Physics
Physics
3rd Edition
ISBN: 9781259233616
Author: GIAMBATTISTA
Publisher: MCG
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Chapter 13, Problem 67P
To determine

The rms speeds of helium atom, nitrogen, hydrogen and oxygen molecules.

Expert Solution & Answer
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Answer to Problem 67P

The rms speeds of helium atom, nitrogen, hydrogen and oxygen molecules are 1360 ms1, 515 ms1, 1920 ms1 and 482 ms1 respectively.

Explanation of Solution

The given temperature is 25°C.

Find the temperature in Kelvin scale

T(K)=273.15K+T(°C)                                (I)

Here, T(K) is the temperature in Kelvin and T(°C) is the temperature in degree Celsius.

Substitute 25 for T(°C) in (I) to find T(K)

T(K)=273.15K+25K=298.15K

Write the expression for rms speed of atoms and molecules

vrms=3kTm                                                           (II)

Here, vrms is the rms speed, k is the Boltzmann constant, T is the temperature and m is the molecular mass.

The atomic mass of helium is 4.002602u.

Substitute 4.002602u for m, 1.38×1023JK1 for k and 298.15K for T in (II) to find vrms of helium atom.

vrms=3(1.38×1023JK1)(298.15K)4.002602u=3(1.38×1023J)(298.15)4.002602 u×(1.66×1027kgu)=3(1.38×1023kgm2s2)(298.15)4.002602×1.66×1027kg=1363 ms11360 ms1

Thus, the rms speed of helium atom is 1360 ms1.

The molecular mass of nitrogen is 2(14.00674u).

Substitute 2(14.00674u) for m, 1.38×1023JK1 for k and 298.15K for T in (II) to find vrms of nitrogen molecule.

vrms=3(1.38×1023JK1)(298.15K)2(14.00674u)=3(1.38×1023J)(298.15)2(14.00674u)×(1.66×1027kgu)=3(1.38×1023kgm2s2)(298.15)2(14.00674)×1.66×1027kg=515 ms1

Thus, the rms speed of nitrogen molecule is 515 ms1.

The molecular mass of hydrogen is 2(1.00794u).

Substitute 2(1.00794u) for m, 1.38×1023JK1 for k and 298.15K for T in (II) to find vrms of hydrogen molecule.

vrms=3(1.38×1023JK1)(298.15K)2(1.00794u)=3(1.38×1023J)(298.15)2(1.00794u)×(1.66×1027kgu)=3(1.38×1023kgm2s2)(298.15)2(1.00794)×1.66×1027kg=1920ms1

Thus, the rms speed of hydrogen molecule is 1920ms1.

The molecular mass of oxygen is 2(15.9994u).

Substitute 2(15.9994u) for m, 1.38×1023JK1 for k and 298.15K for T in (II) to find vrms of oxygen molecule.

vrms=3(1.38×1023JK1)(298.15K)2(15.9994u)=3(1.38×1023J)(298.15)2(15.9994u)×(1.66×1027kgu)=3(1.38×1023kgm2s2)(298.15)2(15.9994)×1.66×1027kg=482 ms1

Thus, the rms speed of oxygen molecule is 482 ms1.

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Chapter 13 Solutions

Physics

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