Question: An air bubble of volume 20 cm³is at the bottom of a lake 40 mdeep, where the temperature is 4. 0 ⁰C. The bubble rises to the surface, which is at a temperature of. Take the temperature of 20 ⁰C the bubble’s air to be the same as that of the surrounding water. Just as the bubble reaches the surface, what is its volume?

1. Volume of the air bubble at the bottom of the lake,$V_1=20{\text{cm}}^3=20\times {10}^{-6}{\text{m}}^3.$
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,

,$p_1=p_0+\rho gh$

Where, $p_0$is the atmospheric pressure and $\rho$is the density of water.

Thus,

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

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:

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

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

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