Temperature and Thermal Expansion

The particles that make up an object can have ordered energy and disordered energy.  The kinetic energy of an object 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 substance.

In a gas the individual atoms and molecules are moving randomly.  The average kinetic energy of the molecules in a gas is proportional to the absolute temperature T of the gas.  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 K = temperature in oC + 273.15.

To convert to temperature in Fahrenheit we can use

temperature in oC = (temperature in oF - 32) ´ 5/9.

Two objects with different temperatures can exchange energy, if they are in thermal contact.  The energy exchanged between object because they are in thermal contact is called heat.  If two objects are in thermal contact and do not exchange heat, then they are in thermal equilibrium.

The zeroth law of thermodynamics states that two object, which are separately in thermal equilibrium with a third object, are in thermal equilibrium with each other.  Two objects in thermal equilibrium with each other are at the same temperature.

How can we heat things up?

You can add thermal energy to an object by doing work on the object.  If you rub an object, the force of sliding friction does work, and changes ordered kinetic energy into thermal energy.  If fuel burns, chemical energy is converted into thermal energy.

Atoms in molecules and solids are held together by chemical bonds.  Chemical bonds are electromagnetic in origin, but can be modeled well by tiny springs.  Two atoms held together by a spring have an equilibrium position.  If they are pushed closer together, they repel each other.  If they are pulled farther apart, they attract each other.  If they are displaced in any way from their equilibrium position and then released, they start vibrating about their equilibrium position.  An atom can form different chemical bonds with a variety of other atoms.  Different bonds are represented by springs with different spring constants.  The stiffer the spring, the more work it takes to pull the atoms apart.  If enough work is done, then the spring is stretched too much and it breaks, i.e. the chemical bond breaks.  

At room temperature, gas molecules have random translational kinetic energy associated with the motion of their center of mass and random vibrational energy and rotational kinetic energy associated with the motion about their center of mass.  Collisions continuously transfer energy between the different degrees of freedom and the average energy in each degree of freedom is the same.  If work is done on the molecules which increases their vibrational energy, the amplitude of the vibrations increases, and eventually the chemical bonds break.  Most free atoms quickly form new bonds.  If the new bonds are stronger, i.e. if the new springs are stiffer, then they do more work pulling the atom towards their new equilibrium positions than was needed to break the old bonds and the atoms will have more kinetic energy as they pass through the equilibrium positions.  This kinetic energy is quickly shared with the other degrees of freedom, the energy of all degrees of freedom increases, i.e. the thermal energy increases.  Thermal energy is released by a chemical reaction.  The temperature increases.

To burn fuel, work must first be done to break the chemical bonds in the fuel.  This work provides the activation energy, the energy needed to start the chemical reaction.  The free atoms and molecules then bond with oxygen. The new bonds with the oxygen atoms are much stronger than the broken bonds.  As the atoms form new bonds, they gain thermal energy.  When you strike a match, you first do work against friction to break the chemical bonds in some of the fuel on the head.  The free atoms and molecules now combine with oxygen from the air, forming stronger bonds and thus releasing thermal energy.  The random kinetic energy of these fast molecules is transferred in collisions to neighboring atoms and molecules, breaking their bonds, etc.

Ideal Gases

If a gas of mass m is confined to a volume V with pressure P and absolute temperature T, the equation of state relates the quantities V, P, and T.   For gases at very low densities and pressures the equation of state is found experimentally to be quite simple.  

Experiments revealed that 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)
These observation are summarized by the following equation of state:
Ideal Gas Law: PV = nRT.

Here n is the number of moles of the gaseous substance, and R is a constant, called the universal gas constant.  R = 8.31 J/(mol K).  (One mole M of a substance is the mass of the substance that contains NA = 6.022´1023 molecules.  NA is Avogadro's number.  The number of moles n is given by n = m/M where m is the mass of the gas in the volume and M is its molar mass.)

PV = nRT is called the ideal gas law.  It holds well for real gases at low densities and pressures, such as atmospheric density and pressure.  If we use T = 0 oC = 273 K and P = 1atm, then we find that one mole of gas occupies a volume of 22.4 liters.  One mole of gas contains NA gas particles.  For all low density gases, NA gas particles occupy 22.4 liters at T = 273K and P = 1atm.

We may write n = N/NA, where N is the total number of molecules in the volume.  We can then rewrite the ideal gas law as

PV = nRT = (N/NA)RT = NkBT

where kB = R/NA = 1.38´10-23 J/K is called the Boltzmann constant.

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.

