Zero'th law of Thermodynamics
P, V, T
SI units include meters / kilograms / seconds .
Force: 1 Newton = 1 N $\equiv$ kg$\cdot$m$\cdot$s${}^{-2}$.
Work (energy): Force * distance= 1 N $\cdot$ m $\equiv$ 1 Joule = 1 J.
Pressure
Need a better definition of pressure.
Pressure @ hyperphysics.
Usually, we'll be concerned with forces arising due to pressure.
Gas inside a piston at a pressure $P$,
- Piston surface area $A$,
The effect of pressure is that there arises a force $F$ on the piston perpendicular to the surface, of magnitude $F = PA$.
The work done by the force if the piston moves a distance $dx$ is... $$\delta W = F \,dx = (PA) dx = P(A\,dx)= P\,dV.$$ So, the units of pressure are also energy / volume
SI unit of pressure:
Force / Area = $1 \ {\rm N} / {\rm m}^2 \equiv 1 \ {\rm Pa}$ = 1 "Pascal".
At sea level, room temperature, the pressure is 1 atmosphere.
A column of mercury (density $\rho$ in kg/m${}^3$) in an evacuated tube, having a cross-sectional area, $A$, rises to a height, $h$, of 76 cm.
weight $= Mg = \rho V g=\rho h A g$
Pressure of this column = weight / A = $\rho h g=
(1.36 \times 10^4 {\rm kg}/{\rm
m}^3)(0.76 {\rm m})(9.8 {\rm m}/{\rm s}^2)=1.01 \times 10^5$ "Pa".
The "torr" is defined as the pressure due to a column of 1 mm of mercury (Hg): $$ 1\ {\rm torr} = \frac{1}{760}{\rm atm} = 133.3 {\rm Pa}$$
When your tire pressure measures zero, is that a vacuum?
No! Tire pressure gauges are measuring the pressure above atmospheric pressure. One atmosphere is ~15 PSI, so a reading of 30 psi on a tire gauge (sometimes denoted as "gauge pressure") corresponds to an absolute pressure of 45 psi.
Temperature
...and the zero'th law
Thermal equilibrium and temperature
When two systems (say, hot coffee / cold mug) are separated by a diathermal boundary, the thermodynamic parameters that characterize each one change for a while. But if left in contact long enough, both systems eventually reach a state of equilibrium, and we say the two systems are in thermal equilibrium with each other.
Transitive property
If systems
A and C (the thermometer) are in thermal equilibrium with each other...
[System C happens to be a "thermometer", but could be any system for the purposes of this argument]
And
if systems B and C are also in thermal equilibrium with each other...
Then
when systems A and B are brought into thermal contact they will already be in thermal
equilibrium with each other as well: No "heat" will flow across a diathermal boundary that separates them.
Zero'th law
Temperature is a thermodynamic parameter of any equilibrium system. For two thermodynamic systems to be in thermal equilibrium with each other on contact, all we need to know is that they both have the same temperature. (No info needed about *how* they got that way.)
Change of State: Heat or Caloric?
What happens when two systems of different temperature are brought into contact?
If the initial states of each system $A$ and $C$ are specified by $(P_{Ai}, V_{Ai}, T_{Ai})$ and $(P_{Ci},V_{Ci},T_{Ci})$, and $T_{Ai} \neq T_{Ci}$, then the zero'th law means that at least the temperatures must change before reaching equilibrium. This means the thermodynamic state of each system must change.
What is it that causes this change of state?
- if the boundary of each system is *fixed, then no mechanical work can be exchanged between the systems. So something else...
- 18th century answer: "caloric"--a conserved substance--flows in (or out) of the system.
- late 19th century: "heat"--which is only one kind of energy-- flows in (or out) of the system. It's energy that is the conserved quantity...not heat alone.
Read Wikipedia's article on Caloric Theory to hear about successes and the ultimate downfall of caloric theory.
Defining a temperature scale
To define a temperature scale, any thermometric property will do (any property that varies with temperature).
Volume
The most familiar thermometers depend on a change in volume of alcohol (or
years ago mercury).
Length
A
bimetallic wire or strip is a sandwich of two wires that have different thermal
expansion coefficients.
Pressure
A constant-volume
gas thermometer is shown schematically at right. The bulb, $B_1$, is put in thermal contact with the system to measure. $B_3$ is lifted or lowered
until the mercury meniscus at the top of $B_2$ returns to its 0 setting. The
thermometric property is the pressure of the gas, which is $h$-mm of Hg (or $\rho g h$ Pascals).
If this thermometer is calibrated to water's freezing and boiling points,
then, extrapolating, as $P \rightarrow 0$, we find $T
\rightarrow -273.15 {\rm C}^o$ for all real gases!
Calibration: phases of water
Here is a phase diagram for water (and nothing else) in a container.
- Historically, 100${}^o$C is the difference between the melting and boiling point of water under terrestrial conditions: $P=1$atm.
- But a more appealing point for calibrating thermometers is the triple point (TP) of water: liquid, vapor, solid in equilibrium at $P=0.006$atm and $T=0.1^o$C.
Carter's book has potentially confusing notation in equation 1.18: $$T(K)=273.16 \lim_{P_{TP}\to 0} \frac{P}{P_{TP}}$$ Wait--you can't change the triple point pressure of water?!
No, $P_{TP}$ refers to the pressure of the gas thermometer when it is in contact with a system consisting of water at its triple point.
The equation refers to using a constant volume gas thermometer. You measure the pressure $P$ when it is in contact with a system of unknown temperature. You measure $P_{TP}$ when it is in contact with the the reference cell: water at its triple point.
Different real gases (oxygen, helium, nitrogen, etc) give slightly different temperatures by this method.
But if you reduce the amount of the gas in the gas thermometer, remeasure your system, and do this many times, each time reducing the amount of gas, you can extrapolate the ratio down to the no gas at all ($P_{TP}\to 0$) limit, for all real gases this limit converges to exactly the same ratio $P/P_{TP}$!
Image credits
Flickr user Jacqueline, Wikipedia on alcohol thermometers, Hustvedt, Komehache 888, David Mogk
