## Volume of Bubbles Released From Seabed at Sea Surface

Suppose for illustrative purposes that the sea is 100 m deep at a certain point. and a bubble of volume

\[1 \: cm^3= 10^{-6} \: m^3\]

is released from the seabed.The pressure on the seabed is due to atmospheric pressure plus the pressure due to the 100m of water above that point. If we take atmospheric pressure as

\[10^5 \: Pa\]

and the density of seawater to be a constant \[1030 \: kg/m^3\]

.\[p= \rho g h = 1030 \times 9.8 \times 100 +10^5 = 1009400 + 10^5 =1109400 Pa\]

.The temperature at the sea bed will be

\[4^{\circ} C \equiv 277^{\circ} K\]

(this is the temperature at which water is densest). When it reaches the water surface, the pressure will be atmospheric pressure. Take the temperature as \[15^{\circ} C \equiv 288^{\circ} C\]

Assuming the gas in the bubble is ideal we can use the ideal gas equation

\[\frac{p_1V_1}{T_1} = \frac{p_2 V_2}{T_2}\]

\[\frac{1109400 \times 10^{-6}}{277} = \frac{10^5 \times V_2}{288} \rightarrow V_2 = \frac{1109400 \times 10^{-6} \times 288}{277 \times 10^5}= 1.153 \: m^3\]

or \[11.53 \: cm^3\]

.