\documentclass[12pt]{article}
\usepackage{graphicx}
%\usepackage{epsfig}
%\topmargin=+0.0in
%\oddsidemargin=0.25in
\textwidth=6.0in
%\textheight=8in
%\footskip=6ex
%\footheight=2ex
\begin{document}
\begin{center}
{\bf Notes on Thermodynamics}
\end{center}
\bigskip
The topic for the last part of our physics class this quarter will
be thermodynamics. Thermodynamics deals with energy transfer
processes. {\bf The key idea is that materials have "internal energy"}.
The internal energy is the energy that the atoms and molecules of
the material possess. For example, in a gas and liquid the molecules are moving
and have kinetic energy. The molecules can also rotate and vibrate,
and these motions also contribute to the gases total internal energy.
In a solid, the atoms can oscillate about their equilibrium position and
also possess energy. The total internal energy is defined as:\\
\noindent {\bf The total internal energy of a substance = the sum of
the energies of the constituents of the substance.}\\
\noindent When two substances come in contact, internal energy from one
substance can decrease while the internal energy of the other increases.
The first law of thermodynamics states that the energy lost by one
substance is gained by the other. That is, that there exists a
quantity called energy that is conserved. We will develop this idea
and more over the next 4 weeks.
We will be analyzing gases, liquids, and solids, so we need to
determine which properties are necessary for an appropriate
description of these materials. Although we will be making
models about the constituents of these materials, what is
usually measured are their {\bf macroscopic quantities} or "large
scale" properties of the systems. We have already some of these
when we studied fluids in the beginning of the course:\\
\noindent {\bf Volume} ($V$): The volume that the material occupies. Volume
is a scalar.\\
\noindent {\bf Pressure} ($P$): The pressure equals the force/area if a surface
is placed in the liquid or gas. Pressure is also a scalar.\\
\noindent Both these quantities are {\bf state variables}. State variables
are quantities (macroscopic) that only depend on the state of the system and
not on how the system was formed. The systems total internal energy ($U$)
defined above is also a state variable. We will be interested in the
relationships between the state variables of a system. One of the most
important concepts in thermodynamics is that of temperature. Our first
task will be to determine a consistent way to quantify it, that is
find a way to measure it and assign a number value to the temperature of
a system.
\bigskip
\centerline{\bf Temperature and Thermometers}
\bigskip
We have a general idea of what temperature is: hot things have a
higher temperature than colder ones. However, Physics is an exact
quantitative science, so we need to be precise with temperature.
In order for a system to have a unique temperature it needs to
be in thermal equilibrium.\\
\noindent {\bf Thermal Equilibrium}: A system is in thermal
equilibrium if the macroscopic state variables do not change
in time.\\
Before we can assign a number value to the temperature we need to
have a qualitative understanding of what temperature is. Temperature
is some sort of measure of which way energy will be transfered when
two systems are in contact. Energy will flow from a system that
has a higher temperature to one that has a lower one. For two
systems to have the same temperature, it means that when they
are placed in contact there will not be any energy transfer
from one to the other. That is, when they are placed in
contact with each other, the macroscopic quantities stay the
same, i.e. they are in thermal equilibrium. The "Zeroth law"
of thermodynamics includes a third system:\\
\noindent {\bf Zero'th Law of Thermodynamics}: If system A and
system B are in thermal equilibrium with system C (i.e. they have
the same temperature as C), they they are in thermal equilibrium
with each other (i.e. they have the same temperature).\\
This property of nature enables one to assign a number value to
temperature. We just need to choose two reference values, and
use a substance with a macroscopic quantity that varies
with "our preception of temperature". The substance we call a
thermometer.
The Celsius choice of reference values is to assign
the triple point of water (where $H_2O$ exists in solid, liquid and vapor
forms) to be zero degrees. The other reference value is assigned to be
100 degrees where water and steam co-exist at one atmosphere pressure.
Suppose we have a third substance at an unknown temperature $T$.
How do we determine $T$? One way is to pick a substance to be
a thermometer and define $T$ to be proportional to a macroscopic
quantity of the substance. Let's demonstrate this with an example.
We will find the value of $T$ of an object using three different substances for
thermometers: liquid mercury, a resistor and a dilute gas. As our
parameter we will choose the volume of mercury, the resistance of the
resistor and the pressure of the dilute gas. Here is our (made-up) data:
\begin{center}
\begin{tabular}{c|c|c|c}
Temperature & Height of Hg & Resistance & Pressure of Dilute Gas \\
\hline
100 & 8.0 cm & 3.40 mV & $1.37 \times 10^5$ Pa \\
\hline
$T$ & 6.5 cm & 3.24 mV & $1.23 \times 10^5$ Pa \\
\hline
0 & 4.0 cm & 3.00 mV & $1.00 \times 10^5$ Pa \\
\hline
\end{tabular}
\end{center}
\noindent The "unknown" temperature $T$ of the object is determined for each thermometer
by having the temperature be proportional to the parameter that is changing.
So using the mercury thermometer,
\begin{equation}
T = 100 ({{6.5 - 4.0} \over {8.0 - 4.0}}) = 62.5^\circ
\end{equation}
\noindent Using the resistance thermometer,
\begin{equation}
T = 100 ({{3.24 - 3.0} \over {3.4 - 3.0}}) = 60.0^\circ
\end{equation}
\noindent Using the dilute gas thermometer,
\begin{equation}
T = 100 ({{1.23 - 1.0} \over {1.37 - 1.0}}) = 62.2^\circ
\end{equation}
\noindent The data was "made-up", however if real data were used
the results would be similar: in general the temperature $T$
can depend on the type of thermometer used. This is because temperature
is defined to make the parameter of the thermometer proportional
to $T$. All thermometers will agree at the two reference temperatures.
