Recall:

Thermometers and Temperature Scales

Thermometers are devices for measuring temperature. The common thermometers consist of a quantity of mercury or alcohol (glycol) in a thin glass tube. As the temperature varies, the fluid expands and contracts, and its height in the tube changes. A scale is supplied to read off the height of the fluid directly in terms of temperature.

The simple thermometers described above have certain drawbacks. They don't operate below the freezing point of the liquid, and thermometers based on different liquids may not agree at all temperatures. Normally the thermometers are calibrated for 0°C (32°F) when ice and water are in equilibrium, and 100°C (212°F) when water and steam are in equilibrium.

There are three temperature scales you need to be familiar with, the Celsius, Fahrenheit, and Kelvin scales. You are probably familiar with the Celsius and Fahrenheit temperature scales. The Kelvin scale (the SI standard) measures temperature relative to absolute zero. Absolute zero is the lowest achievable temperature, corresponding to 0°K, or -273.15°C. One Kelvin unit equals on Celsius unit -- a change of temperature by 1°K equals a change of temperature by 1°C. The following table summarizes the different scales.

Thermal Expansion of Solids and Liquids

Common thermometers make use of the expansion of liquids with increasing temperature. This phenomenon is called thermal expansion.

On a microscopic scale, thermal expansion occurs because the average separation of atoms and molecules increases. The atoms and molecules in a substance are constantly vibrating, but on average they remain a certain distance apart. If the temperature of the substance increases, then the energy of the vibrations increases, meaning that the size (amplitude) of the vibrations increases. The result is that the average separation between atoms and molecules increases.

Empirically, we find that an object of initial length L₀ will undergo a change of length, DL for a temperature change DT, given by:
DL = aL₀DT.
The proportionality constant a is called the coefficient of linear expansion. The coefficients of linear expansion for a variety of substances near room temperature are listed in table 10.1 of the text.

Exapmle: P10.3

The New River Gorge bridge in West Virginia is a 518 m long steel arch. How much will its length change between temperature extremes -20°C and 35°C?

From table 10.1 we find the coefficient of linear expansion for steel to be a = 11x10^-6/°C. If L₀ = 518 m, and DT = 55°C, then
DL = aL₀DT = (11x10^-6/°C)(518 m)(55°C) = 0.31 m = 31 cm.

Similar calculations can be done for area and volume expansion, for example, determining the change in area of a metal plate, or the change in volume of a slab of concrete when it is heated or cooled. The resulting equations are:
DA = gA₀DT, and
DV = bV₀DT
where g is the coefficient of surface expansion and equals 2a, and b is the coefficient of volume expansion and equals 3a.

Macroscopic Description of an Ideal Gas

This long title is basically a way to introduce the ideal gas law into a Physics course. It is found that under many conditions, the pressure P, volume V, temperature T, and moles n, of a gas are related by:
PV = nRT
where R is the universal gas constant, R = 8.31 J/mol K in SI units. Recall that in SI units, pressure is measured in Pa = N/m², volume in m³, and temperature in °K. If the pressure is given in atmospheres (atm) and the volume in liters (L), then you can use the value of R = 0.0821 L atm/mol K.

You may recall the fact that 1 mol of gas at STP (standard temperature and pressure occupies a volume of 22.4L. This result can be obtained by using the above value of R, letting P = 1atm, T = 0°C = 273°K, and n = 1mol. V = nRT/P = (1mol)(0.0821 L atm/mol K)(273°K)/(1atm) = 22.4 L.

Example: P10.21

One mole of gas is at a pressure of 6.00 atm and a temperature of 27.0°C. (a) If the gas is heated at constant volume until the pressure triples, what is the final temperature? (b) If the gas is heated so that both the pressure and volume are doubled, what is the final temperature?

(a) Since the number of moles and the volume are held constant, we have
P₀/T₀ = nR/V = P_fT_f.
Since P_f = 3P₀ and we're given that T₀ = 300°K, then
T_f = (P_f/P₀)T₀ = (3)(300°K) = 900°K = 627°C.
NOTICE that if the temperature is left in Celsius, then the result would be 81°C which is WRONG.

