A bubble of gas, diameter 2 mm, is trapped in a container of liquid at normal atmospheric pressure (1 × 10^5 Pa) and a temperature of 25 °C. The container is opened on board an aircraft where the temperature is 22 °C and the surrounding pressure is 0.9 × 10^5 Pa. What is the diameter of the bubble as it escapes from the liquid?

Question

A bubble of gas, diameter 2 mm, is trapped in a container of liquid at normal atmospheric pressure (1 × 10^5 Pa) and a temperature of 25 °C. The container is opened on board an aircraft where the temperature is 22 °C and the surrounding pressure is 0.9 × 10^5 Pa. What is the diameter of the bubble as it escapes from the liquid?

Answer 1

Volume increases by a ratio

(P1/P2)*(T2/T1)
where 1 denotes initial and 2 final conditions. T must be in Kelvin.
That ratio is therefore
V2/V1 = (1/0.9)*(295/298)= 1.100

Take the cube root of that ratio to get the diameter ratio.
D2/D1 = 1.032

D2 = 2.065 mm

You don't have many significant figures in your numbers.

Answer 2

To find the diameter of the bubble as it escapes from the liquid, we can use the ideal gas law and apply Boyle's law.

First, let's gather the given information:

Initial conditions:
Pressure inside the container (P1) = 1 × 10^5 Pa
Temperature inside the container (T1) = 25 °C = 25 + 273.15 = 298.15 K
Diameter of the bubble inside the liquid (d1) = 2 mm = 2 × 10^-3 m

Final conditions:
Pressure outside the container (P2) = 0.9 × 10^5 Pa
Temperature outside the container (T2) = 22 °C = 22 + 273.15 = 295.15 K

Now, let's calculate the volume of the bubble under the initial conditions.

The volume of the bubble (V1) can be calculated using the formula:

V1 = (4/3) * π * (d1/2)^3

Substituting the given diameter value:

V1 = (4/3) * π * (2 × 10^-3 / 2)^3
= (4/3) * π * (10^-3)^3
= (4/3) * π * 10^-9

Next, let's calculate the new volume of the bubble using Boyle's law, which states that the product of pressure and volume is constant when the temperature is held constant.

P1 * V1 = P2 * V2

Solving for V2:

V2 = (P1 * V1) / P2

Substituting the given values:

V2 = (1 × 10^5 * (4/3) * π * 10^-9) / (0.9 × 10^5)
= (4/3) * π * 10^-9

Now, we can calculate the diameter of the bubble under the new conditions using the new volume calculated.

The volume of a sphere is given by the formula:

V = (4/3) * π * (d/2)^3

Let's solve for d:

(d/2)^3 = V2 / ((4/3) * π)
d/2 = (V2 / ((4/3) * π))^(1/3)
d = 2 * (V2 / ((4/3) * π))^(1/3)

Substituting the given volume value:

d = 2 * ((4/3) * π * 10^-9 / ((4/3) * π))^(1/3)
= 2 * ((10^-9)^(1/3))
= 2 * 10^(-9/3)
= 2 * 10^(-3/3)
= 2 * 10^(-1)
= 0.2 mm (or 2 × 10^-4 m)

Therefore, the diameter of the bubble as it escapes from the liquid will be 0.2 mm (or 2 × 10^-4 m).