Chapter 1: Q11-34E (page 1)

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.

Short Answer

Expert verified

$(a) - 0.0527 m^{3} (b) 1.09 \times 10^{3} k g / m^{3}$

Step by step solution

Given information:

$V_{0} = 1 m^{3}, ∆ p = 1 . 16^{*} 10^{8} P a$

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:

$β = \frac{Δ p}{\frac{Δ V}{V_{0}}}$

Where, $β$ is Bulk modulus

$Δ p$ is change of the pressure or force applied per unit area on the material

$∆ V$ is Change of the volume of the material due to the compression

V is Initial volume of the material

Volume change Calculation

(a)

$\begin{array}{rcl} β & = & \frac{Δ p}{\frac{Δ V}{V_{0}}} \\ Δ V & = & - \frac{(Δ p) V_{0}}{β} \\ = & - \frac{(1.16 \times 10^{8} P a) (1 m^{3})}{(2.2 \times 10^{9} P a)} \\ = & - 0.0527 m^{3} \end{array}$

Density of seawater

(b)

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

$\begin{array}{rcl} V_{0} + Δ V & = & 1 - 0.0527 \\ = & 0.9473 m^{3} \end{array}$

The density is:

$\begin{array}{rcl} ρ & = & \frac{1.03 \times 10^{3} k g}{0.9473 m^{3}} \\ = & 1.09 \times 10^{3} k g / m^{3} \end{array}$

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Short Answer

Step by step solution

Given information:

Concept/Formula used:

Volume change Calculation

Density of seawater

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