However, in general, the parameters for each thermometer do not have exactly
the same behavior between the reference points.
You might ask, which substance is the best to use as a thermometer? Is it possible
to define temperature such that the equations of thermodynamics have a simple
form for all substances? It would be nice if there were a class of substances
that all give the same value for $T$. There is. After trying different types of thermometers we would
discover the following property about the dilute gas thermometer above:
we get the same value for $T$ for any gas that we use! That is, if we use
air, or $H_2$, or $O_2$, or $He$, etc. the value of $T$ for our object,
obtained by measuring the pressure of the gas, would be the same in each case.
There are many proerties that all dilute gases have in common. Thus, if we quantify
temperature using the constant volume dilute gas thermometer, there is a good
chance that this method gives a useful definition of temperature. This will
be our first way to quantify temperature.
You might be thinking: How small should the density of the gas be for it to
be delute enough? The common properties converge only in the limit as the
density of the gas goes to zero. We define an {\bf ideal gas}:\\
\noindent {\bf Ideal Gas}: The limit as the density of the gas approaches
zero of a dilute gas.\\
\noindent We will call the thermometer a constant volume "ideal gas thermometer", and
the parameter is the pressure of the gas. Using the references of the Celsius scale,
we find that the pressure of the gas would be zero at around $-273^\circ$C. It will
be useful to choose the value of temperature when the pressure of an ideal gas is zero
to be zero degrees. With this choice we have:
\begin{equation}
T \propto P
\end{equation}
\noindent So temperature is defined to be proportional to the pressure of an
ideal gas. The constant of proportionality is chosen such that the change of
one degree is the same as the change of one degree Celsius. That is, the difference
in temperature between boiling water and the triple point is $100^\circ$.
For this choice, the proportionality constant becomes $273.16^\circ$ at
the triple point of water.
\begin{equation}
T \equiv {P \over {P_{tp}}} 273.16
\end{equation}
\noindent where $P_{tp}$ is the pressure of the gas at the triple point of
water. It is important to note that {\bf both pressures must be measured at the
same volume of the gas} (Constant Volume Thermometer). The
unit of this definition of temperature is the Kelvin ($K$).
To measure temperature in the lab or in our homes we can't carry around
a dilute gas and water at the triple point. Commercial thermometer are
calibrated against this standard thermometer and have a range of
validity.
Are the units of temperature derivable from other fundamental units such as
length, time, mass, ...? Or is temperature a fundamental physical quantity?
Later in the course we will consider these questions and discuss better
ways to define temperature that
do not require a thermometer substance at all. One way uses a "Carnot cycle"
and the other way is a statistical approach developed by Boltzmann. However,
the ideal gas thermometer gives the same value for $T$ as these other approaches.
Until we develop the physics behind these other approaches, we will define
temperature via the ideal-gas constant-volume thermometer.
\bigskip
\centerline{\bf Thermal Expansion}
\bigskip
When the internal energy of a substance increases, the atoms and
molecules gain energy, tend to vibrate with larger amplitudes and
the substance usually increases in size. In addition to increasing
in size, an increase in internal energy usually is accompanied by
an increase in temperature. To a good approximation the
fractional increase in size is proportional to the change in temperature.
Consider first a long rod made of a solid material. As long as the
change in temperature, $\Delta T$, is not too large, the fractional
change in length is proportional to $\Delta T$:
\begin{equation}
{{\Delta L} \over {L_0}} \; is \; proportional \; to \; \Delta T
\end{equation}
\noindent where $L_0$ is the original length before the increase
in temperature. The proportionality constant for {\bf linear expansion}
is usually called $\alpha$:
\begin{equation}
{{\Delta L} \over {L_0}} = \alpha \Delta T
\end{equation}
\noindent where $\alpha$ depends on the type of material.
If we call $L$ the final length of the rod after the increase
in temperature, we have:
\begin{equation}
{{L - L_0} \over {L_0}} = \alpha \Delta T
\end{equation}
\noindent or
\begin{equation}
L = L_0 (1 + \alpha \Delta T)
\end{equation}
A rectangular object will expand in the same fractional way. If
the original sides of the rectangle are $a_0$ and $b_0$, then
after a temperature increase of $\Delta T$, side $a$ will
increase to $a_0 (1 + \alpha \Delta T)$ and side $b$ will
increase to $b_0 (1 + \alpha \Delta T)$. The new area $A$ will
be:
\begin{equation}
A = ab = a_0 (1 + \alpha \Delta T) b_0 (1 + \alpha \Delta T)
\end{equation}
\noindent multiplying out the terms gives
\begin{equation}
A = a_0 b_0 (1 + 2 \alpha \Delta T + (\alpha \Delta T)^2)
\end{equation}
\noindent Since $\alpha \Delta T$ is small, the last term is
on the right is much smaller than the middle term, so to
a good approximation we have:
\begin{equation}
A \approx a_0 b_0 (1 + 2 \alpha \Delta T)
\end{equation}
\noindent Since the original area $A_0 = a_0 b_0$, the
equation for "area" expansion is
\begin{equation}
A \approx A_0 (1 + 2 \alpha \Delta T)
\end{equation}
Using similar reasoning, one obtains the {\bf volume expansion}
formula
\begin{equation}
V \approx V_0 (1 + 3 \alpha \Delta T)
\end{equation}
These formulas are not fundamental laws of nature, but rather empirical
results and have a limited range of validity. If the temperature
keeps increasing, the solid does not keep getting bigger and bigger,
but eventually melts.
