Question

An air bubble of volume 1.0 \[{\mathbf{c}}{{\mathbf{m}}^{\mathbf{3}}}\] rises from the bottom of a lake 40 m deep at a temperature of \[{\mathbf{1}}{{\mathbf{2}}^ \circ }{\text{C}}\] . To what volume does it grow when it reaches the surface which is at a temperature of \[{\mathbf{3}}{{\mathbf{5}}^ \circ }{\mathbf{C}}\;\]?

Answer 1

Hint: To answer this question, you should recall the concept of transfer of gases. Use the relation of an ideal gas for the two conditions. Use the given values to find the answer to this question.

Formula used:
Ideal gas equation \[{\text{PV = nRT}}\]where ${\text{P}}$is pressure,${\text{V}}$is volume, ${\text{n}}$is the number of moles, ${\text{R}}$is the universal gas constant, ${\text{T}}$is temperature.

Complete Step by step solution:
We know that as the gas is transferred without the change in the number of moles, they will remain constant in both the conditions.
The ideal gas equation for both the states can be written as:\[\dfrac{{{{\text{P}}_{\text{1}}}{{\text{V}}_{\text{1}}}}}{{{{\text{T}}_{\text{1}}}}}{\text{ = }}\dfrac{{{{\text{P}}_{\text{2}}}{{\text{V}}_{\text{2}}}}}{{{{\text{T}}_{\text{2}}}}}\].
Now considering the two situations, we have:
Volume of the air bubble, \[{{\text{V}}_{{\text{1}}}} = 1.0{\text{c}}{{\text{m}}^{\text{3}}} = 1.0 \times {10^{ - 6}}{{\text{m}}^{\text{3}}}\]
Bubble rises to height, d = 40m
Temperature at a depth of 40 m, \[{{\text{T}}_{\text{1}}} = {12^o}{\text{C}} = 285{\text{K}}\]
Temperature at the surface of the lake, \[{{\text{T}}_{\text{2}}} = {35^o}{\text{C}} = 308{\text{K}}\]
The pressure on the surface of the lake: \[{{\text{P}}_{\text{2}}} = 1{\text{atm}} = 1 \times 1.103 \times {10^5}{Pa\;}\]
The pressure at the depth of 40 m: \[{{\text{P}}_1}{\text{ = }}1{\text{atm}} +{d\rho g}\]
Where,
\[\rho \] is the density of water \[ = {10^3}{\text{kg/m3}}\]
\[g\;\]is the acceleration due to gravity \[ = 9.8{\text{m/}}{{\text{s}}^{\text{2}}}\]
\[\therefore {{\text{P}}_{\text{1}}} = 1.103 \times {10^5} + 40 \times {10^3} \times 9.8 = 493300{\text{Pa}}\]
We have, \[\dfrac{{{{\text{P}}_{\text{1}}}{{\text{V}}_{\text{1}}}}}{{{{\text{T}}_{\text{1}}}}}{\text{ = }}\dfrac{{{{\text{P}}_{\text{2}}}{{\text{V}}_{\text{2}}}}}{{{{\text{T}}_{\text{2}}}}}\]
Where, \[{{\text{V}}_{\text{2}}}\] ​ is the volume of the air bubble when it reaches the surface.
\[{{\text{V}}_{\text{2}}}{\text{ = }}\dfrac{{{{\text{P}}_{\text{1}}}{{\text{V}}_{\text{1}}}{{\text{T}}_{\text{2}}}}}{{{{\text{T}}_{\text{1}}}{{\text{P}}_{\text{2}}}}}\]$ = \dfrac{{493300 \times 1 \times {{10}^{ - 6}} \times 308}}{{285 \times 1.013 \times 105}}$
\[ \Rightarrow 5.263 \times {10^{ - 6}}{{\text{m}}^{\text{3}}} = \;5.263{\text{c}}{{\text{m}}^{\text{3}}}\]
Therefore, when the air bubble reaches the surface, its volume becomes \[5.263{\text{c}}{{\text{m}}^{\text{3}}}\].

Note: At 'higher temperature' and 'lower pressure', a gas behaves like an ideal gas, as the potential energy due to intermolecular forces becomes less significant compared with the particles' kinetic energy, and the size of the molecules becomes less significant compared to the space between them. So, the ideal gas equation is preferred in these cases rather than the real gas equation.