An air bubble of volume 1 cm 3 rises from the bottom of a lake 40 m deep to the surface at a temperature of 12 ° C . The atmospheric pressure is 1 × 10 5 Pa , the density of water is 1000 kg m - 3 and

Question

An air bubble of volume 1 cm 3 rises from the bottom of a lake 40 m deep to the surface at a temperature of 12 ° C . The atmospheric pressure is 1 × 10 5 Pa , the density of water is 1000 kg m - 3 and g = 10 m s - 2 . There is no difference of the temperature of water at the depth of 40 m and on the surface. The volume of air bubble when it reaches the surface will be

TestHub · Accepted Answer

The pressure at the bottom of the lake can be calculated as follows:
$\begin{array}{rcl} P_{b o t t o m} & = & P_{0} + ρ g h \\ = & (10^{5} + 1000 \times 10 \times 40) Pa \\ = & 5 \times 10^{5} Pa \end{array}$
As the temperature of water remains constant, it can be written that
$P_{bottom} V_{bottom} = P_{top} V_{top} . . . (1)$
Substitute the values of the known parameters into equation (1) to calculate the required volume of the bubble when it reaches the surface.
$\begin{array}{rcl} 5 \times 10^{5} \times 1 & = & 1 \times 10^{5} \times V_{top} \\ \Rightarrow & V_{top} = \frac{5 \times 10^{5} \times 1}{1 \times 10^{5}} {cm}^{3} \\ = & 5 {cm}^{3} \end{array}$

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