Q11-34E - University Physics with Modern Physics

Question

In the Challenger Deep of the Marianas Trench, the depth of seawater is 10.9 km and the pressure is 1.16^*10⁸Pa (about 1.15^*10³atm) If a cubic meter of water is taken from the surface to this depth, what is the change in its volume? (Normal atmospheric pressure is about 1.0^*10⁵Pa. Assume that   for seawater is the same as the freshwater value given in Table 11.2.) (b) What is the density of seawater at this depth? (At the surface, seawater has a density of 1.03^*10³kg/m³.

Step-by-Step Solution

Answer

$$\begin{gathered} \left(a\right) -0.0527{\text{ m}}^3 \\ \left(b\right) 1.09\times {10}^{3}kg/m^3 \end{gathered}$$

Step 1: Given information:

$V_0=1m^3, ∆p=1.{16}^{*}{10}^{8}Pa$

Step 2: Concept/Formula used:

Bulk modulus is given by the ratio of pressure applied to the corresponding relative decrease in the volume of the material. 

Mathematically, it is represented as follows:

$\mathit{\beta }\mathbf{=}\frac{\mathbf{\Delta }\mathbf{p}}{\frac{\mathbf{\Delta }\mathbf{V}}{{\mathbf{V}}_{\mathbf{0}}}}$

Where, $\mathit{\beta }$ is Bulk modulus

$\mathit{\Delta }\mathit{p}$ is change of the pressure or force applied per unit area on the material

$\mathbf{∆}\mathit{V}$ is Change of the volume of the material due to the compression

V₀ is Initial volume of the material

Step 3: Volume change Calculation

(a)

$\begin{array}{rcl}\beta & =& \frac{\Delta p}{\frac{\Delta V}{V_0}}\\ \Delta V& =& -\frac{\left(\Delta p\right)V_0}{\beta }\\ & =& -\frac{\left(1.16\times {10}^{8}Pa\right)\left(1m^3\right)}{\left(2.2\times {10}^{9}Pa\right)}\\ & =& -0.0527{\text{ m}}^3\end{array}$

Step 4: Density of seawater

(b)

At the given depth mass of seawater is $1.03\times {10}^{3}kg$ having volume

$\begin{array}{rcl}{V}_0+\Delta V& =& 1-0.0527\\ & =& 0.9473{\text{ m}}^3\end{array}$

The density is:

$\begin{array}{rcl}\rho & =& \frac{1.03\times {10}^{3}kg}{0.9473{\text{ m}}^3}\\ & =& 1.09\times {10}^{3}kg/m^3\end{array}$