## Temperature and pressure

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.

### Definition:

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<v2>.  In SI units, the proportional constant is (3/2)kB, where kB = 1.381*10-23 J/K or 1.381*10-23 Pa m3/K is called the Boltzmann constant

½m<v2> = (3/2)kBT

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 (oC) scale, except it is shifted so that 0 degree Celsius equals 273.15 K.  We have

temperature in oC = temperature in K - 273.15.

To convert to temperature in Fahrenheit we can use

temperature in oF = (9/5) * temperature in oC + 32.

#### Problem:

Liquid nitrogen has a boiling point of -195.81 oC at atmospheric pressure.  Express this temperature in
(a) degrees Fahrenheit and
(b) Kelvin.

Solution:
(a) temperature in oF = (9/5) * temperature in oC + 32.
temperature in oF = [(9/5)(-195.81) + 32] oF = -320.5 oF.
(b) temperature in K = (-195.81+ 273.15) K = 77.34 K.

#### Problem:

One of the hottest temperatures ever recorded on the surface of Earth was 134 oF in Death Valley, CA.
(a)  What is this temperature in oC?
(b)  What is this temperature in Kelvin?

Solution:
(a)  (5/9)*(temperature in oF - 32)= temperature in oC
(5/9)*(134 - 32) oC = 56.67 oC
(b)  temperature in oC + 273.15 = temperature in K
(56.67 + 273.15) K = 329.82 K

#### Problem:

(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:
(a) temperature in oF = (9/5) * temperature in oC + 32
X = (9/5) * X + 32,  X - (9/5)X = 32,  -(4/5)X = 32, X = -5*32/4 = -40
-40 oF = -40 oC
(b)  temperature in oC = (5/9)*(temperature in oF - 32) = temperature in K - 273.15.
(5/9)*(temperature in oF - 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 oF = 574.59 K

### What is the relationship between temperature and pressur?

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 |vx| back and forth between two infinitely-massive walls, which are a distance L apart.  When it hits the right wall its momentum changes from p1 = +m|vx| to p2 = -m|vx|.  The change in the molecule's momentum is Δpmol = p2 - p1 = -2m|vx|.  The time interval between successive hits on the right wall is Δt = 2L/|vx|.  So the average force the wall exerts on this molecule is Fmol = Δpmol/Δt = -2m|vx|/(2L/|vx|) = -mvx2/L.  By Newton's third law, the average force that the molecule exerts on the wall is Fwall = mvx2/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 |vx|.  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<vx2>/L, where <vx2> is the average value of vx2.

The pressure is P = F/A  = Nm<vx2>/V, since L*A = V.  With ρparticle = N/V we have

P = F/A = ρparticlemvx2.

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.

<vx2> = <vy2> = <vz2>.

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

<v2> = <vx2> + <vy2> + <vz2>.

<vx2> = (1/3)<v2>.

We may therefore write

P = (1/3)ρparticlem<v2> = (2/3)ρparticle(m<v2>/2)

This equation relates the pressure to the kinetic energy of the atoms or molecules, since m<v2>/2 is the kinetic energy of the center-of-mass or translational motion of an atom or molecule.  Using ½m<v2> = (3/2)kBT and ρparticle = N/V from above we therefore find that

PV = (2/3)N(m<v2>/2)

PV = NkBT.

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 = NkBT 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 = nNAkBT = nRT.

Here n is the number of moles of the gaseous substance, NA is Avogadro's number, NA = 6.022*1023 molecules/mol,  and R = NAkB is a constant, called the universal gas constant,  R = 8.31 J/(mol K).

#### Problem:

The particle density of atmospheric air at 273.15 K at sea level is 2.687*1025/m3.  Calculate the pressure P.

Solution:
P = (2.687*1025/m3)(1.381*10-23 Pa m3/K)(273.15 K) = 1.01*105 Pa  = 101 kPa.

#### Problem:

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

Solution:
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.

#### Problem:

Average atomic and molecular speeds (vrms = <v2>1/2 = root mean square speed) are large, even at low temperatures.  What is vrms for helium atoms at 5.00 K, just one degree above helium's liquefaction temperature?

Solution:
½m<v2> = (3/2)kBT = (3/2)*1.381*10-23 J/K*(5 K) = 1.04*10-22 J
<v2> = (2*1.04*10-22 J)/(4*1.66*10-27 kg) = 3.13*104 m2/s2
vrms = 177 m/s
(The mass of the 4He atom is 4 atomic mass units = 4*1.66*10-27 kg.)

