Question

Check my work Check My Work button is now disabledItem 16Item 16 3 points As a spherical ammonia vapor bubble rises in liquid ammonia, its diameter changes from 1 cm to 3 cm. Calculate the amount of work produced by this bubble, in kJ, if the surface tension of ammonia is 0.07 N/m.

Answer 1

Answer:

W = 1.7593 * 10 ^ (-7) KJ

Explanation:

The work done by the bubble is given:

$W = sigma*\int\limits^2_1 {} \, dA \\\\W = sigma*( {A_{2} - A_{1} } ) \\\\A = pi*D^2\\\\W = sigma*pi*(D^2_{2} - D^2_{1})\\\\W = 0.07 * pi * (0.03^2 - 0.01^2)*10^(-3)\\\\W = 1.7593 *10^(-7) KJ$

Answer: W = 1.7593 * 10 ^ (-7) KJ