Physics For Scientists And Engineers: Foundations And Connections, Extended Version With Modern Physics
Physics For Scientists And Engineers: Foundations And Connections, Extended Version With Modern Physics
1st Edition
ISBN: 9781305259836
Author: Debora M. Katz
Publisher: Cengage Learning
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Chapter 20, Problem 64PQ

(a)

To determine

The number of molecules.

(a)

Expert Solution
Check Mark

Answer to Problem 64PQ

The number of molecules is 1.07×1023molecules_.

Explanation of Solution

Write the equation of ideal gas law.

  PV=NkBT                                                                                               (I)

Here, P is the pressure, V is the volume, N is the number of molecules, kB is the Boltzmann constant and T is the temperature.

Write the expression for the volume.

  V=43πr3                                                                                         (II)

Here, r is the radius.

Rearrange the expression from equation (I) to express in terms of number of molecules.

  N=P[43πr3]kBT                                                                                                    (III)

Conclusion:

Substitute 1.013×105Pa for P, 10.0cm for r, 1.38×1023J/K for kB and 15°C for T in equation (III) to find N.

  N=(1.013×105Pa)[43π{(10.0cm)(1×102m1cm)}3](1.38×1023J/K)(15°C+273.15)=(1.013×105Pa)[43π(0.100m)3](1.38×1023J/K)(288.15K)=1.07×1023molecules

Thus, the number of molecules is 1.07×1023molecules_.

(b)

To determine

The average kinetic energy.

(b)

Expert Solution
Check Mark

Answer to Problem 64PQ

The average kinetic energy is 5.96×1021J_.

Explanation of Solution

Write the equation of average kinetic energy.

  Kav=32kBT                                                                                                             (IV)

Here, Kav is the average kinetic energy.

Conclusion:

Substitute 1.38×1023J/K for kB and 15°C for T in equation (IV) to find Kav.

  Kav=32(1.38×1023J/K)(15°C+273.15)=32(1.38×1023J/K)(288.15K)=5.96×1021J

Thus, the average kinetic energy is 5.96×1021J_.

(c)

To determine

The rms speed of oxygen.

(c)

Expert Solution
Check Mark

Answer to Problem 64PQ

The rms speed of oxygen is 474m/s_.

Explanation of Solution

Write the equation for the rms speed.

  vrms=3kBTm                                                                                                            (V)

Here, vrms is the rms speed, m is the molar mass.

Conclusion:

Substitute 32(1.66×1027kg) for m, 1.38×1023J/K for kB and 15°C for T in equation (V) to find vrms.

  vrms=3(1.38×1023J/K)(15°C+273.15)[32(1.66×1027kg)]=3(1.38×1023J/K)(288.15K)[32(1.66×1027kg)]=474m/s

Therefore, the rms speed of oxygen is 474m/s_.

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Chapter 20 Solutions

Physics For Scientists And Engineers: Foundations And Connections, Extended Version With Modern Physics

Ch. 20 - Prob. 5PQCh. 20 - Prob. 6PQCh. 20 - Prob. 7PQCh. 20 - Prob. 8PQCh. 20 - Particles in an ideal gas of molecular oxygen (O2)...Ch. 20 - Prob. 10PQCh. 20 - Prob. 11PQCh. 20 - Prob. 12PQCh. 20 - Prob. 13PQCh. 20 - Prob. 14PQCh. 20 - The mass of a single hydrogen molecule is...Ch. 20 - Prob. 16PQCh. 20 - The noble gases neon (atomic mass 20.1797 u) and...Ch. 20 - Prob. 18PQCh. 20 - Prob. 19PQCh. 20 - Prob. 20PQCh. 20 - Prob. 22PQCh. 20 - Prob. 23PQCh. 20 - Prob. 24PQCh. 20 - Prob. 25PQCh. 20 - Prob. 26PQCh. 20 - Prob. 27PQCh. 20 - Prob. 28PQCh. 20 - Consider the Maxwell-Boltzmann distribution...Ch. 20 - Prob. 30PQCh. 20 - Prob. 31PQCh. 20 - Prob. 32PQCh. 20 - Prob. 33PQCh. 20 - Prob. 34PQCh. 20 - Prob. 35PQCh. 20 - Prob. 36PQCh. 20 - Prob. 37PQCh. 20 - Prob. 38PQCh. 20 - Prob. 39PQCh. 20 - Prob. 40PQCh. 20 - Prob. 41PQCh. 20 - Prob. 42PQCh. 20 - Prob. 43PQCh. 20 - Prob. 44PQCh. 20 - Figure P20.45 shows a phase diagram of carbon...Ch. 20 - Prob. 46PQCh. 20 - Prob. 47PQCh. 20 - Consider water at 0C and initially at some...Ch. 20 - Prob. 49PQCh. 20 - Prob. 50PQCh. 20 - Prob. 51PQCh. 20 - Prob. 52PQCh. 20 - Prob. 53PQCh. 20 - Prob. 54PQCh. 20 - Prob. 55PQCh. 20 - Prob. 56PQCh. 20 - Consider again the box and particles with the...Ch. 20 - Prob. 58PQCh. 20 - The average kinetic energy of an argon atom in a...Ch. 20 - For the exam scores given in Table P20.60, find...Ch. 20 - Prob. 61PQCh. 20 - Prob. 62PQCh. 20 - Prob. 63PQCh. 20 - Prob. 64PQCh. 20 - Prob. 65PQCh. 20 - Prob. 66PQCh. 20 - Determine the rms speed of an atom in a helium...Ch. 20 - Consider a gas filling two connected chambers that...Ch. 20 - Prob. 69PQCh. 20 - Prob. 70PQCh. 20 - A 0.500-m3 container holding 3.00 mol of ozone...Ch. 20 - Prob. 72PQCh. 20 - Prob. 73PQ
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