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F i = Δ p i Δ t = 2 m v i x Δ t .

(In this equation alone, p represents momentum, not pressure.) There is no force between the wall and the molecule except while the molecule is touching the wall. During the short time of the collision, the force between the molecule and wall is relatively large, but that is not the force we are looking for. We are looking for the average force, so we take Δ t to be the average time between collisions of the given molecule with this wall, which is the time in which we expect to find one collision. Let l represent the length of the box in the x -direction. Then Δ t is the time the molecule would take to go across the box and back, a distance 2 l , at a speed of v x . Thus Δ t = 2 l / v x , and the expression for the force becomes

F i = 2 m v i x 2 l / v i x = m v i x 2 l .

This force is due to one molecule. To find the total force on the wall, F , we need to add the contributions of all N molecules:

F = i = 1 N F i = i = 1 N m v i x 2 l = m l i = 1 N v i x 2 .

We now use the definition of the average, which we denote with a bar, to find the force:

F = N m l ( 1 N i = 1 N v i x 2 ) = N m v x 2 l .

We want the force in terms of the speed v , rather than the x -component of the velocity. Note that the total velocity squared is the sum of the squares of its components, so that

v 2 = v x 2 + v y 2 + v z 2 .

With the assumption of isotropy, the three averages on the right side are equal, so

v 2 = 3 v i x 2 .

Substituting this into the expression for F gives

F = N m v 2 3 l .

The pressure is F / A , so we obtain

p = F A = N m v 2 3 A l = N m v 2 3 V ,

where we used V = A l for the volume. This gives the important result

p V = 1 3 N m v 2 .

Combining this equation with p V = N k B T gives

1 3 N m v 2 = N k B T .

We can get the average kinetic energy of a molecule, 1 2 m v 2 , from the left-hand side of the equation by dividing out N and multiplying by 3/2.

Average kinetic energy per molecule

The average kinetic energy of a molecule is directly proportional to its absolute temperature:

K = 1 2 m v 2 = 3 2 k B T .

The equation K = 3 2 k B T is the average kinetic energy per molecule. Note in particular that nothing in this equation depends on the molecular mass (or any other property) of the gas, the pressure, or anything but the temperature. If samples of helium and xenon gas, with very different molecular masses, are at the same temperature, the molecules have the same average kinetic energy.

The internal energy    of a thermodynamic system is the sum of the mechanical energies of all of the molecules in it. We can now give an equation for the internal energy of a monatomic ideal gas. In such a gas, the molecules’ only energy is their translational kinetic energy. Therefore, denoting the internal energy by E int , we simply have E int = N K , or

E int = 3 2 N k B T .

Often we would like to use this equation in terms of moles:

E int = 3 2 n R T .

We can solve K = 1 2 m v 2 = 3 2 k B T for a typical speed of a molecule in an ideal gas in terms of temperature to determine what is known as the root-mean-square ( rms ) speed of a molecule.

Rms speed of a molecule

The root-mean-square (rms) speed    of a molecule, or the square root of the average of the square of the speed v 2 , is

v rms = v 2 = 3 k B T m .

The rms speed is not the average or the most likely speed of molecules, as we will see in Distribution of Molecular Speeds , but it provides an easily calculated estimate of the molecules’ speed that is related to their kinetic energy. Again we can write this equation in terms of the gas constant R and the molar mass M in kg/mol:

Questions & Answers

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Practice Key Terms 8

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Source:  OpenStax, University physics volume 2. OpenStax CNX. Oct 06, 2016 Download for free at http://cnx.org/content/col12074/1.3
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