Derive the expression for excess pressure for a bubble.
Asked by Arun Iyer | 19th Feb, 2011, 05:24: PM
Expert Answer:
Dear student
Excess pressure inside a liquid drop and a bubble
We know that in small drops and bubbles, the effect of gravity is negligible on account of surface tension. Small drops do not collapse due to surface tension. This means that the pressure inside the drop is greater than the pressure outside. Let this excess pressure i.e., pi - po = p. Due to this pressure difference, let the radius increase from R to R + dR.
Outside force = pressure difference x surface area
Hence, to increase the surface area
Work done by the excess pressure is stored in the form of potential energy.
Increase in P.E. = surface tension x increase in surface area.
dW = increase in P.E.
hope this clarifies your doubt.
Regards
Team
Topperlearning
Excess pressure inside a liquid drop and a bubble
We know that in small drops and bubbles, the effect of gravity is negligible on account of surface tension. Small drops do not collapse due to surface tension. This means that the pressure inside the drop is greater than the pressure outside. Let this excess pressure i.e., pi - po = p. Due to this pressure difference, let the radius increase from R to R + dR.
Outside force = pressure difference x surface area
Hence, to increase the surface area
Work done by the excess pressure is stored in the form of potential energy.
Increase in P.E. = surface tension x increase in surface area.
dW = increase in P.E.
hope this clarifies your doubt.
Regards
Team
Topperlearning
Answered by | 20th Feb, 2011, 02:14: PM
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