a) Radius of a bubble at the bottom of the found to be 1 cm, then find the radius of - Chemistry - States of Matter

Question

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

Vartika Jain · Accepted Answer

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