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

Question

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Answer 1

Question

Chapter 13, Problem 67P

To determine

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

Expert Solution & Answer

Answer to Problem 67P

The rms speeds of helium atom, nitrogen, hydrogen and oxygen molecules are 1360 ms−1, 515 ms−1, 1920 ms−1 and 482 ms−1 respectively.

Explanation of Solution

The given temperature is 25° C.

Find the temperature in Kelvin scale

T(K)=273.15 K+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)=273.15 K+25 K=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 4.002602 u.

Substitute 4.002602 u for m, 1.38×10−23JK−1 for k and 298.15K for T in (II) to find vrms of helium atom.

vrms=3(1.38×10−23JK−1)(298.15K)4.002602 u=3(1.38×10−23J)(298.15)4.002602 u×(1.66×10−27kgu)=3(1.38×10−23kgm2s−2)(298.15)4.002602×1.66×10−27kg=1363 ms−1≈1360 ms−1

Thus, the rms speed of helium atom is 1360 ms−1.

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

Substitute 2(14.00674 u) for m, 1.38×10−23JK−1 for k and 298.15K for T in (II) to find vrms of nitrogen molecule.

vrms=3(1.38×10−23JK−1)(298.15K)2(14.00674 u)=3(1.38×10−23J)(298.15)2(14.00674 u)×(1.66×10−27kgu)=3(1.38×10−23kgm2s−2)(298.15)2(14.00674)×1.66×10−27kg=515 ms−1

Thus, the rms speed of nitrogen molecule is 515 ms−1.

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

Substitute 2(1.00794 u) for m, 1.38×10−23JK−1 for k and 298.15K for T in (II) to find vrms of hydrogen molecule.

vrms=3(1.38×10−23JK−1)(298.15K)2(1.00794 u)=3(1.38×10−23J)(298.15)2(1.00794 u)×(1.66×10−27kgu)=3(1.38×10−23kgm2s−2)(298.15)2(1.00794)×1.66×10−27kg=1920 ms−1

Thus, the rms speed of hydrogen molecule is 1920 ms−1.

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

Substitute 2(15.9994 u) for m, 1.38×10−23JK−1 for k and 298.15K for T in (II) to find vrms of oxygen molecule.

vrms=3(1.38×10−23JK−1)(298.15K)2(15.9994 u)=3(1.38×10−23J)(298.15)2(15.9994 u)×(1.66×10−27kgu)=3(1.38×10−23kgm2s−2)(298.15)2(15.9994)×1.66×10−27kg=482 ms−1

Thus, the rms speed of oxygen molecule is 482 ms−1.

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Answer 2

Question

Chapter 13, Problem 67P

To determine

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

Expert Solution & Answer

Answer to Problem 67P

The rms speeds of helium atom, nitrogen, hydrogen and oxygen molecules are 1360 ms−1, 515 ms−1, 1920 ms−1 and 482 ms−1 respectively.

Explanation of Solution

The given temperature is 25° C.

Find the temperature in Kelvin scale

T(K)=273.15 K+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)=273.15 K+25 K=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 4.002602 u.

Substitute 4.002602 u for m, 1.38×10−23JK−1 for k and 298.15K for T in (II) to find vrms of helium atom.

vrms=3(1.38×10−23JK−1)(298.15K)4.002602 u=3(1.38×10−23J)(298.15)4.002602 u×(1.66×10−27kgu)=3(1.38×10−23kgm2s−2)(298.15)4.002602×1.66×10−27kg=1363 ms−1≈1360 ms−1

Thus, the rms speed of helium atom is 1360 ms−1.

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

Substitute 2(14.00674 u) for m, 1.38×10−23JK−1 for k and 298.15K for T in (II) to find vrms of nitrogen molecule.

vrms=3(1.38×10−23JK−1)(298.15K)2(14.00674 u)=3(1.38×10−23J)(298.15)2(14.00674 u)×(1.66×10−27kgu)=3(1.38×10−23kgm2s−2)(298.15)2(14.00674)×1.66×10−27kg=515 ms−1

Thus, the rms speed of nitrogen molecule is 515 ms−1.

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

Substitute 2(1.00794 u) for m, 1.38×10−23JK−1 for k and 298.15K for T in (II) to find vrms of hydrogen molecule.

vrms=3(1.38×10−23JK−1)(298.15K)2(1.00794 u)=3(1.38×10−23J)(298.15)2(1.00794 u)×(1.66×10−27kgu)=3(1.38×10−23kgm2s−2)(298.15)2(1.00794)×1.66×10−27kg=1920 ms−1

Thus, the rms speed of hydrogen molecule is 1920 ms−1.

