An air bubble starts rising from the bottom of a lake. Its diameter is ` 3.6 mm` at the bottom and `4 mm` at the surface. The depth of the lake is `250 cm` and the temperature at the surface is `40^@ C`. What is the temperature at the bottom of the lake? Given atmospheric pressure = `76 cm of Hg and g = 980 cm//s^2`.

Answer

A

Explanation

At the bottom of the lake, volume of the bubble
`V_1 = 4/3 pi r _1 ^3 = 4/3pi (0.18)^3 cm^3`
Pressure on the bubble `(p_(1)` = Atmospheric pressure + pressure due to a column of `250 cm` of water
`= 76 xx 13.6 xx 980 + 250 xx 1 xx 980`
`= (76 xx 13.6 + 250) 980 "dyne"//cm^2`
At the surface of the lake, Volume of the bubble
`V_(2) = (4)/(3) pi r_(2)^(3) = (4)/(3) pi (0.2)^(3) cm^(3)`
pressure on the bubble, `p_(2) = atmospheric pressure`
= `(76 xx 13.6 xx 980)dyne //cm^2`
`T_2 = 273 + 40^@ C`
= `313^@ K`
Now `(p_1 V_1)/(T_1) = (p_2V_2)/(T_2)`
or `((76 xx 13.6 + 250) 980 xx(4/3)pi (0.18)^3)/(T_1)=((76 xx 13.6) xx 980(4/3) pi (0.2)^3)/(313)`
or `T_1 = 288.37 K`
:. `T_1 = 283.37 -273`
= `10.37^@ C`.