon at atmospheric pressure. [2] The pressure of the air inside the balloon is less at the surface than at a depth of 15 m. Explain in terms of the air molecules inside the balloon, why the pressure is less. [3]  (c) State one difference between the arrangement of the molecules of water in the lake and the molecules of air in the balloon. [1] (d) When the diver releases the tray, the balloon accelerates upwards and reaches a constant speed before it arrives at the surface. Explain how the forces acting on the balloon cause it to behave in this way. [3] Sketch the distance-time graph for the balloon as it travels 15 m to the surface. [3]

Background information:

• The equation for Ideal Gas Law is PV = nRT ==> PV/T = nR where n = number of moles of gas and R = gas constant. For an enclosed cylinder, the amount of gas is constant and so PV/T = constant.
  
• For fixed amount of gas, P₁V₁/T₁ = P₂V₂/T₂. 

Variable to keep constant	Law	Explanation using the Kinetic Theory
Temperature is constant, T1 = T2	P1V1/T1 = P2V2/T2 P1V1 = P2V2 (Boyle's Law)	If the gas is compressed (volume is decreased), the number of molecules per unit volume increases. This leads to increased frequency of collision of molecules on the walls of container. Therefore, the pressure increases.
Pressure is constant, P1 = P2	P1V1/T1 = P2V2/T2 V1/T1 = V2/T2 (Charles's Law)	If the gas is heated, the molecules move faster (the KE of the molecules increases). This leads to increased frequency of collision of molecules on the walls of the container. The gas expands and pushes the piston so as to maintain the initial pressure.
Volume is constant, V1 = V2	P1V1/T1 = P2V2/T2 P1/T1 = P2/T2 (Lussac's Law)	If the gas is heated, the molecules move faster (the KE of the molecules increases). This leads to increased frequency of collision of molecules on the walls of the cylinder. Therefore, the pressure in the fixed volume of container increases.

Note: V₁/T₁ = V₂/T₂ and P₁/T₁ = P₂/T₂ are not in the current syllabus.

Sketch out a diagram to understand the problem.




(a) (i) Pressure due to the water
= hρg
= 15 m x 1000 kg/m³ x 10 N/kg [1]
= 1.5 x 10⁵ Pa (2 significant figures) [1]

(ii) Total pressure
= P_atm + hρg
= 1.0 x 10⁵ Pa + 1.5 x 10⁵ Pa
= 2.5 x 10⁵ Pa (2 significant figures) [1]

(b) (i) P₁V₁ = P₂V₂ (Boyle's Law)
(2.5 x 10⁵ Pa)(0.048 m³) = (1.0 x 10⁵ Pa)V₂ [1]
V₂ = 0.12 m³ (2 significant figures) [1]

(ii) The balloon is larger at the surface than at the bottom of the lake. [1] The number of molecules per unit volume decreases as the balloon rises. [1] The frequency of collision of molecules on the inner surface of the balloon decreases. [1] Therefore, the pressure in the balloon is less at the surface than at the 15-m depth.

(c) The water molecules are very close together while the air molecules are very far apart. [1]

(d) (i) ∑F = ma
U – W – R = ma where U = upward thrust, W = weight, R = water resistance [1]
When U > W + R, a > 0 i.e. The balloon accelerates. [1]
As the balloon speeds up, R increases in magnitude.
Eventually, when U = W + R, a = 0 i.e. The balloon travels with a constant speed. [1]

(ii) (See diagram)




Try these questions before you check with the answers below.

1. (Nov 2013) The conditions at the bottom and at the surface of a lake are given.

	Bottom of lake	Surface of lake
Temperature	10 ºC	10 ºC
Pressure	500 kPa	100 kPa

A bubble of volume 1.0 cm³ forms at the bottom of the lake. What is the volume of the bubble as it reaches the surface?
A 0.20 cm³
B 0.25 cm³
C 4.0 cm³
D 5.0 cm³

2. (Jun 2011) (a) Explain using ideas about molecules

1. why a balloon filled with gas expands when heated
   
2. why a balloon filled with water expands very little when heated
   

(b) (i) A bubble rises from the bottom of a lake to the surface. The pressure at the bottom of the lake is 3.0 x 10⁵ Pa and the pressure at the surface is 1.0 x 10⁵ Pa. The volume of the bubble at the bottom of the lake is 2.0 cm³. Calculate the volume of the bubble at the surface. [2]

2. State one assumption that you have made in your calculation. [1]
   

Answers: 1. P₁V₁ = P₂V₂ ==> (500 kPa)(1.0 cm³) = (100 kPa)V₂ ==> V₂ = 5.0 cm³. 2. (a) (i) When heat is supplied to the gas, the gas molecules move faster (The molecules have greater kinetic energy). The frequency of collision on the inner surface of the balloon increases. Therefore the balloon expands to maintain the initial pressure. (ii) When liquid water is heated, the water molecules push against one another with more force. The liquid water expands. But it expands very little because the water molecules are still very close together. (b) (i) P₁V₁ = P₂V₂ ==> (3.0 x 10⁵ Pa)(2.0 cm³) = (1.0 x 10⁵ Pa)V₂ ==> V₂ = 6.0 cm³. (ii) Assumption: Temperature remains constant when the bubble rises from the bottom of the lake to the surface.