Question

A small bubble rises from the bottom of a lake, where the temperature and pressure are $4{{\text{ }}^ \circ }C$ and ${\text{3}}{\text{.0 atm}}$ the water's surface, where the temperature is $25{{\text{ }}^ \circ }C$ and pressure is ${\text{95 atm}}$ Calculate the final volume of the bubble if its initial volume was ${\text{2}}{\text{.1 mL}}$?

Answer 1

Here we are given pressure, volume and temperature at the bottom of the lake and we will find the volume of the bubble when it comes to the surface given its pressure and temperature. Thus we will use the ideal gas equation for finding the volume of the bubble according to $PV = nRT$.
Formula Used:
$PV = nRT$
Where, $P$ is pressure of gas, $V$ is the volume of gas, $n$ is the number of moles of gas, $R$ is universal gas constant and $T$ is temperature.

Complete Answer:
From the concept of ideal gas equation we know that,
$PV = nRT$
It can be arranged as,
$\Rightarrow \dfrac{{PV}}{T} = nR$
Since we know that while bubbles move upwards there is no change in the number of moles of gas, thus it is constant. Therefore the term $nR$ will be constant. Thus the above equation can be written as:
$\Rightarrow \dfrac{{PV}}{T} = {\text{ constant}}$ _________$(i)$
Let assume the pressure of bubble at bottom of lake be ${P_1}$, the volume of bubble be ${V_1}$ and temperature of gas will be ${T_1}$. Thus the equation $(i)$ can be written as:
$\Rightarrow \dfrac{{{P_1}{V_1}}}{{{T_1}}} = {\text{ constant}}$ $(ii)$
Now when the bubble moves upward then let the pressure at the surface be ${P_2}$, the volume of the bubble be ${V_2}$ and the temperature of the gas be ${T_2}$. Thus equation $(i)$ can be written as:
$\Rightarrow \dfrac{{{P_2}{V_2}}}{{{T_2}}} = {\text{ constant}}$$(iii)$
On comparing equation $(ii)$ and $(iii)$ we get result as:
$\Rightarrow \dfrac{{{P_1}{V_1}}}{{{T_1}}}{\text{ }} = {\text{ }}\dfrac{{{P_2}{V_2}}}{{{T_2}}}$
The temperature must be in kelvin, therefore ${T_1}{\text{ }} = {\text{ }}273 + 4{\text{ }} = {\text{ }}277{\text{ }}K$ and ${T_1}{\text{ }} = {\text{ }}273 + 25{\text{ }} = {\text{ }}298{\text{ }}K$. Now putting the respective values we get the result as:
$\Rightarrow \dfrac{{3{\text{ }} \times {\text{ 2}}{\text{.1}}}}{{277}}{\text{ }} = {\text{ }}\dfrac{{95{\text{ }} \times {\text{ }}{V_2}}}{{298}}$
On rearranging we get the result as,
$\Rightarrow {\text{ }}{V_2} = {\text{ }}\dfrac{{95{\text{ }} \times {\text{ }}277}}{{298{\text{ }} \times {\text{ 6}}{\text{.1}}}}$
$\Rightarrow {\text{ }}{V_2} = {\text{ }}\dfrac{{26315}}{{1817.8}}$
$\Rightarrow {\text{ }}{V_2} = {\text{ 14}}{\text{.47 mL}}$
Thus the final volume will be ${\text{14}}{\text{.47 mL}}$.

Note:
The result $\dfrac{{{P_1}{V_1}}}{{{T_1}}}{\text{ }} = {\text{ }}\dfrac{{{P_2}{V_2}}}{{{T_2}}}$ must be remember for further reference. We can also use this result directly. The temperature which is used must be in Kelvin scale. If we use degrees celsius then the answer would be different and it would be wrong. For converting into kelvin we degree celcius with $273{\text{ K}}$.