\bigskip
\centerline{\bf Energy Transfer Processes}
\bigskip
The study of energy transfer processes assumes that all "systems"
possess "internal energy". By internal energy we mean the
sum of the kinetic and potential energies of the consitituents
of the "system". In this course we will consider systems that
are bulk materials like
gases, liquids, and solids. One can extend these ideas to smaller
systems like molecules, atoms or nuclei. Exeriments verify that
objects are made up of atoms and molecules and contain internal energy.
The constituents of a system can gain energy or lose energy, and
consequently the total interal energy of a system can change.
$U$ is denoted as the total internal energy, and \\
\noindent {\bf $\Delta U$ is defined as the change in the
total internal energy of a system}\\
\noindent $U$ can change in different ways, but we divide the
possibilities into two catagories: those due to mechanical
work and those only due a temperature difference:\\
\noindent {\bf $W$ is defined as the mechanical work done
by a substance}\\
\noindent Note the preposition {\bf by} in the defination of
$W$. {\bf If $W$ is positive then the system loses internal
energy}, since the system is doing work. {\bf If $W$ is
negative then the system gains internal energy}, since work
is done on the system. This is the convention chosen by
scientists. It will make sense when we use a $P-V$ diagram.\\
\noindent {\bf $Q$ is defined as the energy tranfered to the
system by processes that don't involve mechanical work}\\
\noindent Note that $Q$ is the energy transfered {\bf to} the
system, so if $Q$ is positive then so is $\Delta U$. $Q$ is
also defined as energy that is transfered as a result of a
temperature difference only. $Q$ is also called heat. Heat
is a transfer of energy.
Both $W$ and $Q$ have units of energy. The metric unit of energy
is the Joule. However, experiments with heat processes were done
before it was realized that heat processes could be related to
mechanical energy. The unit of the Calorie was used. The
conversion from calorie to the Joule, the mechanical equivalent
of heat, is $4.186$ Joules = $1$ calorie. The "calorie" listed
on food packages is equal to $1000$ calories or $4186$ Joules.
In general both types of processes can occur at the same time,
so the change in $U$ is the sum of these two energy transfer
processes:
\begin{equation}
\Delta U = Q - W
\end{equation}
\noindent The minus sign on $W$ is due to the definition of $W$
as the work done by the system. This equation is called
{\bf the first law of thermodynamics}. It is a statement
recognizing the existence of internal energy and its conservation.
\bigskip
\noindent {\bf Some Heat Processes}
\bigskip
When a system gains (or loses) internal energy, the temperature
of the system usually changes. It is interesting to measure
how much the temperature changes for a certain amount of
internal energy change. Here we consider how much the
temperature changes due to heat processes.
Suppose we have a particular substance, whose mass is $m$.
If we transfer an amount of energy $Q$ into the substance,
keeping its volume constant, the temperature of the substance
will in general change. Call the change in temperature $\Delta T$.
How is $Q$ related to $m$ and $\Delta T$? If $m$ is doubled, we
will need twice the amount of energy transfered ($2Q$). This is
true, since doubling the mass is the same has having two identical
samples. Twice the energy is needed. So for the same $\Delta T$,
$Q \propto m$. Also, if $\Delta T$ is not too large, if we want
twice the temperature change, we will need twice the energy transfer.
So for a fixed mass, $Q \propto \Delta T$. Combining these
two proportions gives:
\begin{equation}
Q \propto m \Delta T
\end{equation}
\noindent We can make this an equal sign by adding a constant of
proportionality:
\begin{equation}
Q = c_v m \Delta T
\end{equation}
\noindent The parameter $c_v$ is called the {\bf specific heat
capacity} for the process in which the volume is held constant.
Note that this is not a new fundamental law of nature, but rather
a phenomenological equation. $c_v$ will depend on the type of
substance and can also depend on temperature. If $\Delta T$ is
so large that $c_v$ changes with $T$, then one has to integrate:
\begin{equation}
Q = \int_{T_i}^{T_f} c_v(T) m \; dT
\end{equation}
\noindent where $T_i$ and $T_f$ are the initial and final temperatures
respectively. An important feature of specific heat capacities
is that they are measureable quantities.
The challenge we have as a physicist is to understand
$c_v$ from the microscopic structure of the substance. In this course
we will do this for gases.
If the heat capacity of a substance is large, it means that it takes a lot
of energy transfer to increase the temperature a little. Likewise, if a
substance has a high heat capacity, it has to lose a lot of internal energy
for a small temperature change. It stays "hot" for a long time. The cheese
on a pizza has a higher heat capacity than the bread crust.
\bigskip
\noindent {\bf Latent Heat}
\bigskip
Latent heat is the amount of energy transfer (per mass) that is needed to change the
phase of a substance. For example, if you heat up water on the stove
the temperature rises until it reaches $100^\circ$C. As more energy
is transfered to the water the temperature does not increase any
more. Rather, the water evaporates and changes its state from liquid
to vapor. The amount of energy transfer that is required for this
phase change (per unit mass) is called the latent heat. Tt takes
twice the amount of energy to change the phase of twice the mass. So,
we have:
\begin{equation}
Q = m L
\end{equation}
\noindent where $Q$ is the amount of energy tranfered (heat) to change the
phase of a substance of mass $m$. The proportionality constant $L$ is
called the {\bf latent heat} and has units of Joules/Kg, or cal/g. If the
phase change is from solid to liquid (or liquid to solid) $L$ is called
the latent heat of fusion. If the phase change is from liquid to vapor
(or vapor to liquid) $L$ is called the latent heat of vaporization.
