The particles that make up an object can have ordered energy and disordered
energy. The kinetic energy of an object as a whole due to its motion with velocity **
v** with respect to an observer is an example of ordered energy. The
kinetic energy of individual atoms, when they are randomly vibrating about their
equilibrium position, is an example of disordered energy. **Thermal energy**** **is disordered energy. The
**temperature** is a measure of this internal,
disordered energy.

The **absolute temperature** of any substance is
proportional to the average kinetic energy associated with the
random motion of the atoms or molecules that make
up the substance.

In a gas, the individual atoms and molecules are moving randomly. The
**absolute temperature** T of the gas is proportional to the average
translational kinetic energy of a gas atom or molecule, ½m<v^{2}>.
In SI units, the proportional constant is (3/2)k_{B}, where k_{B
}= 1.381*10^{-23 }J/K or 1.381*10^{-23}
Pa m^{3}/K is called the **Boltzmann constant**.

½m<v^{2}> = (3/2)k_{B}T

In a solid, the atoms can move randomly about their equilibrium positions.
In addition, the solid as a whole can move with a given velocity and have
ordered kinetic energy. Only the kinetic energy associated with the random
motion of the atoms is proportional to the absolute temperature of the solid.

In ideal gases the disordered energy is all kinetic energy, in molecular
gases and solids it is a combination of kinetic and potential energy. If
we model the atoms in a solid as being held together by tiny springs, then the
random internal energy of each atom constantly switches between kinetic energy
and elastic potential energy.

In classical physics, zero absolute temperature means zero kinetic energy
associated with random motion. The atoms in a substance do not move with
respect to each other. (The uncertainty principle in quantum mechanics
requires that there is some zero-point energy.) Room temperature is **
not** close to absolute zero temperature. At room temperature the atoms and
molecules of all substances have random motion.

In SI units the scale of absolute temperature is
**Kelvin** (K).
The Kelvin scale is identical to the **Celsius** (^{o}C) scale, except it is shifted so that 0 degree
Celsius equals 273.15 K. We have

temperature in ^{o}C = temperature in K - 273.15.

To convert to temperature in **Fahrenheit** we can use

temperature in ^{o}F = (9/5) * temperature in ^{o}C + 32.

Average atomic and molecular speeds (v_{rms} = <v^{2}>^{½}
= root mean square speed) are large, even at low temperatures. What is v_{rms}
for helium atoms at 5.00 K, just one
degree above helium's liquefaction temperature?

Solution:

- Reasoning:

The absolute temperature T of a gas is proportional to the average translational kinetic energy of the gas atoms or molecules.

½m<v^{2}> = (3/2)k_{B}T. - Details of the calculation:

½m<v^{2}> = (3/2)k_{B}T = (3/2)*1.381*10^{-23 }J/K*(5 K) = 1.04*10^{-22}J

<v^{2}> = (2*1.04*10^{-22}J)/(4*1.66*10^{-27}kg) = 3.13*10^{4}m^{2}/s^{2}v_{rms}= 177 m/s

(The mass of the^{4}He atom is 4 atomic mass units = 4*1.66*10^{-27}kg.)

The root-mean-square speed of the atoms or molecule with mass m is v_{rms} = <v^{2}>^{½}
= (3k_{B}T/m)^{½}.

Liquid nitrogen has a boiling point of -195.81 ^{o}C at atmospheric
pressure. Express this temperature in

(a) degrees Fahrenheit and

(b) Kelvin.

Solution:

- Reasoning:

Unit conversion - Details of the calculation:

(a) temperature in^{o}F = (9/5) * temperature in^{o}C + 32.

temperature in^{o}F = [(9/5)(-195.81) + 32]^{o}F = -320.5^{o}F.

(b) temperature in K = (-195.81+ 273.15) K = 77.34 K.

One of the hottest temperatures ever recorded on the surface of Earth was
134 ^{o}F in Death Valley, CA.

(a) What is this temperature in ^{o}C?

(b) What is this temperature in Kelvin?

Solution:

- Reasoning:

Unit conversion - Details of the calculation:

(a) (5/9)*(temperature in^{o}F - 32)= temperature in^{o}C.

(5/9)*(134 - 32)^{o}C = 56.67^{o}C.

(b) temperature in^{o}C + 273.15 = temperature in K.

(56.67 + 273.15) K = 329.82 K.

(a) At what temperature do the Fahrenheit and Celsius scales have the same
numerical value?

(b) At what temperature do the Fahrenheit and Kelvin scales have the same
numerical value?

Solution:

- Reasoning:

Unit conversion - Details of the calculation:

(a) temperature in^{o}F = (9/5) * temperature in^{o}C + 32.

X = (9/5) * X + 32,

X - (9/5)X = 32, -(4/5)X = 32, X = -5*32/4 = -40.

-40^{o}F = -40^{ o}C.

(b) temperature in^{o}C = (5/9)*(temperature in^{o}F - 32) = temperature in K - 273.15.

(5/9)*(temperature in^{o}F - 32) + 273.15 = temperature in K.

(5/9)*(X - 32) + 273.15 = X,

(X - 32) + 491.67 = (9/5)X, 459.67 = (4/5)X, X = 574.59.

574.59^{o}F = 574.59 K.

Assume we have a collection of gas molecules in gravity-free space in a container with volume V at absolute temperature T.

