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A diver 40 m deep in 10 degrees C fresh water exhales a 1.5 cm diameter bubble.
What is the bubble's diameter just as it reaches the surface of the lake, where the water temperature is 20∘C?
Assume that the air bubble is always in thermal equilibrium with the surrounding water.

Answer :

boffeemadrid

Answer:

0.0257259766982 m

Explanation:

[tex]P_2[/tex] = Atmospheric pressure = 101325 Pa

[tex]d_1[/tex] = Initial diameter = 1.5 cm

[tex]d_2[/tex] = Final diameter

[tex]\rho[/tex] = Density of water = 1000 kg/m³

h = Depth = 40 m

The pressure is

[tex]P_1=P_2+\rho gh\\\Rightarrow P_1=101325+1000\times 9.81\times 40\\\Rightarrow P_1=493725\ Pa[/tex]

From ideal gas law we have

[tex]\dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}\\\Rightarrow \dfrac{P_1\dfrac{4}{3\times8}\pi d_1^3}{T_1}=\dfrac{P_2\dfrac{4}{3\times8}\pi d_2^3}{T_2}\\\Rightarrow \dfrac{P_1d_1^3}{T_1}=\dfrac{P_2d_2^3}{T_2}\\\Rightarrow d_2=(\dfrac{P_1d_1^3T_2}{P_2T_1})^{\dfrac{1}{3}}\\\Rightarrow d_2=(\dfrac{493725\times 0.015^3\times (20+273.15)}{101325\times (10+273.15)})^{\dfrac{1}{3}}\\\Rightarrow d_2=0.0257259766982\ m[/tex]

The diameter of the bubble is 0.0257259766982 m

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