I have included a nice schematic of the energy transfers involved for
water at one atmosphere as it gains or loses energy. See the attached
figure. The three different states of water go from left to right, and the
temperature goes up and down. Similar schematics can be drawn for other
substances.
\begin{figure}
\includegraphics[width=14cm]{fig1323a.png}
\end{figure}
\bigskip
\noindent {\bf Thermal Conductivity}
\bigskip
Before we carry out a microscopic analysis of a dilute gas, we mention
one relationship that pertains to (one dimensional) thermal conduction.
Consider a rod of length $l$ and area $A$. Let the left end be in contact with
a large object that is at a
temperature $T_A$. Let the right end be in contact with a large object that
is at a temperature $T_B$. What is the rate $H$ at which energy is transfered
through the rod? The units of $H$ are energy/time. What does $H$ depend on?\\
\noindent Area: The larger the area $A$, the faster energy can be transfered, so
$H \propto A$.\\
\noindent Length: The longer the length $l$ that the rod is, the slower is the
energy transfer, so we will guess that $H \propto 1/l$.\\
\noindent Temperature difference: The larger that $T_A - T_B$ is, the faster
the energy will be transfered, so $H \propto (T_A - T_B)$.\\
\noindent Putting these ideas together we have
\begin{equation}
H \propto {{(T_A - T_B) A} \over l}
\end{equation}
\noindent Changing the proportionality by an equal sign brings in a
constant $k$:
\begin{equation}
H = k {{(T_A - T_B) A} \over l}
\end{equation}
\noindent where $k$ is called the thermal conductivity.
It is interesting to include both the properties of heat capacity and
thermal conductivity in one application. Suppose the object at the
left end of the rod has a mass $m$ and a specific heat capacity of
$c_v$. Let the temperature on the right, $T_B$, be held constant, but
let the temperature of the object on the left decrease as it loses
energy. Let $T$ be the temperature of the object on the left at
time $t$. Then we have:
\begin{equation}
H = k {{(T - T_B) A} \over l}
\end{equation}
\noindent where $H$ is the energy loss per second of the left object.
The amount of energy, $Q$, the object loses in a time $\Delta t$ is
\begin{equation}
Q = H \Delta t = k {{(T - T_B) A} \over l} \Delta t
\end{equation}
\noindent Since $Q = m c_v \Delta T$, we have
\begin{equation}
m c_v {{\Delta T} \over {\Delta t}} = - k {{(T - T_B) A} \over l}
\end{equation}
\noindent Taking the limit as $\Delta t$ goes to zero results in
a simple differential equation:
\begin{equation}
{{d T} \over {d t}} = - k {{(T - T_B) A} \over {m c_v l}}
\end{equation}
\noindent Since $T_B$ is a constant, $dT/dt=d(T-T_B)/dt$. If we
define the temperature difference between the left and right sides,
$(T - T_B)$, as $\delta T$, we have:
\begin{equation}
{{d (\delta T)} \over {d t}} = - {{k A} \over {m c_v l}} (\delta T)
\end{equation}
\noindent This is a simple differential equation whose solution is an
exponentially decaying function:
\begin{equation}
\delta T = (\delta T)_0 e^{- {{kA} \over {m c_v l}} t}
\end{equation}
\noindent where $(\delta T_0$ is the initial temperature difference between
the left and right side. Thus the temperature decreases exponentially to
its final value. This type of cooling is called Newton's cooling.
\bigskip
\noindent {\bf Convection}: Energy transfered by a movement of mass
is called convection. Some demonstrations of convection will be
presented in class, but we will not do a quantitative analysis
of convection.
\bigskip
\noindent {\bf Radiation} Energy can also be tranfered through a vacuum (as well as air).
Objects naturally "radiate" electro-magnetic energy. The electro-magnetic interaction will
be covered next quarter. Here we present the formula, without derivation, of the amount
of electro-magnetic energy radiated from an object at a temperature $T$:
\begin{equation}
P = \sigma A e T^4
\end{equation}
\noindent where $P$ is the energy/sec (power) radiated, $A$ is the area of the object,
$e$ is the emissivity, and $T$ is the temperature in $K$. $\sigma$ is a constant,
the Stefan-Boltzmann constant and is equal to $5.669 \times 10^{-8}$ K$^4$ W/m$^2$.
The emissivity depends on the properties of the object and takes values between
zero and one.
\bigskip
\centerline{\bf Ideal Gases}
\bigskip
\centerline{\bf Experiments on an Ideal Gas}
\bigskip
Our discussion so far has been phenominological. That is, we have been
finding empirical formulas that relate (and define) certain quantities.
Now we would like to do some physics. We would like to understand
the relationships between macroscopic quantities from a microscopic
model of the substance. In physics we usually start with the simplest
systems first, then extend our ideas to more complicated situations.
Due to the universality of ideal gases in defining temperature, we
suspect that an ideal gas might be a system for which a simple model
will explain many properties. For the rest of the course we will
mainly analyze ideal gases and see what they can teach us about
energy transfer processes.
Let's first summarize the experimental data on ideal gases. Remember
an ideal gas is a dilute gas in the limit that the density approaches
zero. Think of an ideal gas as a very dilute gas. The main macroscopic
measureable quantities that describe an ideal gas are:\\
\noindent {\bf Volume} (V): the volume of the container that confines the
gas.\\
\noindent {\bf Pressure} (P): The pressure in the gas. If the gas is in
equilibrium then the pressure is the same everywhere (neglecting gravity).\\
\noindent {\bf Temperature} (T): We have used ideal gases to define temperature.
