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An air bubble of volume 1.0 cm^(3) rises...

An air bubble of volume `1.0 cm^(3)` rises from the bottom of a lake 40m 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` ?

Text Solution

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`V_(1) = 1.0 cm^(3) = 1.0 xx 10^(-6) m^(3)`,
`T_(1) = 12^(@)C = 12 + 273 = 285K`,
`P_(1) = 1 atm. + 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)/(285xx1.01xx10^(5))`
`=5.276 xx 10^(-6) m^(3)`.
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