The root-mean-square speed of the atoms or molecule swith mass m is vrms = <v2>1/2 = (3kBT/m)1/2.

### How do we measure temperature?

Assume we have a gas in a container with a movable piston under atmospheric pressure.  As we increase the temperature of the gas it expands, its volume increases.  The piston moves out.  As we cool the gas the piston moves in.  The ideal gas law gives us the volume of the gas as a function of temperature.

PV = NkBT,    V = (NkB/P)T.

The volume is proportional to the absolute temperature of the gas.  Assume the temperature change from 0 oC to 10oC.  We have

(ΔV/V) = (ΔT/T) = 10/273 = 0.037 = 3.7%.

By monitoring the volume of the gas we can monitor its temperature.  We have a gas thermometer.

Solids and liquids also expand as the temperature increases.  For most solids we can define an average coefficient of linear expansion, α.  The change in length Δl of a solid is proportional to its length l and the change in temperature ΔT.  The proportional constant is α.

Δl = αlΔT.

The average volume expansion coefficient β is defined through ΔV = βVΔT.  We have β = 3α.

Linear expansion coefficients α: (per oC)

Aluminum 23 * 10-6 17 * 10-6 9 * 10-6 3.2 * 10-6 11 * 10-6 0.7 * 10-6

For some applications it is important to choose materials that have a small coefficient of linear expansion.  Temperature variations can destroy a precision alignment.  For other applications it is important that all materials have similar coefficients of linear expansion.  When different parts of a structure expand by different amounts, the structure buckles.

#### Problem:

How much taller does the Eiffel Tower become at the end of a day when the temperature has increased by 15 oC?  Its original height is 321 m and you can assume it is made of steel.

Solution:
Δl = αlΔT = (11*10-6/oC)*(321 m)*(15 oC) = 5.3*10-2 m = 5.3 cm.
The height of the tower increases by 5.3 cm.

#### Problem:

How large an expansion gap should be left between steel railroad rails if they may reach a maximum temperature 35.0 oC greater than when they were laid?  Their original length is 10.0 m.

Solution:
Δl = αlΔT = (11*10-6/oC)*(10 m)*(35 oC) = 3.85*10-3 m ~ 4 mm.

If we want to build a thermometer, we want to use a material with a large coefficient of volume expansion.  For most liquids, the coefficient of volume expansion is on the order of 10-4.  For each degree Celsius temperature change, the volume changes by (ΔV/V) = β = 10-4 = 0.0001 = 0.01%.  If we tried to observe this change by monitoring the height of water in a glass, we would probably be unsuccessful.

Assume you have 1 liter = 1000 cm3 of liquid with β = 10-4 in a container with bottom area A = 100 cm2
The height of the liquid is 1 cm.  (Volume = area * height.)
You increase the temperature of the liquid by 20 oC.
ΔV = βVΔT = 10-4 * 1000 cm3 * 20 = 2 cm3
The change in height is Δh = ΔV/A = 2 cm3/(100 cm2) = 0.02 cm = 0.2 mm.

If, however, you put a tight lid on the container at a height of 10 cm with a hole connected to a fine capillary, then the liquid will rise in the capillary to a much greater height.  Assume you connected a tube with a cross sectional area of 1 cm2.  Then Δh = ΔV/A = 2 cm3/(1 cm2) = 2 cm.  You have constructed a usable thermometer.

Thermometers can also be constructed from bimetallic strips.  Two strips of metal, one of steel and one of aluminum, that have the same length at 0 oC, will have different length a 200 oC.  Assume the strips are 10 cm long at 0 oC.  The change in length of the steel strip at 200 oC is Δl = αlΔT = 11*10-6*10 cm*200 = 0.022 cm = 0.22 mm and the change in length of the aluminum strip at 200 oC is Δl = αlΔT = 23*10-6*10 cm*200 = 0.046 cm = 0.46 mm.  If the two strips are bonded together, the bimetallic strip will buckle or curl.  Long bimetallic coils will uncoil slightly as the temperature rises if the inner strip has the larger α.  If one end is fixed and a pointer is attached to the other end the pointer will move.  The position of the pointer indicates the temperature.