Question

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?

Answer 1

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

$P_{1}=P_{2}+\rho gh$

Where, $P_{2}$ = Pressure at the surface

$P_{1}$ = Pressure at the depth

Put the value into the formula

$P_{1}=101325+(1000\times9.8\times52.0)$

$P_{1}=610925\ N/m^2$

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

Using equation of ideal gas

$PV=RT$

$\dfrac{PV}{T}=constant$

Now, The equation of at bottom and top

$\dfrac{P_{1}V_{1}}{T_{1}}=\dfrac{P_{2}V_{2}}{T_{2}}$

$V_{2}=\dfrac{P_{1}V_{1}T_{2}}{P_{2}T_{1}}$

Put the value into the formula

$V_{2}=\dfrac{610925\times1.50\times(18.5+273)}{101325\times(5.5+273)}$

$V=9.47\ m^3$

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