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

Question

Want to see more full solutions like this?

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.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Students have asked these similar questions

When current is flowing through the coil, the direction of the torque can be thought of in two ways. Either as the result of the forces on current carrying wires, or as a magnetic dipole moment trying to line up with an external field (e.g. like a compass). Note: the magnetic moment of a coil points in the direction of the coil's magnetic field at the center of the coil. d) Forces: We can consider the left-most piece of the loop (labeled ○) as a short segment of straight wire carrying current directly out of the page at us. Similarly, we can consider the right-most piece of the loop (labeled ) as a short segment straight wire carrying current directly into the page, away from us. Add to the picture below the two forces due to the external magnetic field acting on these two segments. Then describe how these two forces give a torque and determine if the torque acts to rotate the loop clockwise or counterclockwise according to this picture? B

In each of the following, solve the problem stated. Express your answers in three significant figures. No unit is considered incorrect. 1. For the circuit shown, determine all the currents in each branch using Kirchhoff's Laws. (3 points) 6 5V 2 B C 4 A www 6 VT ww T10 V F E 2. Compute for the total power dissipation of the circuit in previous item. (1 point) 3. Use Maxwell's Mesh to find Ix and VAB for the circuit shown. (3 points) Ix 50 V 20 ww 21x B 4. Calculate all the currents in each branch using Maxwell's Mesh for the circuit shown. (3 points) www 5ი 10 24V 2A 2002 36V

If the mass of substance (1 kg), initial temperature (125˚C), the final temperature (175˚C) and the total volume of a closed container (1 m3) remains constant in two experiments, but one experiment is done with water ( ) and the other is done with nitrogen ( ). What is the difference in the change in pressure between water and nitrogen?

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.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

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.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

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.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

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