Answer:
The volume of the bubble near the surface will be 9.47 m³
Explanation:
Given that,
Depth = 52.0 m
Volume = 1.50 m³
Temperature at bottom = 5.5°C
Temperature at the top = 18.5°C
We need to calculate the pressure at the depth 52.0 m
The pressure is
[tex]P_{1}=P_{2}+\rho gh[/tex]
Where, [tex]P_{2}[/tex] = Pressure at the surface
[tex]P_{1}[/tex] = Pressure at the depth
Put the value into the formula
[tex]P_{1}=101325+(1000\times9.8\times52.0)[/tex]
[tex]P_{1}=610925\ N/m^2[/tex]
We need to calculate the volume of the bubble just before it reaches the surface
Using equation of ideal gas
[tex]PV=RT[/tex]
[tex]\dfrac{PV}{T}=constant[/tex]
Now, The equation of at bottom and top
[tex]\dfrac{P_{1}V_{1}}{T_{1}}=\dfrac{P_{2}V_{2}}{T_{2}}[/tex]
[tex]V_{2}=\dfrac{P_{1}V_{1}T_{2}}{P_{2}T_{1}}[/tex]
Put the value into the formula
[tex]V_{2}=\dfrac{610925\times1.50\times(18.5+273)}{101325\times(5.5+273)}[/tex]
[tex]V=9.47\ m^3[/tex]
Hence, The volume of the bubble near the surface will be 9.47 m³