Problems:

A tank having a volume of 0.1 m3 contains helium gas at 150 atm.  How many balloons can the tank blow up, if each filled balloon is a sphere 0.3 m in diameter at an absolute pressure of 1.2 atm?
Solution:
At constant temperature P1V1 = P2V2. (Boyle's law)
P1 = 150 atm = 1.515´107 Pa. V1 = 0.1m3.
P2 = 1.2 atm = 1.212´105 Pa.
V2 = P1V1/P2 = 12.5 m3.
Let n = number of balloons and Vb = the volume of each blown-up balloon.
Vb = (4p/3)r3 = 1.414´10-2 m3.
n = V2/Vb = 884.  The tank can blow up 884 balloons.
The mass of a hot air balloon and its cargo (not including the air inside) is 200 kg.  The air outside is at 10 oC and 101 kPa.  The volume of the balloon is 400 m3.  To what temperature must the air in the balloon be heated before the balloon will lift off.  (Air density at 10 oC is 1.25 kg/m3.)
Solution
For the balloon to lift off, the buoyant force B must be greater than its weight.  The buoyant force is equal to the weight of the displaced air at 10 oC = 283 K.  B = (1.25 kg/m3)(400 m3)(9.8 m/s2) = 4900 N.  The weight of the balloon is 200 kg(9.8 m/s2) + weight of hot air.  The hot air therefore must weigh less than 4900 N - 1960 N = 2940 N.  Its mass must be less than 2940 N/(9.8 m/s2) = 300 kg.  Its density must be less than r = 300 kg/(400 m3) = 0.75 kg/m3.
At constant pressure, the volume of a gas is proportional to the absolute temperature.  (Law of Gay-Lussac)  The pressures on the inside and outside of the inflated balloon are nearly equal.  The pressure on the outside is the constant atmospheric pressure.  The Law of Gay-Lussac therefore applies.
Since the volume of a gas at constant pressure is proportional to its temperature, its density r = m/V is proportional to 1/T.  We have r1/r2 = T2/T1r1T1/r2 = T2.   (1.25 kg/m3)(283 K)/(0.75 kg/m3) = 472 K = T2.  The air in the balloon must be heated to more than 472 K = 199 oC.
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.

Balloons

The earth atmosphere is a gas in the presence of gravity.   Gravity prevents the atmosphere from escaping.  Near the Earth's surface the acceleration due to gravity is approximately constant (g = 9.8 m/s2).  The atmosphere is a fluid, and the pressure in this fluid is a function of the altitude.  For a liquid, i.e. an incompressible fluid, we found that P = P0 + rgy, where y is the distance below the reference point.  But the atmosphere is a compressible gas.  Let us consider a column of gas containing rparticle particles of mass m per unit volume near the Earth's surface.  

The pressure times the area (i.e. the force) at height y must exceed that at height y + dy by the weight of the intervening gas.  We need 

Py+dyA - PyA = -mrparticleVg = -mrparticlegAdy,  or 
dP = -mrparticlegdy.   

But from the ideal gas law we know that P = rparticlekT, with k being the Boltzmann constant.  We may therefore write

 dP = -P(mg/kT)dy, or 
dP/P = -(mg/kT)dy.  

If the temperature T is constant, then this integrates to 

P = P0e-mgy/(kT).  

The pressure decreases exponentially with altitude.

Near Earth's surface the temperature T is not constant, it generally decreases with altitude y.  In the standard atmosphere T = T0 - ay, with a = 0.0065 K/m.  We then have 

dP/P = -mgdy/(k(T0 - ay)).  

This integrates to 

ln(P/P0) =( mg/ka)ln(1 - ay/T0), or 
(P/P0)(ka/mg) = (1 - ay/T0).  

Since the Earth's atmosphere consist mostly of nitrogen molecules, we have 

(P/P0)0.19 = (1 - (0.0065 K/m)y/T0).

T0 and P0 are the temperature (in Kelvin) and the pressure at sea level.  The pressure, density, and temperature of the atmosphere decrease as the altitude increases. 

Links:

Standard Atmosphere Calculator
The Ideal Atmosphere

How can we make a balloon rise?

We have to make its weight smaller than the weight of an equal shaped volume of surrounding air.  The skin of the balloon and the basket certainly are heavier than an equal volume of the surrounding air.  The inside of the skin therefore has to be filled with something appreciably lighter than the surrounding air.  Ideally we would empty it out completely.  But then the pressure inside would be zero.  The air pressure from the outside would collapse the skin. To keep the skin stretched, the inside and outside pressures on each section of skin must be equal, but the weight of the inside material must be less than that of an equal volume of surrounding air.  The pressure is proportional to the particle density and the temperature, while the weight is proportional to the particle density and the weight of each particle.

