Physics, asked by Darkalin, 1 month ago

An air bubble of radius 8mm is formed inside water . Surface tension of water is
72 x 10⁻³ N/m . Calculate the excess pressure inside the air bubble .

Answers

Answered by Jonathan242009
1

Answer:

answer is down

Explanation:

Coalescence of air bubbles is important in gas-liquid reactors and food processing operations. Bubbles can be stabilized by using non-ionic surfactants. Binary coalescence of air bubbles in ethylene glycol and aqueous glycerol solutions were studied in this work in presence of Span 80. A novel set-up was developed to study long coalescence times. Coalescence time was observed to follow broad stochastic distributions in all systems. The distributions were fitted with a stochastic model developed earlier. The surface tension of ethylene glycol and glycerol solutions was measured at various concentrations of Span 80. These data were fitted using a surface equation of state derived from the Langmuir isotherm. The effect of surfactant concentration on coalescence time was explained in terms of the surface excess of the surfactant and the repulsive force generated at the air-liquid interface. The results from this work illustrate the stochastic nature of bubble coalescence in viscous liquids. This work also demonstrates how non-ionic surfactants can stabilize bubbles in such liquids.

Answered by gargs4720
0

Answer:

24Nm −2

24Nm −2

Explanation:

Given : S=0.072 N/m

Diameter of water drop D=1.2mm=0.012 m

Thus radius of drop r= 0.012/2 =0.006 m

Excess pressure inside water drop

ΔP = r /2S

∴ ΔP= 2×0.072 / 0.006 =24 N/

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