Question

(II) At depths of 2000 m in the sea, the pressure is about 200 times atmospheric pressure\(\left( {{{\bf{P}}_{\bf{a}}}{\bf{ = 1}}{\bf{.0 \times 1}}{{\bf{0}}^{\bf{5}}}\;{\bf{N/}}{{\bf{m}}^{\bf{2}}}} \right)\). By what percentage does the interior space of an iron bathysphere’s volume change at this depth?

Short answer

The volume of the interior space of an iron bathysphere decreases by \(2 \times {10^{ - 2}}\;\% \).

Step-by-step solution

Bulk modulus

If pressure acts on an object from all the sides in an inward direction, its volume decreases by an amount directly proportional to the original volume (\({V_0}\)) and the change in the pressure\(\left( {\Delta P} \right)\), i.e., \(\Delta V = - \frac{1}{B}{V_0}\Delta P\).

Here, the minus sign indicates the volume decreases on increasing the pressure, and B is the constant of proportionality, termed the bulk modulus.

In this problem, the change in volume of the interior space of an iron bathysphere can be calculated using the above expression. The value of bulk modulus of the iron is\(B{\bf{ = 90 \times 1}}{{\bf{0}}^{\bf{9}}}\;{\bf{N/}}{{\bf{m}}^{\bf{2}}}\).

Given information

The value of atmospheric pressure is\({P_{\rm{a}}} = 1.0 \times {10^5}\;{\rm{N/}}{{\rm{m}}^{\rm{2}}}\).

The original pressure on the iron bathysphere is equal to the atmospheric pressure, i.e., \({P_0} = {P_{\rm{a}}}\).

At a depth of 2000 m in the sea, the pressure on the iron bathysphere is\(P = 200{P_{\rm{a}}}\).

Thus, change in pressureis calculated as follows:

\(\begin{array}{c}\Delta P = P - {P_0}\\ = {P_{\rm{a}}} - 200{P_{\rm{a}}}\\ = 199{P_{\rm{a}}}\end{array}\)

Determination of percentage change in the volume of interior space of iron bathysphere

If\({V_0}\)is the original volume of interior space of bathysphere, and\(\Delta V\)is the change in volume at a depth of 2000 m in the sea, the percentage chain in the volume of the iron bathysphere is

\(\frac{{\Delta V}}{{{V_0}}} \times 100\).

The change in volumeof the interior space of an iron bathysphere can be written as follows:

\(\begin{array}{c}\Delta V = - \frac{1}{B}{V_0}\Delta P\\\frac{{\Delta V}}{{{V_0}}} = - \frac{1}{B}\Delta P\end{array}\)

Thus, the percentage change in volume is as follows:

\(\begin{array}{c}\frac{{\Delta V}}{{{V_0}}} \times 100 = - \frac{1}{B}\Delta P \times 100\\ = - \frac{{199{P_{\rm{a}}}}}{B} \times 100\\ = - \frac{{199 \times \left( {1.0 \times {{10}^5}\;{\rm{N/}}{{\rm{m}}^{\rm{2}}}} \right)}}{{90 \times {{10}^9}\;{\rm{N/}}{{\rm{m}}^{\rm{2}}}}} \times 100\\ = - 2.21 \times {10^{ - 2}}\;\% \\ \approx - 2 \times {10^{ - 2}}\;\% \end{array}\)

Here, the negative sign shows that the volume of the bathysphere has decreased.

Thus, the volume of interior space of an iron bathysphere decreases by \(2 \times {10^{ - 2}}\;\% \).