Wednesday, April 04, 2007

Surface Tension–Questions involving excess of pressure inside a bubble

Here is a question which is popular among question setters:

Two spherical soap bubbles of radii r1 and r2 in vacuum combine under isothermal conditions. The resulting bubble has a radius equal to

(a) (r1+r2)/2 (b) √(r1+r2) (c) r1r2/(r1+r2) (d) (r1+r2)/√2 (e) √(r12+r22)

As the temperature is constant, we have P1V1+ P2V2 =.PV where P1 and P2 are the pressures inside the separate bubbles, V1 and V2 are their volumes, P is the pressure inside the combined bubble and V is its volume.

Since the bubbles are located in vacuum, the pressure inside the bubble is equal to the excess of pressure 4T/r, where T is the surface tension and ‘r’ is the radius so that we have (4T/r1)×[(4/3)πr13]+ (4T/r2)×[(4/3)πr23] = (4T/R)×[(4/3)πR3] where R is the radius of the combined bubble.

This yields R = √(r12+r22).

[You can work out this problem by equating the surface energies: 4πr12T+4πr22T= 4πR2T, from which R = √(r12+r22)]

Now, consider the following MCQ:

Excess of pressure inside one soap bubble is four times that inside another. Then the ratio of the volume of the first bubble to that of the second is

(a) 1:16 (b) 1:32 (c) 1:64 (d) 1:2 (e) 1:4

If r1 and r2 are the radii of the bubbles and T is the surface tension of soap solution, we have (4T/r1)/(4T/r2) = 4.

Therefore, r1/r2 = ¼. Since the volume is directly proportional to the cube of the radius, the ratio of volumes V1/V2 = (r1/r2)3 = (¼)3 = 1/64 [Option (d)].

The following MCQ appeared in AIEEE 2004 question paper. This question is popular among question setters and has appeared in other entrance test papers as well:

If two soap bubbles of different radii are connected by a tube,

(a) air flows from the bigger bubble to the smaller bubble till the sizes become equal

(b) air flows from the bigger bubble to the smaller bubble till the sizes are interchanged

(c) air flows from the smaller bubble to the bigger

(d) there is no flow of air

The excess of pressure inside the smaller bubble is greater than that inside the bigger bubble. (Remember, P = 4T/r and hence excess of pressure ‘P’ is inversely proportional to the radius ‘r’). Therefore, air will flow from the smaller bubble to the bigger bubble [Option (c)].

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