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An air bubble at the bottom of a lake 52.0 m deep has a volume of 1.50m^3. If the temperature at the bottom is 5.5 degree's Celsius and at the top is 18.5 degree's Celsius, what is the volume of the bubble just before it reaches the surface?

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Answer:

The volume of the bubble near the surface will be 9.47 m³

Step-by-step 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


P_(1)=P_(2)+\rho gh

Where,
P_(2) = Pressure at the surface


P_(1) = Pressure at the depth

Put the value into the formula


P_(1)=101325+(1000*9.8*52.0)


P_(1)=610925\ N/m^2

We need to calculate the volume of the bubble just before it reaches the surface

Using equation of ideal gas


PV=RT


(PV)/(T)=constant

Now, The equation of at bottom and top


(P_(1)V_(1))/(T_(1))=(P_(2)V_(2))/(T_(2))


V_(2)=(P_(1)V_(1)T_(2))/(P_(2)T_(1))

Put the value into the formula


V_(2)=(610925*1.50*(18.5+273))/(101325*(5.5+273))


V=9.47\ m^3

Hence, The volume of the bubble near the surface will be 9.47 m³

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User MadhuP
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