If the gas is in equilibrium, the temperature is the same everywhere.\\
\noindent {\bf Mass or Number of Molecules} (N): The mass of a gas can be
measured, and we assume that $N$, the number of molecules, is proportional
to its mass. We will use the parameter $N$, although difficult to measure,
could be determined using Avogadro's number.\\
How are these four quantities related to each other. We need to keep two
fixed and experimentally determine how the other two depend on each other.\\
\noindent {\it N and V held constant}\\
How does $T$ depend on $P$? We defined temperature to be proportional to
$P$, so
\begin{equation}
T \propto P
\end{equation}
\noindent Although we defined $T$ so this proportionality is true, the fact
that we get the same value for $T$ no matter what the gas is something
special. It is a nice experimental result.\\
\noindent {\it N and T held constant}\\
\noindent How does $P$ depend on $V$? If the volume of a gas is
increases with constant temperature, the pressure is reduced.
We will show in lecture that:
\begin{equation}
P \propto {1 \over V}
\end{equation}
\noindent This is a really nice and simple experimental result! Why is it
so simple? What can we learn from this relationship?\\
\noindent {\it T and V held constant}\\
\noindent How does $P$ depend on $N$? We will show in lecture that:
\begin{equation}
P \propto N
\end{equation}
\noindent We can combine the results of these three experiments into one
equation:
\begin{equation}
P \propto {{NT} \over V}
\end{equation}
\noindent {\it N and P held constant}\\
\noindent How does $V$ depend on $T$ in this case? From the above equation
we must have $V \propto T$, which is verified experimentally.\\
We can replace the proportionality sign with an equal sign by adding a
proportionality constant:
\begin{equation}
P = {{NkT} \over V}
\end{equation}
\noindent where the constant $k$ is called Boltzmann's constant. We will discuss
its value later. This equation is also written in terms of the number of moles
of the gas by multiplying and dividing by Avogadro's number, $N_A$:
\begin{eqnarray*}
Nk & = & ({N \over N_A})(N_A k)\\
& = & n R
\end{eqnarray*}
\noindent where $n$ is the number of moles of the gas, and $R$ is called
the gas constant. Note that $R = N_A k$. The {\bf equation of state
for an ideal gas} is often written as:
\begin{equation}
PV = nRT
\end{equation}
\noindent The constant $R$ is measureable, once we have decided on how much
mass a mole is and our temperature scale (i.e. $\Delta T$ between boiling
water and the triple point). We need to measure the pressure, volume and Temperature
for a known number of moles. The resuls is:
\begin{equation}
R \approx 8.34 {{Joules} \over {mole \; ^\circ K}}
\end{equation}
\noindent Once Avagadro's number is known to be $6.02 \times 10^{23}$ then
$k = 1.38 \times 10^{-23} \; J/^\circ K$.
Our task now is to come up with a microscopic model to explain this equation
of state. There are other experimental quantites to compare with, $c_v$ being
one of them. First, we will try to understand the equation of state.
\bigskip
\centerline{\bf Kinetic Theory: A model for ideal gases}
\bigskip
Consider a dilute gas that is contained in a box with sides of equal length
$L$. Our model of a gas is that it consists of a large number $N$ of molecules
that are "bouncing around" inside the container. We will assume that the
molecules bounce off each other and the sides of the container elastically.
That is, the molecules do not lose kinetic energy in the collisions. We will also
neglect the gravitational potential energy of the molecules, $mgy$. This will
be OK if the kinetic energy of each molecule is much larger than $mgL$. As
we will see, this is a good approximation.
Pressure is force/area. How are the molecules applying a force to the sides?
By bouncing off, their momentum is changed. Force is change in momentum
per time. Let's determine the change in momentum of a molecule as it bounces
off a side of the box, a side in the y-z plane.\\
\noindent 1. {\bf Momentum Change per bounce}: Momentum is $m \vec{v}$.
When a molecule bounces {\it elastically} off a side (in the y-z plane), the x-component of its
velocity gets reversed. The x-component of its momentum changes from
$m v_x$ to $-m v_x$. This is a {\it net change} of\\
\centerline{$2 m v_x$}
\bigskip
\noindent 2. {\bf Force of one molecule on side}: The force the
side feels from this one molecule will be: (the number of times the molecule hits
the side per second) times $2m v_x$, since force is (momentum change)/time. The number of
times the molecule hits the side per second is $v_x/(2L)$. So the force of one molecule on
the side is: $(2m v_x)(v_x/(2L)$ or\\
\centerline{$m v_x^2/L$}
\bigskip
\noindent 3. {\bf Average Pressure one molecule exerts on one side}: Since pressure is force/area,
the pressure is just $(m v_x^2/L)/L^2$ or $m v_x^2/L^3$. As the molecules bounce off each other they
will have collisions at different angles and their speeds will change. At any time the speeds of the
molecules in a gas will have a range of values. To find the average pressure, we should take the
average value of $v_x^2$ which we write as $\bar{v_x^2}$. So the average pressure that one molecule
exerts on a side (in the y-z plane) is\\
\begin{equation}
{{m \bar{v_x^2}} \over L^3}
\end{equation}
\noindent 4. {\bf Average Pressure for N molecules}: If there are $N$ molecules in the gas, then
the average pressure on a side is:
\begin{equation}
P = {{N m \bar{v_x^2}} \over L^3}
\end{equation}
\noindent Similarly, the average pressure on a side in the x-y plane is
$P = N m \bar{v_z^2}/L^3$ and on a side in the x-z plane is $P=N m \bar{v_y^2}/L^3$.
