We\'ve all spent some time staring at the CO2 bubbles rising in a soda or a beer

Question

We've all spent some time staring at the CO2 bubbles rising in a soda or a beer. Take the density
of beer to be 1.10 x 10^3 kg/m^3 and the viscosity to be n = 1.80 mPa * s. Use Stokes' law for the drag force on a sphere due to viscosity to find the terminal velocity of ascent for a 1 mm diameter CO2 bubble. (Hint: To find the terminal velocity, balance the buoyant force on the bubble against the drag force, ignoring the weight of the bubble. (Why is this justiable?)) Roughly how long should it take for such a bubble to rise to the top of a typical beer glass?

Explanation / Answer

ignoring the weight of the buble,

B = viscous force ;

the mass*9.8 is verrry less in comparison to viscous forces and buoyancy force , so this is justifiable .

4 / 3 r^3 * w * g = 6 r v ;

or v =( 2 / 9 ) r^2 g ;

so v = ( 2 / 9*1.8e-3 ) ( 0.5 e -3 ) ^2 * ( 1.1 e 3 ) * 9.8 = 0.332716 m / s ;

this is the velocity of the bubble , let us assume that beer glass height that has to be travelled is 10cm to 15 cm , then

time = [ 10e-2 ( or 15e-2 ) ] / 0.33271 = 0.333 s or 0.450s it roughly takes 1 /4 th to 1/2 of a second .

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