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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?

User Jarederaj
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1 Answer

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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

User MadhuP
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