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InterviewSolution
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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`. |
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Answer» At the bottom of the lake, volume of the bubble `V_(1) = (4)/(3)pi (0.18)^(3)cm^(3)` Pressure on the bubble `P_(1) =` Atmospheric pressure `+` Pressure due to a column of `250cm` of water `= (76 xx 13.6 + 250)980 "dyne"//cm^(2)` At the surface of the lake, volume of the bubble `V_(2) = (4)/(3)pi(0.2)^(3)cm^(3)` Pressure on the bubble , `P_(2)` = atm. pressure `= (76 xx 13.6 xx 980)"dyne"//cm^(2)` `T_(2) = 273 + 40^(0)C = 313 K` Now `(P_(1)V_(1))/(T_(1)) = (P_(2)V_(2))/(T_(2))` `rArr ((76 xx 13.6 + 250)680xx ((4)/(3))pi(0.18)^(3))/(T_(1))` `= ((76 xx 13.6 + 250)680xx ((4)/(3))pi(0.2)^(3))/(313)` or `T_(1) = 283.37 K , T_(1) = 283.37 - 273 = 10.37^(0)C` |
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