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  • Express the ideal gas law in terms of molecular mass and velocity.
  • Define thermal energy.
  • Calculate the kinetic energy of a gas molecule, given its temperature.
  • Describe the relationship between the temperature of a gas and the kinetic energy of atoms and molecules.
  • Describe the distribution of speeds of molecules in a gas.

We have developed macroscopic definitions of pressure and temperature. Pressure is the force divided by the area on which the force is exerted, and temperature is measured with a thermometer. We gain a better understanding of pressure and temperature from the kinetic theory of gases, which assumes that atoms and molecules are in continuous random motion.

A green vector v, representing a molecule colliding with a wall, is pointing at the surface of a wall at an angle. A second vector v primed starts at the point of impact and travels away from the wall at an angle. A dotted line perpendicular to the wall through the point of impact represents the component of the molecule’s momentum that is perpendicular to the wall. A red vector F is pointing into the wall from the point of impact, representing the force of the molecule hitting the wall.
When a molecule collides with a rigid wall, the component of its momentum perpendicular to the wall is reversed. A force is thus exerted on the wall, creating pressure.

[link] shows an elastic collision of a gas molecule with the wall of a container, so that it exerts a force on the wall (by Newton’s third law). Because a huge number of molecules will collide with the wall in a short time, we observe an average force per unit area. These collisions are the source of pressure in a gas. As the number of molecules increases, the number of collisions and thus the pressure increase. Similarly, the gas pressure is higher if the average velocity of molecules is higher. The actual relationship is derived in the Things Great and Small feature below. The following relationship is found:

PV = 1 3 Nm v 2 ¯ , size 12{ ital "PV"= { {1} over {3} } ital "Nm" {overline {v rSup { size 8{2} } }} ,} {}

where P size 12{P} {} is the pressure (average force per unit area), V size 12{V} {} is the volume of gas in the container, N size 12{N} {} is the number of molecules in the container, m size 12{m} {} is the mass of a molecule, and v 2 ¯ size 12{ {overline {v rSup { size 8{2} } }} } {} is the average of the molecular speed squared.

What can we learn from this atomic and molecular version of the ideal gas law? We can derive a relationship between temperature and the average translational kinetic energy of molecules in a gas. Recall the previous expression of the ideal gas law:

PV = NkT . size 12{ ital "PV"= ital "NkT"} {}

Equating the right-hand side of this equation with the right-hand side of PV = 1 3 Nm v 2 ¯ size 12{ ital "PV"= { {1} over {3} } ital "Nm" {overline {v rSup { size 8{2} } }} } {} gives

1 3 Nm v 2 ¯ = NkT . size 12{ { {1} over {3} } ital "Nm" {overline {v rSup { size 8{2} } }} = ital "NkT"} {}

Making connections: things great and small—atomic and molecular origin of pressure in a gas

[link] shows a box filled with a gas. We know from our previous discussions that putting more gas into the box produces greater pressure, and that increasing the temperature of the gas also produces a greater pressure. But why should increasing the temperature of the gas increase the pressure in the box? A look at the atomic and molecular scale gives us some answers, and an alternative expression for the ideal gas law.

The figure shows an expanded view of an elastic collision of a gas molecule with the wall of a container. Calculating the average force exerted by such molecules will lead us to the ideal gas law, and to the connection between temperature and molecular kinetic energy. We assume that a molecule is small compared with the separation of molecules in the gas, and that its interaction with other molecules can be ignored. We also assume the wall is rigid and that the molecule’s direction changes, but that its speed remains constant (and hence its kinetic energy and the magnitude of its momentum remain constant as well). This assumption is not always valid, but the same result is obtained with a more detailed description of the molecule’s exchange of energy and momentum with the wall.

Diagram representing the pressures that a gas exerts on the walls of a box in a three-dimensional coordinate system with x, y, and z components.
Gas in a box exerts an outward pressure on its walls. A molecule colliding with a rigid wall has the direction of its velocity and momentum in the x size 12{x} {} -direction reversed. This direction is perpendicular to the wall. The components of its velocity momentum in the y size 12{y} {} - and z size 12{z} {} -directions are not changed, which means there is no force parallel to the wall.

