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Q13.5 An air bubble of volume1.0 cm^{3}  rises from the bottom of a lake 40m deep at a temperature of 12^{0}C. To what volume does it grow when it reaches the surface, which is at a temperature of 35^{0}C?

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Initial Volume of the bubble, V1 = 1.0 cm3

Initial temperature, T1 = 12 oC = 273 + 12 = 285 K

Density of water is \rho _{w}=10^{3}\ kg\ m^{-3}

Initial Pressure is P1

Depth of the bottom of the lake = 40 m

\\P_{1}=Atmospheric\ Pressure+Pressure\ due\ to\ water\\ P_{1}=P_{atm}+\rho _{w}gh\\ P_{1}=1.013\times 10^{5}+10^{3}\times 9.8\times 40\\ P_{1}=4.93\times 10^{5}Pa

Final Temperature, T2 = 35 oC = 35 + 273 = 308 K

Final Pressure = Atmospheric Pressure =1.013\times 10^{5}Pa

Let the final volume be V2

As the number of moles inside the bubble remains constant we have

\\\frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}}\\ V_{2}=\frac{P_{1}T_{2}V_{1}}{P_{2}T_{1}}\\ V_{2}=\frac{4.93\times 10^{5}\times 308\times 1}{1.013\times 10^{5}\times 285}\\ V_{2}=5.26\ cm^{3}

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