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

Substitute 2(15.9994 u) for m, 1.38×10−23JK−1 for k and 298.15K for T in (II) to find vrms of oxygen molecule.

vrms=3(1.38×10−23JK−1)(298.15K)2(15.9994 u)=3(1.38×10−23J)(298.15)2(15.9994 u)×(1.66×10−27kgu)=3(1.38×10−23kgm2s−2)(298.15)2(15.9994)×1.66×10−27kg=482 ms−1

Thus, the rms speed of oxygen molecule is 482 ms−1.

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Answer 3

Question

Chapter 13, Problem 67P

To determine

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

Expert Solution & Answer

Answer to Problem 67P

The rms speeds of helium atom, nitrogen, hydrogen and oxygen molecules are 1360 ms−1, 515 ms−1, 1920 ms−1 and 482 ms−1 respectively.

Explanation of Solution

The given temperature is 25° C.

Find the temperature in Kelvin scale

T(K)=273.15 K+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)=273.15 K+25 K=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 4.002602 u.

Substitute 4.002602 u for m, 1.38×10−23JK−1 for k and 298.15K for T in (II) to find vrms of helium atom.

vrms=3(1.38×10−23JK−1)(298.15K)4.002602 u=3(1.38×10−23J)(298.15)4.002602 u×(1.66×10−27kgu)=3(1.38×10−23kgm2s−2)(298.15)4.002602×1.66×10−27kg=1363 ms−1≈1360 ms−1

Thus, the rms speed of helium atom is 1360 ms−1.

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

Substitute 2(14.00674 u) for m, 1.38×10−23JK−1 for k and 298.15K for T in (II) to find vrms of nitrogen molecule.

vrms=3(1.38×10−23JK−1)(298.15K)2(14.00674 u)=3(1.38×10−23J)(298.15)2(14.00674 u)×(1.66×10−27kgu)=3(1.38×10−23kgm2s−2)(298.15)2(14.00674)×1.66×10−27kg=515 ms−1

Thus, the rms speed of nitrogen molecule is 515 ms−1.

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

Substitute 2(1.00794 u) for m, 1.38×10−23JK−1 for k and 298.15K for T in (II) to find vrms of hydrogen molecule.

vrms=3(1.38×10−23JK−1)(298.15K)2(1.00794 u)=3(1.38×10−23J)(298.15)2(1.00794 u)×(1.66×10−27kgu)=3(1.38×10−23kgm2s−2)(298.15)2(1.00794)×1.66×10−27kg=1920 ms−1

Thus, the rms speed of hydrogen molecule is 1920 ms−1.

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

Substitute 2(15.9994 u) for m, 1.38×10−23JK−1 for k and 298.15K for T in (II) to find vrms of oxygen molecule.

vrms=3(1.38×10−23JK−1)(298.15K)2(15.9994 u)=3(1.38×10−23J)(298.15)2(15.9994 u)×(1.66×10−27kgu)=3(1.38×10−23kgm2s−2)(298.15)2(15.9994)×1.66×10−27kg=482 ms−1

Thus, the rms speed of oxygen molecule is 482 ms−1.

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Answer 4

Question

Chapter 13, Problem 67P

To determine

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

Expert Solution & Answer

Answer to Problem 67P

The rms speeds of helium atom, nitrogen, hydrogen and oxygen molecules are 1360 ms−1, 515 ms−1, 1920 ms−1 and 482 ms−1 respectively.

Explanation of Solution

The given temperature is 25° C.

Find the temperature in Kelvin scale

T(K)=273.15 K+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)=273.15 K+25 K=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 4.002602 u.

Substitute 4.002602 u for m, 1.38×10−23JK−1 for k and 298.15K for T in (II) to find vrms of helium atom.

vrms=3(1.38×10−23JK−1)(298.15K)4.002602 u=3(1.38×10−23J)(298.15)4.002602 u×(1.66×10−27kgu)=3(1.38×10−23kgm2s−2)(298.15)4.002602×1.66×10−27kg=1363 ms−1≈1360 ms−1

Thus, the rms speed of helium atom is 1360 ms−1.

