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InterviewSolution
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An air bubble of volume `1.0 cm^(3)` rises from the bottom of a take 40 m deep at a temperature of `12^(@) C`. To what volume does it grow when it reaches the surface, which is at a temperature of `35^(@) C`. ? Given `1 atm = 1.01 xx 10^(5) Pa`. |
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Answer» `V_(1) = 1.0 cm^(3) = 1.0 xx 10^(-6) m^(3) , T_(1) = 12^(@)C = 12+273 = 285 K`, `P_(1) = 1 at, + h_(1) rho g = 1.01 xx 10^(5) + 40 xx 10^(3) xx 9.8 = 493000 Pa`. when the air bubble reaches at the surface of lake , then `V_(2) = ? , T_(2) = 35^(@)C = 35 + 273 = 308 K, P_(2) = 1 atm , = 1.01 xx 10^(5) Pa` Now , `(P_(1)V_(1))/(T_1) = (P_(2)V_(2))/(T_2)` or `V_(2) = (P_(1)V_(1)T_(2))/(T_(1)P_(2))` `:. V_(2) = ((493000) xx (1.0xx10^(-6))xx308)/(285 xx 1.01 xx 10^(5)) = 5.275 xx 10^(-6)m^(3)`. |
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