If the molecule’s velocity changes in the x size 12{x} {} -direction, its momentum changes from mv x size 12{– ital "mv" rSub { size 8{x} } } {} to + mv x size 12{+ ital "mv" rSub { size 8{x} } } {} . Thus, its change in momentum is Δ mv = + mv x mv x = 2 mv x size 12{Δ ital "mv""=+" ital "mv" rSub { size 8{x} } – left (– ital "mv" rSub { size 8{x} } right )=2 ital "mv" rSub { size 8{x} } } {} . The force exerted on the molecule is given by

F = Δ p Δ t = 2 mv x Δ t . size 12{F= { {Δp} over {Δt} } = { {2 ital "mv" rSub { size 8{x} } } over {Δt} } "." } {}

There is no force between the wall and the molecule until the molecule hits the wall. During the short time of the collision, the force between the molecule and wall is relatively large. We are looking for an average force; we take Δ t size 12{Dt} {} to be the average time between collisions of the molecule with this wall. It is the time it would take the molecule to go across the box and back (a distance 2 l ) size 12{2l \) } {} at a speed of v x size 12{v rSub { size 8{x} } } {} . Thus Δ t = 2 l / v x size 12{Δt=2l/v rSub { size 8{x} } } {} , and the expression for the force becomes

F = 2 mv x 2 l / v x = mv x 2 l . size 12{F= { {2 ital "mv" rSub { size 8{x} } } over { {2l} slash {v rSub { size 8{x} } } } } = { { ital "mv" rSub { size 8{x} } rSup { size 8{2} } } over {l} } "." } {}

This force is due to one molecule. We multiply by the number of molecules N size 12{N} {} and use their average squared velocity to find the force

F = N m v x 2 ¯ l , size 12{F=N { {m {overline {v rSub { size 8{x} } rSup { size 8{2} } }} } over {l} } ,} {}

where the bar over a quantity means its average value. We would like to have the force in terms of the speed v size 12{v} {} , rather than the x size 12{x} {} -component of the velocity. We 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 ¯ . size 12{ {overline {v rSup { size 8{2} } }} = {overline {v rSub { size 8{x} } rSup { size 8{2} } }} + {overline {v rSub { size 8{y} } rSup { size 8{2} } }} + {overline {v rSub { size 8{z} } rSup { size 8{2} } }} "." } {}

Because the velocities are random, their average components in all directions are the same:

v x 2 ¯ = v y 2 ¯ = v z 2 ¯ . size 12{ {overline {v rSub { size 8{x} } rSup { size 8{2} } }} = {overline {v rSub { size 8{y} } rSup { size 8{2} } }} = {overline {v rSub { size 8{z} } rSup { size 8{2} } }} "." } {}

Thus,

v 2 ¯ = 3 v x 2 ¯ , size 12{ {overline {v rSup { size 8{2} } }} =3 {overline {v rSub { size 8{x} } rSup { size 8{2} } }} ,} {}

or

v x 2 ¯ = 1 3 v 2 ¯ . size 12{ {overline {v rSub { size 8{x} } rSup { size 8{2} } }} = { {1} over {3} } {overline {v rSup { size 8{2} } }} } {}

Substituting 1 3 v 2 ¯ size 12{ { {1} over {3} } {overline {v rSup { size 8{2} } }} } {} into the expression for F size 12{F} {} gives

F = N m v 2 ¯ 3 l . size 12{F=N { {m {overline {v rSup { size 8{2} } }} } over {3l} } "." } {}

The pressure is F / A , size 12{F/A,} {} so that we obtain

P = F A = N m v 2 ¯ 3 Al = 1 3 Nm v 2 ¯ V , size 12{P= { {F} over {A} } =N { {m {overline {v rSup { size 8{2} } }} } over {3 ital "Al"} } = { {1} over {3} } { { ital "Nm" {overline {v rSup { size 8{2} } }} } over {V} } ,} {}

where we used V = Al size 12{V= ital "Al"} {} for the volume. This gives the important result.

PV = 1 3 Nm v 2 ¯ size 12{ ital "PV"= { {1} over {3} } ital "Nm" {overline {v rSup { size 8{2} } }} } {}

This equation is another expression of the ideal gas law.

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Source:  OpenStax, College physics. OpenStax CNX. Jul 27, 2015 Download for free at http://legacy.cnx.org/content/col11406/1.9
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