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

Substitute 2(14.00674 u) for m, 1.38×10−23JK−1 for k and 298.15K for T in (II) to find vrms of nitrogen molecule.

vrms=3(1.38×10−23JK−1)(298.15K)2(14.00674 u)=3(1.38×10−23J)(298.15)2(14.00674 u)×(1.66×10−27kgu)=3(1.38×10−23kgm2s−2)(298.15)2(14.00674)×1.66×10−27kg=515 ms−1

Thus, the rms speed of nitrogen molecule is 515 ms−1.

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

Substitute 2(1.00794 u) for m, 1.38×10−23JK−1 for k and 298.15K for T in (II) to find vrms of hydrogen molecule.

vrms=3(1.38×10−23JK−1)(298.15K)2(1.00794 u)=3(1.38×10−23J)(298.15)2(1.00794 u)×(1.66×10−27kgu)=3(1.38×10−23kgm2s−2)(298.15)2(1.00794)×1.66×10−27kg=1920 ms−1

Thus, the rms speed of hydrogen molecule is 1920 ms−1.

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

Substitute 2(15.9994 u) for m, 1.38×10−23JK−1 for k and 298.15K for T in (II) to find vrms of oxygen molecule.

vrms=3(1.38×10−23JK−1)(298.15K)2(15.9994 u)=3(1.38×10−23J)(298.15)2(15.9994 u)×(1.66×10−27kgu)=3(1.38×10−23kgm2s−2)(298.15)2(15.9994)×1.66×10−27kg=482 ms−1

Thus, the rms speed of oxygen molecule is 482 ms−1.

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Answer 5

Chapter 13, Problem 67P

To determine

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

Expert Solution & Answer

Answer to Problem 67P

The rms speeds of helium atom, nitrogen, hydrogen and oxygen molecules are 1360 ms−1, 515 ms−1, 1920 ms−1 and 482 ms−1 respectively.

Explanation of Solution

The given temperature is 25° C.

Find the temperature in Kelvin scale

T(K)=273.15 K+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)=273.15 K+25 K=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 4.002602 u.

Substitute 4.002602 u for m, 1.38×10−23JK−1 for k and 298.15K for T in (II) to find vrms of helium atom.

vrms=3(1.38×10−23JK−1)(298.15K)4.002602 u=3(1.38×10−23J)(298.15)4.002602 u×(1.66×10−27kgu)=3(1.38×10−23kgm2s−2)(298.15)4.002602×1.66×10−27kg=1363 ms−1≈1360 ms−1

Thus, the rms speed of helium atom is 1360 ms−1.

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

Substitute 2(14.00674 u) for m, 1.38×10−23JK−1 for k and 298.15K for T in (II) to find vrms of nitrogen molecule.

vrms=3(1.38×10−23JK−1)(298.15K)2(14.00674 u)=3(1.38×10−23J)(298.15)2(14.00674 u)×(1.66×10−27kgu)=3(1.38×10−23kgm2s−2)(298.15)2(14.00674)×1.66×10−27kg=515 ms−1

Thus, the rms speed of nitrogen molecule is 515 ms−1.

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

Substitute 2(1.00794 u) for m, 1.38×10−23JK−1 for k and 298.15K for T in (II) to find vrms of hydrogen molecule.

vrms=3(1.38×10−23JK−1)(298.15K)2(1.00794 u)=3(1.38×10−23J)(298.15)2(1.00794 u)×(1.66×10−27kgu)=3(1.38×10−23kgm2s−2)(298.15)2(1.00794)×1.66×10−27kg=1920 ms−1

Thus, the rms speed of hydrogen molecule is 1920 ms−1.

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

Substitute 2(15.9994 u) for m, 1.38×10−23JK−1 for k and 298.15K for T in (II) to find vrms of oxygen molecule.

vrms=3(1.38×10−23JK−1)(298.15K)2(15.9994 u)=3(1.38×10−23J)(298.15)2(15.9994 u)×(1.66×10−27kgu)=3(1.38×10−23kgm2s−2)(298.15)2(15.9994)×1.66×10−27kg=482 ms−1

Thus, the rms speed of oxygen molecule is 482 ms−1.

Answer 6

Chapter 13, Problem 67P

To determine

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

Expert Solution & Answer

Answer to Problem 67P

The rms speeds of helium atom, nitrogen, hydrogen and oxygen molecules are 1360 ms−1, 515 ms−1, 1920 ms−1 and 482 ms−1 respectively.

Explanation of Solution

The given temperature is 25° C.

Find the temperature in Kelvin scale

T(K)=273.15 K+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)=273.15 K+25 K=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 4.002602 u.

