a) Radius of a bubble at the bottom of the found to be 1 cm, then find the radius of the bubble at the surface of water considering the temperature at the surface& bottom being same.The height of water level in a tank is 10.336 m.
b) If the absolute temperature at the surface is 4 times that at the bottom , then find radius of bobble at the surface.

Dear User,
(a)Pressure at the bottom of water, P1 =Atmospheric pressure+Hydrostatic pressure                                                               = Pa   + hρg Now, Depth,h = 10.336 m and, Density of water,ρ =1000kg/m2 , g= 9.8kg , Pa=105 atmTherefore, P1= 105 + (10.336 ×1000×9.8) = 20.129 ×104 PaPressure at the surface, P2 = Pa=105 PaVolume of the bubble at the bottom,V1 = 43πr3=43×3.14×(1)3= 4.186 cm3 = 4.186 ×10-3 m3Volume of the bubble at surface, V2 =43πr3Now, Temperature at surface and bottom is same i.e. T1=T2=TWe assume that the gas inside the bubble behaves like an ideal gas, and therefore satisfies the ideal gas equation i.e.,P1V1T1=P2V2T2or, 20.129 ×104× 4.186 ×10-3T=105×43πr3 Tor, 842.599= 43(3.14) r3 ×105or, r3 =842.5994.1866 ×105=201.260 ×10-5=2.0126×10-3or, r =2.0126×10-33= 0.1262 m Therefore, radius of bubble at the surface is 0.1262 m 
(b) If the absolute temperature at the surface is 4 times at the bottom, i.e. 
                   T2= 4T1
Then,
P1V1T1=P2V2T2or, 20.129 ×104× 4.186 ×10-3T=105×43πr3 4Tor, 842.599= 43×4(3.14) r3 ×105or, r3 =842.599×44.1866 ×105=8.0504 × 10-3 mor, r =8.0504 × 10-33= 0.2004 m Therefore, radius of bubble at the surface will be 0.2004 m

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