If a molecule is equally likely to move in any direction, then $\bar{v_x^2} =
\bar{v_y^2} = \bar{v_z^2}$. Since the magnitude of the velocity is
$v^2 = v_x^2 + v_y^2 + v_z^2$, we have $\bar{v_x^2} = \bar{v^2}/3$. Substituting
this into the equation above we have:
\begin{equation}
P = {{N m \bar{v^2}} \over {3 L^3}}
\end{equation}
\noindent However, $V = L^3$, so the equation can be written
\begin{equation}
P = {N \over V}({{m \bar{v^2}} \over 3})
\end{equation}
\bigskip
Wow, the model is consistent with the data. The equation states that
pressure is proportional to $N$ and inversely proportional to $V$ as
demonstrated by experiment. The terms on the right side deserve some
comment.\\
\noindent {\it Why $1/L^3$}? Pressure is Force/Area, and Area goes as $L^2$.
The other factor of $L$ in the denominator is because the larger $L$ is,
the molecules will have fewer collisions/sec with the sides. Hence, the pressure
is proportional to $1/L^2$ times $1/L$.\\
\noindent {\it Why $\bar{v^2}$}?: One power of $v$ is because momentum is
proportional to velocity. The other power of $v$ is because the faster the
molecule travels the more often it will hit the side. Hence, pressure is
proportional to the average of $v^2$.\\
\noindent {\it Why $N$}? The pressure is proportional to $N$ because the
number of collisions with the wall per sec is proportional to the number of
molecules in the container.\\
In addition to understanding the data, the above equation gives us an
expression relating temperature and the average kinetic energy of
a molecule in the gas. The Kinetic Theory gives
$(PV/N) = m \bar{v^2} /3$, and experiment yields $(PV/N) = kT$. Equating
the right sides of these equations gives:
\begin{equation}
kT = {{m \bar{v^2}} \over 3}
\end{equation}
\noindent Remembering that the kinetic energy equals $m v^2 /2$, we can write
the right side in terms of the average kinetic energy of a gas molecule:
\begin{equation}
kT = {2 \over 3} \; \overline{K.E.}
\end{equation}
\noindent where $\overline{K.E}$ is the average kinetic energy of a molecule in
the gas.\\
\noindent {\bf Temperature Revisited}: The temperature $T$ used in these equations was
defined using an ideal-gas constant-volume thermometer.
The kinetic theory model gives $PV = (2/3) N \overline{K.E.}$. Thus, if $V$ is held constant
then $P$ is proportional to the average kinetic energy of a molecule. The constant
volume thermometer defines $T$ proportional to $P$. So, using an ideal-gas
constant-volume thermometer is essentially defining temperature to be proportional
to the average $K.E.$ per molecule. With this definition of $T$, it doesn't matter
which type of gas is used. This explains the universal nature of an ideal-gas
thermometer.
\noindent {\it Note}: For the ideal gas, it is OK to think of temperature as the average
kinetic energy per molecule. However, this is not true in general. We will discuss a broader
concept of temperature later. Nonetheless, $kT$ has units of energy, and temperature
does not require a new unit. It can always be coupled with $k$ and hence the
units of energy.\\
\noindent {\bf R.M.S. Velocity}: From our model, we can calculate the average speed of a gas molecule. The easiest
"speed" to calculate is the root-mean-square speed or RMS speed. The RMS speed
is the square-root of the average speed-squared: $v_{rms} \equiv \sqrt{\bar{v^2}}$.
From our equation above,
\begin{equation}
v_{rms} = \sqrt{ {{3kT} \over m} } = \sqrt{ {{3RT} \over M}}
\end{equation}
\noindent where $M$ is the {\it molar mass} of the gas. Remember {\bf T must
be in Kelvin}. Let's calculate
$v_{rms}$ for $N_2$ at room temperature of $T = 293^\circ$K.
$v_{rms} = \sqrt{3(8.34)293/(0.028 \; kg)} \approx 510$m/s. The air molecules
are hitting you with an average speed of $510$m/s. Our assumption that
the kinetic energy is much greater than $mgL$ is OK.
\bigskip
\centerline{\bf Internal Energy of an Ideal Gas}
\bigskip
From the kinetic theory model we can obtain an expression for the total
internal energy $U$ of an ideal gas. The simplest form would be the
sum of the kinetic energies of the molecules. If the gas contains $N$
molecules:
\begin{equation}
U = N {{m \bar{v^2}} \over 2}
\end{equation}
\noindent Using $m \bar{v^2} = 3 kT$, $U$ can be expressed in terms of $N$ and $T$:
\begin{equation}
U = {3 \over 2} NkT = {3 \over 2} nRT
\end{equation}
\noindent This is result is due to our definition of $T$ and the
properties of an ideal gas. If the number of molecules are
fixed, the total energy $U$ only depends on the temperature $T$. Another
way of writing the equation is to use $P$ and $V$. Since $NkT = PV$,
we have $U = (3/2) PV$. However, it is nicer to recognize that $U$ depends
on only one parameter $T$.\\
\noindent {\bf Heat Capacity of an Ideal Gas at Constant Volume}:
Heat capacity relates the temperature increase to the energy transfered
to the substance. For gases
it is convenient to consider the {\bf heat capacity per mole} of the
gas. If the volume is held fixed during the process, no work is
done by the gas, so $Q = \Delta U$, so we have
\begin{equation}
Q = \Delta U = {3 \over 2} n R \Delta T
\end{equation}
\noindent So, $c_v = (3/2) R = 12.5$ Joules/$^\circ$C per mole of the gas.
The measured values of $c_v$ for the nobel gases (e.g. $He$, $Ne$, $A$,
and $Kr$) are very close to this value for all temperatures where the
elements are gases. The measured values of $c_v$ per mole for other gases are
also close to $12.5$ Joules/$^\circ$C {\it for low temperatures}.\\
Since the kinetic theory explains many properties of gases, we
are now ready to examine some thermodynamic processes
for ideal gases. We first consider monatomic gases, since they are the
simplest.
