Question

(II) One liter of alcohol \(\left( {{\bf{1000}}\;{\bf{c}}{{\bf{m}}^{\bf{3}}}} \right)\) in a flexible container is carried to the bottom of the sea, where the pressure is \({\bf{2}}{\bf{.6 \times 1}}{{\bf{0}}^{\bf{6}}}\;{\bf{N/}}{{\bf{m}}^{\bf{2}}}\). What will be its volume there?

Short answer

The volume of the alcohol at the bottom of the sea is \(997.5\;{\rm{c}}{{\rm{m}}^3}\).

Step-by-step solution

Bulk modulus

When pressure acts on an object from all sides, 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, B is the constant of proportionality, termed the bulk modulus or Young’s modulus, and the minus sign indicates that volume decreases on increasing the pressure.

In this problem, when the container of alcohol is carried to the bottom of the sea, the water exerts pressure on the object in all directions.

The value of the bulk modulus of the water is\(B{\bf{ = 1}}{\bf{.0 \times 1}}{{\bf{0}}^{\bf{9}}}\;{\bf{N/}}{{\bf{m}}^{\bf{2}}}\).

Given information

The original volume of alcohol placed in a flexible container is as follows:

\(\begin{array}{c}{V_0} = 1000\;{\rm{c}}{{\rm{m}}^3}\\ = 1000 \times {10^{ - 6}}\;{{\rm{m}}^3}\\ = 1.0 \times {10^{ - 3}}\;{{\rm{m}}^3}\end{array}\)

Original pressure on the surface of the sea is equal to the atmospheric pressure, i.e.,\({P_0} = 1.0 \times {10^5}\;{\rm{N/}}{{\rm{m}}^{\rm{2}}}\).

Pressure at the bottom of the sea is\(P = 2.6 \times {10^6}\;{\rm{N/}}{{\rm{m}}^{\rm{2}}}\).

Let V be the volume of alcohol at the bottom of the sea such that the change in the volume is \(\Delta V = V - {V_0}\).

Determination of change in volume of the alcohol

The change in volume of the alcohol when it is carried at the bottom of the sea is as follows:

\(\begin{array}{c}\Delta V = - \frac{1}{B}{V_0}\Delta P\\ = - \frac{1}{B}{V_0}\left( {P - {P_0}} \right)\\ = - \frac{1}{{\left( {1.0 \times {{10}^9}\;{\rm{N/}}{{\rm{m}}^{\rm{2}}}} \right)}} \times \left( {1.0 \times {{10}^{ - 3}}\;{{\rm{m}}^3}} \right) \times \left( {2.6 \times {{10}^6} - 1.0 \times {{10}^5}\;} \right){\rm{N/}}{{\rm{m}}^{\rm{2}}}\\ = - 2.5 \times {10^{ - 6}}\;{{\rm{m}}^3}\end{array}\)

Thus, the volume of alcohol will get decreased by \(2.5 \times {10^{ - 6}}\;{{\rm{m}}^3}\).

Determination of the volume of the alcohol at the bottom of the sea

The change in volume of the alcohol is given by the following:

\(\begin{array}{c}\Delta V = V - {V_0}\\V = {V_0} + \Delta V\\ = 1.0 \times {10^{ - 3}}\;{{\rm{m}}^3} - 2.5 \times {10^{ - 6}}\;{{\rm{m}}^3}\\ = 997.5 \times {10^{ - 6}}\;{{\rm{m}}^3}\\ = 997.5\;{\rm{c}}{{\rm{m}}^3}\end{array}\)

Thus, the volume of the alcohol at the bottom of the sea is \(997.5\;{\rm{c}}{{\rm{m}}^3}\).