P = krparticleT,

weight = rparticle ´ volume ´ weight per particle.

To get the weight down and keep the pressure the same we have two options.

Option 1:

We can decrease the weight per particle.  Air is composed of 77% nitrogen, 21% oxygen and 2% other gases by weight.  The density of air at room temperature and atmospheric pressure is approximately 1.3 kg/m3.  If we replace the air with Helium gas at the same temperature, the same particle density will yield the same pressure.  But the mass density of Helium gas at room temperature and atmospheric pressure is only approximately 0.18 kg/m3.  If we make the balloon big enough, the total volume of air displaced by the balloon will weigh more than the balloon itself, and the net force on the balloon will be upward.  Since the density of air decreases with altitude, the weight of the volume of air displaced by the balloon will decrease as the Helium balloon rises.  Eventually it will equal the weight of the balloon and the net force on the balloon will be zero.  As the balloon is rising it is also acted on by the drag force.  The direction of the drag force is opposite the direction of the balloon's velocity.  Because the drag force the the acceleration of the balloon will quickly decrease to close to zero and the balloon will rise with nearly constant velocity.  You can explore this behavior in an exercise associated with this module.

Option 2:

We can increase the temperature of the air in the balloon.  Then we can obtain the same pressure with a smaller particle density.  Again the weight of the balloon decreases and the net force on the balloon is upward.  The hot air balloon will rise.

Links

How a helium balloon works
Experiments with a helium balloon

Exercise  (You can earn up to 5 points extra credit by completing this exercise.)

Thermometers

Assume we have a gas in a container with a movable piston under atmospheric pressure.

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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 10 oC. We have

(DV/V) = (DT/T) = 10/273 = 0.037 = 3.7%.

By monitoring the volume of the gas we can monitor its temperature.

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

The average volume expansion coefficient b is defined through DV = bVDT.  We have b = 3a.

Linear expansion coefficients a: (per oC)

Aluminum 23 ´ 10-6
Copper 17 ´ 10-6
Glass (ordinary) 9 ´ 10-6
Glass (Pyrex) 3.2 ´ 10-6
Steel 11 ´ 10-6
Invar 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.)

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 (DV/V) = b = 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 = 1000cm3 of liquid with b = 10-4 in a container with bottom area A = 100cm2.  The height of the liquid is 1 cm.  (Volume = area ´ height.)  You increase the temperature of the liquid by 20 oC. DV = bVDT = 10-4 ´ 1000 cm3 ´ 20 = 2 cm3.  The change in height is Dh = DV/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 Dh = DV/A = 2 cm3/(1 cm2) = 2 cm.  You have constructed a usable thermometer.

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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 Dl = alDT = 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 Dl = alDT = 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 a.  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.

Some liquid crystals change color when their temperature changes.  They selectively only reflect one color of the white light falling onto them.  The color that is reflected back depends on the temperature and can be used as a temperature indicator.

Link:

The battery tester
Water and Ice

Problems:

In a constant volume gas thermometer the pressure at 20 oC is 0.98 atm.
(a) What is the pressure at 45o C?
(b) What is the temperature if the pressure is 0.5 atm?
Solution:
(a)  PV/T = constant.  Since V is constant, P/T = constant. P1/T1 = P2/T2. P1T2/T1 = P2
P2 = (0.98 atm)(318 K/293 K) =1.064 atm.
(b)  P2T1/P1 = T2. T2 = (0.5 atm)(293 K)/(0.98 atm) = 149.5 K.
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 oC = (temperature in oF - 32) ´ 5/9.
-195.81 C = ((9/5)(-195.81) + 32) oF = -320.5 oF.
(b)  -195.81 oC = (-195.81+273.15) K = 77.34 K.
A copper telephone wire has essentially no sag between poles 35 m apart on a winter day when the temperature is -20 oC.  How much longer is the wire on a summer day, when the temperature is 35 oC?
Solution:
Dl = alDT.  For copper a = 17´10-6 per oC.  Dl = (17´10-6)(35 m)55 = 0.0327 m = 3.27 cm.
A square hole 8 cm along each side is cut in a sheet of copper.  Calculate the change in the area of the hole if the temperature of the sheet is increased by 50 K.
DA = 2aADT.  For copper a = 17´10-6 per oC.  DA = (34´10-6)(64 cm2)50 = 0.109 cm2.

Link to other Web material:

How Thermometers Work
Temperature and Thermal Expansion
About Atmospheric Models
Standard Atmospheric Computations
Air Pressure, Height, and Temperature

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