\bigskip
\centerline{\bf Processes for a Monatomic Ideal Gas}
\bigskip
By a "process" we mean that some of the parameters of a gas ($V$, $P$, and/or $T$)
are varied as the gas changes from one state to another. For most of
the processes we consider, $N$ will not change. Many kinds of
processes are possible. Here we will consider
some simple ones. We will mainly analyze processes that are "reversible"
or equivalently "quasi-static":\\
\noindent {\bf Reversible Processes}: A slow or "quasi-static" process
in which the system is always in equilibrium.\\
\noindent In a process, the internal energy $U$ can change. This
change is due to work done {\bf by} the gas $W$, and energy transfered due
to temperature differences $Q$. Processes can be graphically represented
by a plot in the $P-V$ plane. The pressure $P$ is plotted on the vertical
axis, and the volume $V$ is plotted on the horizontal axis. The work done
by the gas is easily determined in the $P-V$ plane.\\
\noindent {\bf Work done by a gas}\\
Work is force times displacement. The work done {\bf by} a gas is the
force it exerts times the distance the force acts. However,
force is pressure times area. Thus, the work done by the gas is
the pressure $P$ times area times the distance the area moves.
Area times distance is the change in the volume of the gas, $\Delta V$.
So if the volume of a gas changes by an amount $\Delta V$, the gas
does work $W$ equal to $P \Delta V$:
\begin{equation}
W_{by \; gas} = P \Delta V
\end{equation}
\noindent If the pressure changes as the volume changes, then one needs to
integrate $P dV$:
\begin{equation}
W_{by} = \int P dV
\end{equation}
\noindent In the $P-V$ plane, {\bf the work is just the area under the curve}
representing the process.
Using this equation for $W_{by}$, and the relationship $\Delta U = (3/2) n R \Delta T$,
we can determine these two quantites and $Q$ for any process involving ideal
monatomic gases.\\
\noindent {\bf Isometric Process}: An isometric process is a process in which the
{\bf volume is constant}, $\Delta V = 0$. $P$ and $T$ will change. In this case, $W = 0$.
$Q = \Delta U = (3/2) nR \Delta T$. If the initial pressure is $P_i$ and the final
pressure is $P_f$, $Q = (3/2) nR(T_f - T_i) = (3/2) (P_f V - P_i V)$. So
$Q = (3/2) V (P_f - P_i)$. The chart below summarizes our results:
\begin{center}
\begin{tabular}{c|c|c}
$W_{by}$ & $Q$ & $\Delta U$ \\
\hline
$0$ & ${3 \over 2} nR \Delta T$ or ${3 \over 2} V(P_f - P_i)$ & ${3 \over 2} nR \Delta T$\\
\end{tabular}
\end{center}
\noindent {\bf Isobaric Process}: An isobaric process is one in which the {\bf pressure is
held constant}. $V$ and $T$ will change. So the work done by the gas is: $W_{by} = P \Delta V$, or $W_{by} = P(V_f - V_i)$.
Since $PV=nRT$, the work done by the gas can also be written in terms of the initial and final
temperatures: $W_{by} = nR(T_f - T_i) = nR \Delta T$ for the isobaric process. The
change in internal energy $\Delta U$ is determined by the change in temperature,
so $\Delta U = (3/2) nR \Delta T$. $Q$ for the isobaric process can be determined from:
\begin{equation}
\Delta U = Q - W_{by}
\end{equation}
\noindent or
\begin{eqnarray*}
Q & = & \Delta U + W_{by}\\
& = & {3 \over 2} nR \Delta T + nR \Delta T\\
Q & = & {5 \over 2} nR \Delta T
\end{eqnarray*}
\noindent The factor $(5/2) nR$ is a {\bf heat capacity for a constant pressure
process}. The heat capacity at constant pressure per mole of a monatomic gas
is $c_p = (5/2) R = 20.8$J/$^\circ$K. Note: $c_p > c_v$. It takes more
heat to raise the temperature for an isobaric process than an isometric process.
This is because in an isobaric process some energy is used by the gas
as it does work expanding. The experimental data for
$c_p$ agree well with this value for the nobel gases at all temperatures.\\
\begin{center}
\begin{tabular}{c|c|c}
$W_{by}$ & $Q$ & $\Delta U$ \\
\hline
$P(V_f-V_i)$ or $nR\Delta T$ & ${5 \over 2} nR \Delta T$ & ${3 \over 2} nR \Delta T$\\
\end{tabular}
\end{center}
\noindent {\bf Isothermal Process}: An isothermal process is one in which the
{\bf temperature is held constant}. $P$ and $V$ will change. Since $U$ is
proportional to $T$, if $\Delta T =0$ then $\Delta U = 0$. This means that
\begin{equation}
Q = W_{by}
\end{equation}
\noindent Consider an isothermal process in which the volume increases from
$V_i$ to $V_f$. From the ideal gas law: $P = nRT/V$, so the
pressure decreases as $V$ increases if $T$ is constant. Since the pressure
changes, to find the work done by the gas, we need to integrate:
\begin{equation}
W_{by} = \int_{V_i}^{V_f} P dV
\end{equation}
\noindent Since $P = nRT/V$ during the isothermal volume expansion,
\begin{equation}
W_{by} = \int_{V_i}^{V_f} {{nRT} \over V} dV
\end{equation}
\noindent the $nRT$ factors out of the integral and we have
\begin{equation}
W_{by} = nRT \int_{V_i}^{V_f} {{ dV} \over V}
\end{equation}
\noindent The integral is easily solved to give
\begin{equation}
W_{by} = nRT ln({{V_f} \over {V_i}})
\end{equation}
\noindent and so $Q = W_{by} = nRT ln(V_f / V_i)$.\\
\begin{center}
\begin{tabular}{c|c|c}
$W_{by}$ & $Q$ & $\Delta U$ \\
\hline
$ nRT \; ln(V_f / V_i)$ & $nRT \; ln(V_f / V_i)$ & $0$\\
\end{tabular}
\end{center}
\noindent {\bf Adiabatic Process}: An adiabatic process is one
in which $Q=0$. In this case, $\Delta U = - W_{by}$. Suppose a gas
starts out with a initial volume $V_i$ and pressure $P_i$. If the
volume is then increased slowly to a value $V > V_i$, what is the pressure?
