Question

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$\mathbf{1024}\mathbf{}\mathbf{kg}\mathbf{/}{\mathbf{m}}^{\mathbf{3}}$, 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

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

Step-by-step solution

The given data

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

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:

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

Hence, the value of hydrostatic pressure is$1.08\times {10}^3\mathrm{atm}$