Then each molecule moves along with constant velocity in a straight line, until it hits another molecule, or a container wall. A collision between two molecules is similar to a collision between two balls. The molecules exchange momentum, but the total momentum of the two molecules is conserved. When a molecule hits a wall, it bounces back. Its momentum changes. To change the molecule's momentum, the wall must exert a force on the molecule. Newton's third law tells us that the molecule exerts a force on the wall. The greater the number of molecules hitting a wall, the greater is the force on the wall. In a container with different size walls, the bigger walls will receive more hits than the smaller walls and therefore experience a greater force. The pressure in the container is the magnitude of the normal force F on a wall divided by the surface area A of the wall.

P = F/A

The faster the molecules move in the container, the greater is the change in
momentum when they bounce off a wall, and the more often do they hit the walls.
Assume a molecule moves horizontally with speed |v_{x}| back and forth
between two infinitely-massive walls, which are a distance L apart. When
it hits the right wall its momentum changes from p_{1 }= +m|v_{x}|
to p_{2 }= -m|v_{x}|. The change in the molecule's
momentum is Δp_{mol }= p_{2 }
- p_{1 }= -2m|v_{x}|. The time interval between successive
hits on the right wall is Δt = 2L/|v_{x}|. So the average force
the wall exerts on this molecule is F_{mol
}= Δp_{mol}/Δt = -2m|v_{x}|/(2L/|v_{x}|) = -mv_{x}^{2}/L.
By Newton's third law, the average force that the molecule exerts on the wall is
F_{wall
}= mv_{x}^{2}/L, it is proportional to the square of the
speed of the molecule or its kinetic energy.

Assume that there are N molecules in the volume V, moving horizontally with
speed |v_{x}|. Not all the molecules have the same
kinetic energy. The force
exerted by the molecules on the walls of a container is therefore F = Nm<v_{x}^{2}>/L,
where <v_{x}^{2}> is the average value of v_{x}^{2}.

The pressure is P = F/A = Nm<v_{x}^{2}>/V,
since L*A = V. With ρ_{particle} = N/V we have

P = F/A = ρ_{particle}mv_{x}^{2}.

There is nothing special about the x-direction. The atoms can move up and down, back and forth, in and out. The average velocity components in all directions are all going to be equal to each other.

<v_{x}^{2}> = <v_{y}^{2}> = <v_{z}^{2}>.

They are each equal to one-third of their sum, which is the square of the magnitude of the average velocity.

<v^{2}> = <v_{x}^{2}> + <v_{y}^{2}>
+ <v_{z}^{2}>.

<v_{x}^{2}> = (1/3)<v^{2}>.

We may therefore write

P = (1/3)ρ_{particle}m<v^{2}> = (2/3)ρ_{particle}(m<v^{2}>/2)

This equation relates the pressure to the kinetic energy of the
atoms or molecules, since m<v^{2}>/2 is the kinetic energy of the center-of-mass or
translational motion of an atom or molecule. Using ½m<v^{2}> =
(3/2)k_{B}T and ρ_{particle} = N/V from above we therefore find that

PV = (2/3)N(m<v^{2}>/2)

PV = Nk_{B}T.

The pressure in a container is proportional to the average kinetic energy of the molecules and therefore to the absolute temperature T of the gas.

If all the molecules in a container would be at rest, their kinetic energy would be zero and the pressure would be zero.

PV = Nk_{B}T is called the **ideal gas law**. Most real gases at ordinary temperatures and
pressures obey the ideal gas law. The ideal gas law can be rewritten as

PV = nN_{A}k_{B}T = nRT.

Here n is the number of moles of the gaseous substance, N_{A}
is Avogadro's number, N_{A }= 6.022*10^{23}
molecules/mol, and R = N_{A}k_{B} is a constant, called
the **universal gas constant**, R = 8.31 J/(mol K).

The number of moles n is given by n = m/M where m is the average mass of the gas particles in the volume and M is the molar mass of the gas.

In the 17^{th} and 18^{th} century experiment with gases at
very low temperature and pressures revealed three relations that are generalized
by the ideal gas law.

- For a low-density gas at constant temperature, P is inversely proportional
to V.
**Boyle's law:**PV = constant (at constant T). - For a low-density gas at constant volume the pressure is proportional to
the temperature.
**Law of Gay-Lussac:**P = constant * T (at constant V). - For a low-density gas at constant pressure the volume is proportional to
the temperature.
**Charles's Law:**V = constant * T (at constant P).

The ideal gas law holds well for real gases at low
densities and pressures, such as atmospheric density and pressure. If we use T
= 0 ^{o}C = 273 K and P = 1 atm, then we find that one mole of gas
occupies a volume of 22.4 liters. One mole of gas contains N_{A} gas
particles. For all low density gases, N_{A} gas particles occupy
22.4 liters at T = 273K and P = 1 atm.

Nothing in the ideal gas law depends on the nature of the gas particles. The value of PV/T depends only on the number of gas particles and a universal constant.

Note: ** **In all gas laws T denote the absolute temperature, measured in Kelvin
in SI units.

The particle density of atmospheric air at 273.15 K at sea level is 2.687*10^{25}/m^{3}.
Calculate the pressure P.

Solution:

- Reasoning:

The ideal gas law, PV = Nk_{B}T, P = (N/V)k_{B}T = ρ_{particle}k_{B}T. - Details of the calculation:

P = (2.687*10^{25}/m^{3})(1.381*10^{-23}Pa m^{3}/K)(273.15 K) = 1.01*10^{5}Pa = 101 kPa.

If a helium-filled balloon initially at room temperature is placed in a freezer, will its volume increase, decrease, or remain the same?

Solution:

- Reasoning:

The ideal gas law states that PV/T is constant. The pressure in the freezer is atmospheric pressure, the temperature in the freezer is lower that the outside temperature, so the volume of the balloon decreases when it is placed into the freezer.

External link: Balloons in Liquid Nitrogen (Youtube)