1. What is Temperature?
Observation: When objects are placed near each other, they may change, even if no work is done. (Example: when you put water from the “hot tap” next to water from the “cold tap”, they both change.)
In 18th century, it was assumed that a substance called “caloric” flowed from one material to the other. It is now known that energy (called “heat”) flows between them.
When they stop changing, we say that they are in “thermal equilibrium”.

2. Zeroth Law of Thermodynamics: If objects A and B are separately in thermal equilibrium with a third object C, then A and B are in thermal equilibrium with each other. [This is an empirical law, i.e. based on our observations.]
Suppose object C has some property, LC, that changes when it is brought into contact with different objects. After C is brought into contact with A, C changes until LC reaches its thermal equilibrium value LC(A). If C is brought into contact with B but doesn’t change (because it is already in thermal equilibrium with B), then LC(B) = LC(A). That is, C is telling us that there is some property of A that has the same value as that property of B. This property is called temperature.

3. Or put another way: Heat (energy) flows between two objects if they are at different temperatures. It stops flowing when they are in thermal equilibrium – i.e. at the same temperature.
If they are initially at the same temperature, no heat will flow between them.

4. If object C has a convenient property that can be measured when its temperature changes, it is called a thermometer. One wants the thermometer to be:
a) small (so it doesn’t disturb the object being measured very much),
b) have a property that can be measured fairly easily, and
c) be reproducible.
Common thermometers include volumes of liquids in a column (mercury, alcohol), resistance of metals or semiconductors, pressure of a gas, color of a liquid crystal.

5. To establish a temperature scale, one chooses two “fixed points”: materials in conditions that can be reproduced. The “Celsius scale” chooses a) ice-water mixture at 1 atmosphere (0 oC) and b) water-steam mixture at one atmosphere (100 oC). Then if one is using a thermometer with property L, the temperature of the material being measured (x) is:
TC(x) = 100oC [L(x) – L(i-w)] / [L(s-w) – L(i-w)]
[Fahrenheit: TF(x) = 32oF + 180oF [L(x) – L(i-w)] / [L(s-w) – L(i-w)]
so TF(x) = 32 + (9/5) TC(x) and TC(x) = 5/9 (TF(x)-32)]
Problem: No guarantee that two different thermometers (e.g. alcohol and mercury) will exactly agree at any points besides i-w and s-w.
 Need to agree on a universal thermometer.

6. TC(x) = 100oC [L(x) – L(i-w)] / [L(s-w) – L(i-w)]
Observations:
1) The pressure of a gas in a fixed volume is proportional to the density (mass/volume) if the density is low. Therefore, for a given gas, [P(x)-P(i-w)] / [P(s-w) – P(i-w)] is independent of the density (as long as the density is small).
Density(1)
= 6/3.6 Density(2)
= 3 Density(3)

7. TC(x) = 100oC [L(x) – L(i-w)] / [L(s-w) – L(i-w)]
Observations:
The pressure of a gas in a fixed volume will be proportional to the density (mass/volume) if the density is low. Therefore, for a given gas, [P(x)-P(i-w)] / [P(s-w) – P(i-w)] is independent of the density (as long as the density is small).
[P(x)-P(i-w)] / [P(s-w) – P(i-w)] is also independent of type of gas (as long as the density is small).
All gases give same temperature if density is small.

8. Universal Thermometer: Low density gas = “Ideal Gas”:
Measure pressure in a fixed volume:
TC(x) = 100oC [P(x) – P(i-w)] / [P(s-w) – P(i-w)]
P  0 at – oC for
any gas
any low density
Suggests shifting temperatures by oC: T = TC
This is the “absolute” or Kelvin scale of temperature.

9. Absolute (Kelvin) temperature scale: TK(x) = 273 K [P(x)/P(i-w)]
T(K) = T(oC)
Since cannot have P < 0, cannot have temperatures < 0K.
The lowest temperature ever achieved  0.3 nK.
Absolute (Kelvin) temperature scale: TK(x) = 273 K [P(x)/P(i-w)]
[ More accurately, TK(x) = K [P(x)/P(TP)] ]
Triple point of water: temperature and pressure ( atm.) where ice, liquid, and steam coexist.

