2.5 Calculate the numerical value for using . · OlympiadBench #1016

Physics · thermodynamics

Problem

2.5 Calculate the numerical value for

u

using

\eta=1.8 \times 10^{-5} \mathrm{kgm}^{-1} \mathrm{~s}^{-1}

. Given constants: An Electrified Soap Bubble A spherical soap bubble with internal air density

\rho_{i}

, temperature

T_{i}

and radius

R_{0}

is surrounded by air with density

\rho_{a}

, atmospheric pressure

P_{a}

and temperature

T_{a}

. The soap film has surface tension

\gamma

, density

\rho_{s}

and thickness

t

. The mass and the surface tension of the soap do not change with the temperature. Assume that

R_{0} \gg t

. The increase in energy,

d E

, that is needed to increase the surface area of a soap-air interface by

d A

, is given by

d E=\gamma d A

where

\gamma

is the surface tension of the film. Context question: 2.1 Find the ratio

\frac{\rho_{i} T_{i}}{\rho_{a} T_{a}}

in terms of

\gamma, P_{a}

and

R_{0}

. Context answer:

1+\frac{4 \gamma}{R_{0} P_{a}}

Context question: 2.2 Find the numerical value of

\frac{\rho_{i} T_{i}}{\rho_{a} T_{a}}-1

using

\gamma=0.0250 \mathrm{Nm}^{-1}, R_{0}=1.00 \mathrm{~cm}

, and

P_{a}=1.013 \times 10^{5} \mathrm{Nm}^{-2}

. Context answer: 1.0001 Context question: 2.3 The bubble is initially formed with warmer air inside. Find the minimum numerical value of

T_{i}

such that the bubble can float in still air. Use

T_{a}=300 \mathrm{~K}, \rho_{s}=1000 \mathrm{kgm}^{-3}

\rho_{a}=1.30 \mathrm{kgm}^{-3}, t=100 \mathrm{~nm}

and

g=9.80 \mathrm{~ms}^{-2}

. Context answer: 307.1 Extra Supplementary Reading Materials: After the bubble is formed for a while, it will be in thermal equilibrium with the surrounding. This bubble in still air will naturally fall towards the ground. Context question: 2.4 Find the minimum velocity

u

of an updraught (air flowing upwards) that will keep the bubble from falling at thermal equilibrium. Give your answer in terms of

\rho_{s}, R_{0}, g, t

and the air's coefficient of viscosity

\eta

. You may assume that the velocity is small such that Stokes's law applies, and ignore the change in the radius when the temperature lowers to the equilibrium. The drag force from Stokes' Law is

F=6 \pi \eta R_{0} u

. Context answer:

\frac{4 R_{0} \rho_{s} t g}{6 \eta}+\frac{\frac{4}{3} R_{0}^{2} \rho_{a} g\left(\frac{4 \gamma}{R_{0} P_{a}}\right)}{6 \eta}

Intuitive explanation

The bubble at thermal equilibrium is denser than the surrounding air and experiences a net downward force. To keep it suspended, an updraught must generate an upward drag force (via Stokes's law) that exactly balances this weight difference. Substitute the given numerical values into the expression from 2.4—the soap film mass and the extra mass from the compressed internal air both contribute to the total weight that the drag force must support.

Intuitive explanation

Related theory

thermodynamics

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