Chapter 14: Q13P (page 407)

At a depth of 10.9 km, the Challenger Deep in the Marianas Trench of the Pacific Ocean is the deepest site in any ocean. Yet, in 1960, Donald Walsh and Jacques Piccard reached the Challenger Deep in the bathyscaph Trieste. Assuming that seawater has a uniform density of $1024 kg / m^{3}$ , approximate the hydrostatic pressure (in atmospheres) that the Trieste had to withstand. (Even a slight defect in the Trieste structure would have been disastrous.)

Short Answer

Expert verified

Hydrostatic pressure in atmospheres is $1.08 \times 10^{3} atm$

Step by step solution

The given data

Density of seawater, $p = 1024 kg / m^{3}$
Depth, $h = 10.9 km or 10.9 \times 10^{3} m$

Understanding the concept of hydrostatic pressure

To solve for pressure, we can use the formula for hydrostatic pressure difference, which depends on density and height. Hydrostatic pressure is that, which is exerted by a fluid at any point of time at equilibrium due to the force applied by gravity.

Formula:

Pressure applied on a fluid surface, p = pgh (i)

Calculation of hydrostatic pressure in atmospheres

Using equation (i) and the given values, the pressure can be written as:

$p = 1024 kg / m^{3} \times 9.8 m / s^{2} \times 10.9 \times 10^{3} m p = 1.09 \times 10^{8} Pa (∵ 1 atm = 1.01 \times 10^{5} Pa) = 1.09 \times 10^{8} \times \frac{1 atm}{1.01 \times 10^{5}} = 1.08 \times 10^{3} atm$