10. Item
Temperature
Boiling pt. of liquid helium (1 atm.)
4.2 K
Boiling pt. of liquid nitrogen ( 1 atm.)
77 K
Freezing pt. of water (1 atm.) K
Triple pt. of water K
Boiling pt. of water (1 atm.) K
Melting point of gold
1337 K
Sun’s surface
5780 K
Sun’s interior (where fusion occurs)
15,000,000 K = 15 MK

11. Problem: The melting point of gold = 1337 K
Problem: The melting point of gold = 1337 K. What is this temperature in Celsius and Fahrenheit?
T(oC) = T(K) – 273  1337 K = 1064 oC
T(oF) = 32 + (9/5) T(oC)  1337 K = 1947 oF
Problem: The “normal” body temperature of a human = 98.6 oF. What is this temperature in Celsius and Kelvin?
T(oC) = (5/9) (T(oF) - 32)  98.6 oF = 37 oC
T(K) = T(oC)  98.6 oF = 310 K

12. Low-Density (Ideal) Gases
For a of a given material gas with low density m/V:
P  T(K)
For a gas of a given material at a particular temperature:
P  m/V
AND: Comparing different gases, for a given (low) m/V and T:
P  1/M (molecular weight)
 P = constant * (m/M) T/V

13. Problem: The pressure of 2
Problem: The pressure of 2.4 g hydrogen gas (H2, 2 g/mole) in 1 m3 at 300 K = 0.03 atm.
a) What is its pressure at 450 K?
b) What would be its pressure at 300 K if twice as much hydrogen gas was in the
container?
c) What would be the pressure of 2.4 g of nitrogen (N2, 28 g/mole) in 1 m3 at 300 K?

16. a) What is its pressure at 450 K?
Problem: The pressure of 2.4 g hydrogen gas (H2, 2 g/mole) in 1 m3 at 300 K = 0.03 atm.
a) What is its pressure at 450 K?
b) What would be its pressure at 300 K if twice as much hydrogen gas was in the
container?
c) What would be the pressure of 2.4 g of nitrogen (N2, 28 g/mole) in 1 m3 at 300 K?
Pressure  temperature, so atm.
Pressure  density, so 0.06 atm.
Pressure  1/molecular weight, so P = 0.03 atm (2/28) = atm.

17. PV = NkBT, here kB = R/NA = 1.38 x 10-23 J/K.
P = constant (m/M) T / V
The experimental value of the constant = J/moleK  R = universal gas constant
i.e. PV = (m/M) RT
But mass /molecular weight = m/M = [g/ (g/mole)] = n (number of moles):
PV = nRT
1 mole = NA molecules, where Avogadro’s number NA = 6.02 x 1023/mole,
i.e. n = N/NA, so PV = N (R/NA)T
# moles # molecules
PV = NkBT, here kB = R/NA = 1.38 x J/K.
= Boltzman’s constant
Ideal Gas Law

18. PV = NkBT
How many molecules of air are in a room V = 5m x 6m x 3m at room temperature (T = 295 K) and atmospheric pressure (P = 1.01 x 105 N/m2 = 1.01 x 105 Pa)?
N = PV/kBT = (1.01 x 105 N/m2) (90 m3)/[(1.38 x 1023 J/K)(295 K) = 2.23 x 1027
i.e. n = N/NA = (2.23 x 1027molecules) / (6.02 x 1023 molecules/mole) = 3709 moles

19. PV = nRT Ideal Gas Law When is a gas (approximately) an “ideal gas”?
a) Density low
i) volume physically occupied by the molecules << volume of container
ii) on the average, the molecules are far apart, so they don’t feel each other
(i.e. interact) except when they collide.
b) T >> boiling point:
iii) average kinetic energy of the molecules >> average intermolecular potential
energy (so they don’t stick together)

20. n
Molecules repel each other when they get very close
Molecules weakly attract each other at large distances
For air (N2 & O2 & Ar) at room temperature and atmospheric density (n/V ~ 44 moles/m3), the pressure deviates from the ideal gas law value by ~ 0.1%.
For CO2 at room temperature and atmospheric density, the pressure deviates from the ideal gas law value by ~ 0.3%.