   Chapter 13.5, Problem 3.1ACP

Chapter
Section
Textbook Problem

Calculate the external pressure experienced by a diver at a depth of 10.0 m. If a scuba diver is using compressed air at this depth, what are the partial pressures of oxygen and nitrogen? (Assume the density of water is 1.00 g/mL.)

Interpretation Introduction

Interpretation: The external pressure experienced by the diver and the partial pressures of oxygen and nitrogen used by the diver has to be determined.

Concept introduction:

• Raoult’s law: the vapor pressure of a solvent above a solution (Psolution)is equal to the vapor pressure of the pure solvent(P0solvent ) at the same temperature scaled by the mole fraction of the solvent (XSolvent )present.

Psolution=Xsolvent×P0solvent

• Density=MassVolume
• Pressureinpascal=density×accelartionduetogravity×depth
• Pressureinatm=pressureinpascal×1atm101325Pascal
Explanation

Given,

Density  of water=1g/mLDensityofwaterinSIunit=1.0g/cm3(1kg1000g)×(100cm1m)3 =1.00×103kg/m3

Pressure can be calculated by the equation given below:

P=ρghρdensityggravitationalconstanthheight

Pressureofwaterinpascal=1.00×103kg/m3×(9.81m/s2)×(10.0m) =9.81×104PaPressureofwaterinatm=9.81×104Pa×(1atm101325Pa) =0

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