Physics
Physics
3rd Edition
ISBN: 9781259233616
Author: GIAMBATTISTA
Publisher: MCG
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Chapter 13, Problem 84P

(a)

To determine

The number of O2 molecules per cubic meter in the surface air at a temperature 20.0°C and pressure 1.00atm.

(a)

Expert Solution
Check Mark

Answer to Problem 84P

The number of O2 molecules per cubic meter in the surface air at a temperature 20.0°C and pressure 1.00atm is 5.2×1024m-3.

Explanation of Solution

The percentage of oxygen in dry surface air is 21%. The percentage of nitrogen is 78%. The percentage of argon is 1%.

Write the formula for the number of oxygen molecule for cubic meter of air.

NOV=0.21ma/ρa (I)

Here, NO is the number of oxygen molecule, V is the volume, ma is the mass of air, ρa is the density of air.

Write the formula for the mass of air.

ma=MNA (II)

Here, M is the molar mass, NA is the Avogadro’s number.

Substitute equation (II) in equation (I).

NOV=0.21ρaNAM (III)

Write the formula for the molar mass of air.

M=(0.78)(2×MN)+(0.21)(2×MO)+(0.01)(MA) (IV)

Here, MN is the molar mass of nitrogen, MO is the molar mass of oxygen, MA is the molar mass of argon.

Substitute equation (IV) in equation (III).

NOV=0.21ρaNA(0.78)(2×MN)+(0.21)(2×MO)+(0.01)(MA) (V)

Conclusion:

Substitute 1.20×103g/m3 for ρa, 6.022×1023mol-1 for NA, 14.00674g/mol for MN, 15.9994g/mol for MO, 39.948g/mol for MA,NOV=0.21(1.20×103g/m3)(6.022×1023mol-1)(0.78)(2×(14.00674g/mol))+(0.21)(2×(15.9994g/mol))+(0.01)(39.948g/mol)=5.2×1024m-3

The number of O2 molecules per cubic meter in the surface air at a temperature 20.0°C and pressure 1.00atm is 5.2×1024m-3.

(b)

To determine

The percentage of oxygen molecule in the tank of the diver who goes to a depth of 100.0m.

(b)

Expert Solution
Check Mark

Answer to Problem 84P

The percentage of oxygen molecule in the tank of the diver who goes to a depth of 100.0m is 1.9%

Explanation of Solution

The percentage of oxygen in dry surface air is 21%. The percentage of nitrogen is 78%. The percentage of argon is 1%. The density of sea water is 1025kg/m3.

Write the ideal gas equation.

PV=nRT

Here, P is the pressure, V is the volume, n is the number of moles, R is the universal gas constant, T is the temperature.

When number of moles and temperature is constant, the volume is inversely proportional to the pressure.

Write the equation for the ratio of initial and final volume.

VfVi=PiPf

Here, Vi is the initial volume, Pi is the initial pressure, Pf is the pressure of the pump as air starts to flow into the tire.

Write the equation for the percentage of oxygen molecules in the tank.

VfVi(21%)=PiPf(21%) (V)

The initial pressure is at atmospheric pressure.

Pi=Patm (VI)

Here, Patm is the atmospheric pressure.

Write the formula for the final pressure.

Pf=Patm+ρgh . (VII)

Here, ρ is the density, g is the acceleration due to gravity, h is the depth.

Substitute equation (VI) and (VII) in equation (V).

VfVi(21%)=PatmPatm+ρgh(21%) (VIII)

Conclusion:

Substitute 1025kg/m3 for ρ, 1.013×105Pa for Patm, 9.80m/s2 for g, 100.0m for h.

VfVi(21%)=1.013×105Pa1.013×105Pa+(1025kg/m3)(9.80m/s2)(100.0m)(21%)=1.9%

The percentage of oxygen molecule in the tank of the diver who goes to a depth of  100.0m is 1.9%

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Chapter 13 Solutions

Physics

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