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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 $1 = 105 Pa$ , the density of water is $1000 kg/m^{3}$ and $g = 10 m/s2$ . 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:
Option: 1
$3 cm ^{3}$

Option: 2
$4cm^{3}$

Option: 3
$2cm^{3}$

Option: 4
$5cm ^{3}$

Answers (1)

best_answer

Pressure at surface $= P _{atm}= 1\times 10^{5 }Pa$
$V _{surface}= ?$
Pressure at $h = 40 m$ depth
$P= P_{atm}+pgh$
$\begin{aligned} & \mathrm{P}=\mathrm{P}_{\text {atm }}+\rho g h \\ & \mathrm{P}=10^5+10^3 \times 10 \times 40 \\ & \mathrm{P}=5 \times 10^5 \mathrm{~Pa} \\ & \mathrm{v}=1 \mathrm{~cm}^3 \end{aligned}$
Temp. is constant
$\begin{aligned} & \mathrm{P}_1 \mathrm{~V}_1=\mathrm{P}_2 \mathrm{~V}_2 \\ & 10^5 \times \mathrm{v}=5 \times 10^5 \times 1 \\ & \mathrm{v}=5 \mathrm{~cm}^3 \end{aligned}$

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