(b) Now only the number of moles remain constant, so
P₀V₀/T₀ = nR = P_fV_f/T_f. This time we're told that P_f = 2P₀ and V_f = 2V₀ so
T_f = T₀(P_f/P₀)(V_f/V₀) = (300°K)(2)(2) = 1200°K = 927°C.
Once again, forgetting to use the kelvin scale yields an incorrect result.

Avogadro's Number

Avogadro's number is
N_A = 6.02 x 10²³
This is the number of molecules in one mole of gas (or any substance). If we have n moles of gas, then the number of molecules is N = nN_A, or n = N/N_A. Thus we can rewrite the ideal gas law as:
PV = (N/N_A)RT = N(R/N_A)T = Nk_BT
where k_B is Boltzmann's constant. The value of k_B is
k_B = R/N_A = (8.31 J/mol K)/(6.02 x 10²³/mol) = 1.38 x 10^-23 J/K.

The Kinetic Theory of Gases

We can relate the properties of an ideal gas to the mechanics that we studied in earlier chapters. To do this, we must picture the gas as a collection of molecules moving about inside a container. To be specific, we make the following assumptions (from the text):

1. The number of molecules is large, and the average separation between them is large compared with their dimensions. This means that the molecules occupy a negligible volume in the container.
2. The molecules obey Newton's laws of motion, but as a whole they move randomly.
3. The molecules undergo elastic collisions with each other and with the walls of the container. Thus in the collisions kinetic energy is constant.
4. The forces between molecules are negligible except during a collision.
5. The gas under consideration is a pure substance; that is all molecules are identical.

In this model, the pressure of the gas is due to the molecules hitting the walls of the container. If there were only a few molecules, then this would be more like banging. But because there are many molecules (of order 10²³), the collisions of molecules with the walls is constant and very uniform across the wall. Imagine the difference between holding an umbrella in a slow rain versus a downpour and I think you've got the picture.

A calculation of the average force on a wall yields the following expression for the pressure in terms of the rms velocity of the molecules:
P = (2/3)(N/V)(\frac12;mv²). The pressure is proportional to the number of molecules per unit volume (the number density of molecules) and the average kinetic energy of the molecules, KE_avg = ½mv²_avg. This is truly an amazing result. It relates the mechanics of microscopic particles (their average kinetic energy) to a macroscopic property (pressure).

In this model, the temperature of a gas is related to the average kinetic energy of the molecules of the gas. To see this, rewrite the above expression as:
PV = (2/3)N(½mv²_avg).
Since the ideal gas law tells us that PV=nRT -- the same PV -- then we can equate nRT = (2/3)N(½mv²_avg). Use the relations n = N/N_A and R/N_A = k_B to get:
T = (2/3k_B)(½mv²_avg)
or, in the more standard form:
KE_avg = ½mv²_avg = (3/2)k_BT
This is another amazing result. It relates the average kinetic energy of a molecule of gas to the temperature of the gas as a whole!

Finally, a comment about the quantity v²_avg. This is the average of the square of the velocities. Often we want to speak of an average velocity (not squared). But the average velocity is zero, because there are as many molecules moving to the right (positive) as to the left (negative) on average, and they cancel to give an average velocity of zero. So instead we use the rms velocity. The letters rms stand for root-mean-square, which translates to the square root of the average of the square of the velocities. The rms velocity is
v_rms = Sqrt{v²_avg} = Sqrt{3k_BT/m} = Sqrt{3RT/M}
where M is the molar mass of the molecule.

Example: P10.38

A cylinder contains a mixture of helium and argon gas in equilibrium at a temperature of 150°C. (a) What is the average kinetic energy of each type of molecule? (b) What is the rms speed of each type of molecule?

(a) KE_avg = (3/2)k_BT = (3/2)(1.38x10^-23 J/K)(150+273°K) = 876x10^-23 J = 8.76x10^-21 J. Since the kinetic energy doesn't depend on the type of atom or molecule (i.e. no mass dependence), the result is the same for both gases.

(b) v_rms = Sqrt{3RT/M}
This does depend on the molecular mass. M_He = 4.0x10^-3 kg/mol and M_Ar = 39.948x10^-3 kg/mol. The resulting rms speeds are:
v_{rms He} = Sqrt{3(8.31 J/mol K)(423K)/(4.0x10^-3 kg/mol) = 1620 m/s
v_{rms Ar} = Sqrt{3(8.31 J/mol K)(423K)/(40.x10^-3 kg/mol) = 513 m/s.