Question

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 0.04 N/m.

Answer 1

So amount of work produced will be
`10{-4}J`

Step-by-step explanation:

We have given diameter of ammonia bubble is changes from 1 cm to 3 cm

So radius changes from 0.5 cm to 1.5 cm

Surface area of bubble
`=4\pi r^2`

So change in area of bubble
`=4\pi (0.015^2-0.005^2)=8* 3.14* (0.015^2-0.005^2)=0.00251m^2`

Surface tension of ammonia = 0.04 N/m

So work done will be
`Work\ done=surface\ tension* change\ in\ area=0.04* 0.00251= 10^(-4)J`