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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 \(8.14\;{\rm{J/mol}} \cdot {\rm{K}}\).

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 \(\rho = 1.00 \times {10^3}\;{\rm{kg/}}{{\rm{m}}^3}\).

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.

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

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.

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

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.