91Ó°ÊÓ

1. Volume of the air bubble at the bottom of the lake,$V_1=20{\text{cm}}^3=20×{10}^{-6}{\text{m}}^3.$
2. The depth of the lake,h = 40 m
3. Temperature of the air bubble at the bottom of the lake,$T_1=4.0°\text{C}=277\text{K}.$

Temperature of the air bubble at the surface of the lake,$T_0=20°\text{C}=293\text{K}.$

Find thenumber of moles using the gas law. Using this value of number of moles, find the volume of the bubble as it reaches the surface by applying the gas law to the air bubble at the surface.

Formula is as follow:

$p_i{v}_i=nR{T}_i$

Here, p is pressure, v is volume, T is temperature, R is universal gas constant and n is number of moles.

According to the gas law,

pV = n RT

Pressure at the bottom of the lake is given by,

,$p_1=p_0+\mathrm{ÒÏ}gh$

Where, $p_0$is the atmospheric pressure and $\mathrm{ÒÏ}$is the density of water.

Thus,

$$\begin{gathered} n=\frac{(p_0+\mathrm{ÒÏ}gh)V_1}{R{T}_1} \\ n=\frac{\left(1.013×{10}^5+998\left(9.8\right)\left(40\right)\right)\left(20×{10}^{-6}\right)}{8.314×277} \\ n=4.277×{10}^{-3}\text{moles} \end{gathered}$$

Consider the bubble reaches the surface the gas, the law becomes:

$$\begin{gathered} p_0{V}_0=nR{T}_0 \\ {V}_0=\frac{nR{T}_0}{P_0} \end{gathered}$$

Substitute the values and solve as:

$$\begin{gathered} V_0=\frac{4.277×{10}^{-3}\left(8.314\right)\left(293\right)}{1.013×{10}^5} \\ {V}_0=1.02×{10}^{-4}{\text{m}}^3 \\ {V}_0=1.02×{10}^2{\text{cm}}^3 \end{gathered}$$

Hence,thevolume of the bubble as it reaches the surface is$1.02×{10}^2{\text{cm}}^3$ .

Therefore, the volume of the bubble as it reaches the surface can be found using the gas law.