Question

A gas filled freely collapsible balloon is pushed from the surface level of a lake to a depth of $50{\text{ m}}$. Approximately what percent of its original volume will the balloon finally have, assuming that the gas behaves ideally and temperature is the same at the surface and at $50{\text{ m}}$ depth?

Answer 1

A gas filled freely collapsible balloon means that if pressure is applied to the balloon the balloon will shrink in its volume. When a balloon is pushed from the surface level, the pressure on the balloon increases as the balloon goes down and as a result the balloon shrinks in its volume. The gas filled inside the balloon behaves ideally and thus, Boyle’s law can be applied.

Complete step by step answer:
Step 1:
Calculate the pressure on the balloon at the surface level as follows:
Let the volume of the balloon at the surface level be ${V_{{\text{surface}}}}$.
The pressure on the balloon at the surface level is equal to the atmospheric pressure. Thus,
${P_{{\text{surface}}}} = 1{\text{ atm}} = 1 \cdot 013 \times {10^5}{\text{ N }}{{\text{m}}^{ - 2}}$.
Step 2:
Calculate the pressure on the balloon at the depth of $50{\text{ m}}$ as follows:
Let the volume of the balloon at the depth of $50{\text{ m}}$ be ${V_{50{\text{ m}}}}$.
The pressure on the balloon at the depth of $50{\text{ m}}$ is,
${P_{50{\text{ m}}}} = {P_{{\text{surface}}}} + {P_{{\text{water}}}}$
${P_{50{\text{ m}}}} = {P_{{\text{surface}}}} + h\rho g$
Where, $h$ is the depth,
$\rho$ is the density of water,
$g$ is the acceleration due to gravity.
Substitute $50{\text{ m}}$ for the depth, $1000{\text{ kg }}{{\text{m}}^{ - 3}}$for the density of water, $9 \cdot 8{\text{ N k}}{{\text{g}}^{ - 1}}$ for the acceleration due to gravity. Thus,
${P_{50{\text{ m}}}} = 1 \cdot 013 \times {10^5}{\text{ N }}{{\text{m}}^{ - 2}} + 50{\text{ m}} \times 1000{\text{ kg }}{{\text{m}}^{ - 3}} \times 9 \cdot 8{\text{ N k}}{{\text{g}}^{ - 1}}$
${P_{50{\text{ m}}}} = 591300{\text{ N }}{{\text{m}}^{ - 2}}$
Thus, the pressure on the balloon at the depth of $50{\text{ m}}$ is $591300{\text{ N }}{{\text{m}}^{ - 2}}$.
Step 3:
Calculate what percent of its original volume will the balloon finally have as follows:
The gas filled in the balloon behaves as an ideal gas. Thus, Boyle’s law is applicable. Thus,
${P_{{\text{surface}}}}{V_{{\text{surface}}}} = {P_{{\text{50 m}}}}{V_{{\text{50 m}}}}$
Thus,
$1 \cdot 013 \times {10^5}{\text{ N }}{{\text{m}}^{ - 2}} \times {V_{{\text{surface}}}} = 591300{\text{ N }}{{\text{m}}^{ - 2}} \times {V_{{\text{50 m}}}}$
$\dfrac{{{V_{{\text{50 m}}}}}}{{{V_{{\text{surface}}}}}} = \dfrac{{1 \cdot 013 \times {{10}^5}{\text{ }}\not{{{\text{N }}{{\text{m}}^{ - 2}}}}}}{{591300{\text{ }}\not{{{\text{N }}{{\text{m}}^{ - 2}}}}}}$
$\dfrac{{{V_{{\text{50 m}}}}}}{{{V_{{\text{surface}}}}}}\% = 0 \cdot 1713 \times 100\%$
$\dfrac{{{V_{{\text{50 m}}}}}}{{{V_{{\text{surface}}}}}}\% = 17 \cdot 13\%$
Thus, the balloon will have $17\%$ of its original volume.

Note:
The gas filled inside the balloon behaves as an ideal gas. Thus, Boyle’s law is applicable. Boyle’s law states that the pressure of gas varies inversely with its volume at a constant temperature.