Question

Black smokers are hot volcanic vents that emit smoke deep in the ocean floor. Many of them teem with exotic creatures, and some biologists think that life on earth may have begun around such vents. The vents range in depth from about 1500 m to 3200 m below the surface. What is the gauge pressure at a 3200-m deep vent, assuming that the density of water does not vary? Express your answer in Pascal and atmospheres.

Short answer

Density of water, in Pa =$3.23\times {10}^{7}Pa$

Density of water ,in atm = 320atm

Step-by-step solution

 Step 1: Identification of given value

Lower depth of vent from surface = 1500 m

Upper depth of vent from surface = 3200 m

Concept of density

Density is a measure of how densely or loosely a particular material is packed into a given volume.

It is directly proportional to mass and inversely proportional to volume of object.

$\rho =\frac{m}{V}$

Determination of density of water in Pa

$P_{\text{gauge}}=h\times \rho \times g=3200\mathrm{m}\times 1030\mathrm{kg}/{\mathrm{m}}^3\times 9.81\mathrm{m}/{\mathrm{s}}^2=3.23\times {10}^{17}\mathrm{Pa}$

Determination of density of water in atm

$P_{\text{gauge}}=3.23\times {10}^7\mathrm{Pa}\times \frac{1\mathrm{latm}}{1.01\times {10}^5\mathrm{Pa}}=320.13\mathrm{atm}$

Thus, density of water in Pa = $3.23\times {10}^{7}Pa$

Density of water in atm = 320atm