Substitute 4.002602 u for m, 1.38×10−23JK−1 for k and 298.15K for T in (II) to find vrms of helium atom.

vrms=3(1.38×10−23JK−1)(298.15K)4.002602 u=3(1.38×10−23J)(298.15)4.002602 u×(1.66×10−27kgu)=3(1.38×10−23kgm2s−2)(298.15)4.002602×1.66×10−27kg=1363 ms−1≈1360 ms−1

Thus, the rms speed of helium atom is 1360 ms−1.

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

Substitute 2(14.00674 u) for m, 1.38×10−23JK−1 for k and 298.15K for T in (II) to find vrms of nitrogen molecule.

vrms=3(1.38×10−23JK−1)(298.15K)2(14.00674 u)=3(1.38×10−23J)(298.15)2(14.00674 u)×(1.66×10−27kgu)=3(1.38×10−23kgm2s−2)(298.15)2(14.00674)×1.66×10−27kg=515 ms−1

Thus, the rms speed of nitrogen molecule is 515 ms−1.

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

Substitute 2(1.00794 u) for m, 1.38×10−23JK−1 for k and 298.15K for T in (II) to find vrms of hydrogen molecule.

vrms=3(1.38×10−23JK−1)(298.15K)2(1.00794 u)=3(1.38×10−23J)(298.15)2(1.00794 u)×(1.66×10−27kgu)=3(1.38×10−23kgm2s−2)(298.15)2(1.00794)×1.66×10−27kg=1920 ms−1

Thus, the rms speed of hydrogen molecule is 1920 ms−1.

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

Substitute 2(15.9994 u) for m, 1.38×10−23JK−1 for k and 298.15K for T in (II) to find vrms of oxygen molecule.

vrms=3(1.38×10−23JK−1)(298.15K)2(15.9994 u)=3(1.38×10−23J)(298.15)2(15.9994 u)×(1.66×10−27kgu)=3(1.38×10−23kgm2s−2)(298.15)2(15.9994)×1.66×10−27kg=482 ms−1

Thus, the rms speed of oxygen molecule is 482 ms−1.

Answer 7

Chapter 13, Problem 67P

To determine

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

Answer 8

Expert Solution & Answer

Answer to Problem 67P

The rms speeds of helium atom, nitrogen, hydrogen and oxygen molecules are 1360 ms−1, 515 ms−1, 1920 ms−1 and 482 ms−1 respectively.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

The given temperature is 25° C.

Find the temperature in Kelvin scale

T(K)=273.15 K+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)=273.15 K+25 K=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 4.002602 u.

Substitute 4.002602 u for m, 1.38×10−23JK−1 for k and 298.15K for T in (II) to find vrms of helium atom.

vrms=3(1.38×10−23JK−1)(298.15K)4.002602 u=3(1.38×10−23J)(298.15)4.002602 u×(1.66×10−27kgu)=3(1.38×10−23kgm2s−2)(298.15)4.002602×1.66×10−27kg=1363 ms−1≈1360 ms−1

Thus, the rms speed of helium atom is 1360 ms−1.

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

Substitute 2(14.00674 u) for m, 1.38×10−23JK−1 for k and 298.15K for T in (II) to find vrms of nitrogen molecule.

vrms=3(1.38×10−23JK−1)(298.15K)2(14.00674 u)=3(1.38×10−23J)(298.15)2(14.00674 u)×(1.66×10−27kgu)=3(1.38×10−23kgm2s−2)(298.15)2(14.00674)×1.66×10−27kg=515 ms−1

Thus, the rms speed of nitrogen molecule is 515 ms−1.

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

Substitute 2(1.00794 u) for m, 1.38×10−23JK−1 for k and 298.15K for T in (II) to find vrms of hydrogen molecule.

vrms=3(1.38×10−23JK−1)(298.15K)2(1.00794 u)=3(1.38×10−23J)(298.15)2(1.00794 u)×(1.66×10−27kgu)=3(1.38×10−23kgm2s−2)(298.15)2(1.00794)×1.66×10−27kg=1920 ms−1

Thus, the rms speed of hydrogen molecule is 1920 ms−1.

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

Substitute 2(15.9994 u) for m, 1.38×10−23JK−1 for k and 298.15K for T in (II) to find vrms of oxygen molecule.

vrms=3(1.38×10−23JK−1)(298.15K)2(15.9994 u)=3(1.38×10−23J)(298.15)2(15.9994 u)×(1.66×10−27kgu)=3(1.38×10−23kgm2s−2)(298.15)2(15.9994)×1.66×10−27kg=482 ms−1