Question

Question 37:(III) An air bubble at the bottom of a lake 41.0 m deep has a volume of\({\bf{1}}{\bf{.00}}\;{\bf{c}}{{\bf{m}}^{\bf{3}}}\). If the temperature at the bottom is 5.5°C and at the top 18.5°C, what is the radius of the bubble just before it reaches the surface?

Short answer

The radius of the air bubble just before it reaches the surface is 1.07 cm.

Step-by-step solution

Step 1:Statement of the ideal gas law 

The ideal gas law is the equation of state for an ideal gas. This equation gives the relation between the pressure (P), volume (V), and temperature (T) of n moles of the ideal gas by the following expression:

\(PV = nRT\)

Here,R is the universal gas constant whose value is \(8.14\;{\rm{J/mol}} \cdot {\rm{K}}\).

Step 2:Given information 

Depth of the lake is \(h = 41.0\;{\rm{m}}\).

The volume of the air bubble at the bottom of the lake is \(V' = 1.00\;{\rm{c}}{{\rm{m}}^{\rm{3}}} = 1.00 \times {10^{ - 6}}\;{{\rm{m}}^3}\).

Pressure at the top of the lake is equal to the atmospheric pressure,\(P = 1\;{\rm{atm}}. = 1.013 \times {10^5}\;{\rm{Pa}}\).

The temperature at the bottom of the lake is \(T' = 5.5^\circ \;{\rm{C}} = \left( {5.5 + 273} \right)\;{\rm{K}} = 278.5\;{\rm{K}}\).

The temperature at the top of the lake is \(T = 18.5^\circ \;{\rm{C}} = \left( {18.5 + 273} \right)\;{\rm{K}} = 291.5\;{\rm{K}}\).

The density of water in the lake is \(\rho = 1.00 \times {10^3}\;{\rm{kg/}}{{\rm{m}}^3}\).

Determination of the pressure at the bottom of the lake                          

The pressure at the bottom of the lake is equal to the sum of the pressure at the surface and the gauge pressure, i.e.,

\(P' = P + h\rho g\), whereg is the acceleration due to gravity.

Substitute the values in the above equation.

\(\begin{array}{c}P' = \left( {1.013 \times {{10}^5}\;{\rm{Pa}}} \right){\rm{ + }}\left( {41.0\;{\rm{m}}} \right)\left( {1.00 \times {{10}^3}\;{\rm{kg/}}{{\rm{m}}^3}} \right)\left( {9.8\;{\rm{m/}}{{\rm{s}}^2}} \right)\\ = 5.031 \times {10^5}\;{\rm{Pa}}\end{array}\)

Determination of the volume of the air bubble just before it reaches the top of the lake

According to the ideal gas law for the air bubble,

\(\begin{array}{c}PV = nRT\\\frac{{PV}}{T} = nR.\end{array}\)

The number of moles (n) of the gas in the air bubble remains constant on rising. Therefore,

\(\frac{{PV}}{T} = {\rm{constant}}\).

Thus,this ratio for the air bubble at the bottom of the lake is equal to that at the top.

\(\begin{array}{c}\frac{{PV}}{T} = \frac{{P'V'}}{{T'}}\\V = V'\frac{{P'}}{P}\frac{T}{{T'}}\end{array}\)

Substitute the values in the above equation.

\(\begin{array}{c}V = \left( {1.00\;{\rm{c}}{{\rm{m}}^3}} \right)\left( {\frac{{5.031 \times {{10}^5}\;{\rm{Pa}}}}{{1.013 \times {{10}^5}\;{\rm{Pa}}}}} \right)\left( {\frac{{291.5\;{\rm{K}}}}{{278.5\;{\rm{K}}}}} \right)\\ = 5.198\;{\rm{c}}{{\rm{m}}^3}\end{array}\)

Determination of the radius of the bubble just before it reaches the surface of the lake

If the radius of the bubble just before reaching the surface is r, then the volume of the bubble just before reaching the surface is

\(V = \frac{4}{3}\pi {r^3}\).

Substitute the values in the above equation.

\(\begin{array}{c}\frac{4}{3}\pi {r^3} = 5.198\;{\rm{c}}{{\rm{m}}^3}\\{r^3} = \frac{3}{{4\pi }} \times 5.198\;{\rm{c}}{{\rm{m}}^3}\\r = \sqrt[3]{{1.242\;{\rm{c}}{{\rm{m}}^3}}}\\ = 1.07\;{\rm{cm}}\end{array}\)

Thus, the radius of the air bubble just before it reaches the surface is 1.07 cm.