Simple isn't it??#10

Classical Mechanics Level 5

What is the work done to blow a soap bubble of radius, R = 4 m R=4m and having surface tension, T = 50 N m − 1 T=50Nm^{-1} .

Take atmospheric P 0 = 1 0 5 P a {P}_{0} = 10^{5}Pa

Use SI units only

Consider the expansion as isothermal .

This problem is of this set.


The answer is 33513.67.

Aditya Kumar by Aditya Kumar , 22, India

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1 solution

Aditya Kumar
Jul 18, 2015

Work Done = Surface energy + work done in isothermal expansion. Surface energy = 8 π ∗ R 2 T 8\pi*R^{2}T work done in isothermal expansion = ( P 0 + 4 T / R ) ( 4 / 3 ∗ π ∗ R 3 ) ( l n ( ( P 0 + 4 T / R ) / P 0 ) ) ({P}_{0}+4T/R)(4/3*\pi*R^{3})(ln(({P}_{0}+4T/R)/{P}_{0}))

3 Helpful 0 Interesting 0 Brilliant 1 Confused

What unit ? Work done is in Joules ? Please post the Units.

Vijay Simha - 4 years, 7 months ago

If we consider the work done by the internal pressure we get the work done for a small increase in the radius dx is equals dW=4π(p_0+4T/x)x^2 dx. After putting the values we get 26828363.5036 Joules.

Alapan Das - 2 years, 2 months ago

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