Login/Sign Up
Home » An air bubble of volume rises from the bottom of a lake deep at a temperature of To what volume does it grow when it reaches the surface, which is at a temperature of 35 C?
An air bubble of volume 1.0 \mathrm{~cm}^{3} rises from the bottom of a lake 40 \mathrm{~m} deep at a temperature of 12{ }^{\circ} \mathrm{C} . To what volume does it grow when it reaches the surface, which is at a temperature of 35 C?
CBSE | Class 11 | Kinetic Theory | NCERT | Physics
An air bubble of volume 1.0 \mathrm{~cm}^{3} rises from the bottom of a lake 40 \mathrm{~m} deep at a temperature of 12{ }^{\circ} \mathrm{C} . To what volume does it grow when it reaches the surface, which is at a temperature of 35 C?

Volume of the air bubble is given as \mathrm{V}_{1}=1.0 \mathrm{~cm}^{3}

=1.0 \times 10^{-6} \mathrm{~m}^{3}

Air bubble rises to height given as d=40 \mathrm{~m}

Temperature at a depth of 40m is given as \mathrm{~T}_{1}=12^{0} \mathrm{C}=285 \mathrm{~K}

Temperature at the surface of the lake is given as T_{2}=35^{0} \mathrm{C}=308 \mathrm{~K}

The pressure on the surface of the lake will be,

P_{2}=1 \mathrm{~atm}=1 \times 1.013 \times 10^{5} \mathrm{~Pa}

The pressure at the depth of 40 \mathrm{~m} will be then,

\mathrm{P}_{1}=1 \mathrm{~atm}+\mathrm{d} \rho \mathrm{g}

where,

\rho= density of water having value 10^{3} \mathrm{~kg} / \mathrm{m}^{3}

\mathrm{g}= acceleration due to gravity having value 9.8 \mathrm{~m} / \mathrm{s}^{2}

Hence,

P_{1}=1.013 \times 10^{5}+40 \times 10^{3} \times 9.8

\mathrm{O}_{1} \mathrm{~V}_{1} / \mathrm{T}_{1}=\mathrm{P}_{2} \mathrm{~V}_{2} / \mathrm{T}_{2}

=493300 \times 1 \times 10^{-6} \times 308 /\left(285 \times 1.013 \times 10^{5}\right)

=5.263 \times 10^{-6} \mathrm{~m}^{3} \text { or } 5.263 \mathrm{~cm}^{3}

As a result, when the air bubble reaches the surface. its volume becomes 5.263 \mathrm{~cm}^{3}.

i

More Material