Since the temperature will change in this process, $P$ does not equal
$V_i P_i / V$. That is, although $PV = nRT$ at any moment of the process,
$T$ is changing and so is the product $PV$. To determine how $P$ changes
with $V$, we start with:
\begin{equation}
\Delta U = - W_{by} = - P \Delta V
\end{equation}
\noindent Since $U = (3/2)nRT=(3/2)PV$ for a monatomic gas, we have
\begin{equation}
{3 \over 2} \Delta (PV) = - P \Delta V
\end{equation}
\noindent Using the product rule for differentials
\begin{equation}
{3 \over 2} (P \Delta V + V \Delta P ) = -P \Delta V
\end{equation}
\noindent Rearrainging terms:
\begin{equation}
{3 \over 2} {{\Delta P} \over P} = - {5 \over 2} {{\Delta V} \over V}
\end{equation}
\noindent Taking the limit at $\Delta P \rightarrow 0$ and $\Delta V \rightarrow 0$
gives
\begin{equation}
{3 \over 2} \int {{d P} \over P} = - {5 \over 2} \int {{dV} \over V}
\end{equation}
\noindent After integrating, the equation becomes
\begin{equation}
{3 \over 2} lnP + {5 \over 2} lnV = ln(P^{3/2} V^{5/2}) = Const
\end{equation}
\noindent If we define the parameter $\gamma \equiv 5/3$, then during the adiabatic process
\begin{equation}
P V^\gamma = P_i V_i^\gamma
\end{equation}
\noindent The work is found by integrating $\int P dV$:
\begin{equation}
W_{by} = \int_{V_i}^{V_f} {{P_i V_i^\gamma} \over {V^\gamma}} dV
\end{equation}
\noindent which, after some algebra, is
\begin{equation}
W_{by} = {{P_i V_i} \over {\gamma - 1}} (1 - ({V_i \over V_f})^{\gamma - 1})
\end{equation}
This work done by the gas will decrease the internal energy. It is interesting to
calculate the change in temperature in adiabatic process. Since $PV^\gamma = const$,
\begin{equation}
P_i V_i^\gamma = P_f V_f^\gamma
\end{equation}
\noindent Using $P = (nRT)/V$ we have
\begin{equation}
T_i V_i^{\gamma -1} = T_f V_f^{\gamma - 1}
\end{equation}
\noindent or
\begin{equation}
T_f = T_i ({V_i \over V_F})^{\gamma - 1}
\end{equation}
For a monatomic gas, $\gamma = 5/3$, so $\gamma - 1 = 2/3$.
If the volume decreases by a factor of $2$, then the final
temperature in Kelvin increases by a factor of $2^{2/3} = 1.59$.
A volume change by $5$ gives $5^{2/3} = 2.92$, and a ten-fold
change gives $10^{2/3} = 4.64$. The work done "on" the gas
increases the internal energy. That {\it the temperature is correctly
predicted by our model supports the hypothesis} of the first
law of thermodynamics: that {\it mechanical work can be directly
converted into internal energy}.
\bigskip
\noindent {\bf Closed Cycle Processes}
\bigskip
By a closed cycle process we mean a process which ends at the same state
that it started. If the final state is the same as the initial state,
then $U_f = U_i$ and $\Delta U = 0$ for the closed cycle. However, the
work done in the closed cycle will not be zero, but rather the area
enclosed by the process in the $P-V$ plane. We will demonstrate these
ideas with an example, which uses specific values for the states.\\
\noindent Our system will be {\bf 2 moles of $He$}, which is a monatomic gas.\\
\noindent {\it State a}: Volume = $2$ liters; Pressure = $400 \times 10^3$Pa.\\
\noindent {\it State b}: Volume = $4$ liters; Pressure = $200 \times 10^3$Pa.\\
\noindent {\it State c}: Volume = $2$ liters; Pressure = $200 \times 10^3$Pa.\\
\noindent Note: With this information, we can calculate the temperatures of
the different states. Since $T = (PV)/(nR)$, we have $T_a = (400 \times 10^3)
(2 \times 10^{-3})/((2)(8.314)) = 48^\circ$K. Here we have used the
conversion $1 \; liter = 10^{-3}m^3$. Similarily, $T_b = 48^\circ$K and
$T_c = 24^\circ$K. The gas changes from state $a \rightarrow b \rightarrow c
\rightarrow a$. We will calculate $\Delta U$, $Q$, and $W_{by}$ for each leg
of the process. The results are:
\begin{center}
\begin{tabular}{c|c|c|c}
Process & $W_{by}$ (Joules) & $Q$ (Joules) & $\Delta U$ (Joules)\\
\hline
$a \rightarrow b$ (isothermal) & $553$ & $553$ & $0$ \\
\hline
$b \rightarrow c$ (isobaric) & $-400$ & $-998$ & $-598$ \\
\hline
$c \rightarrow a$ (isometric) & $0$ & $598$ & $598$\\
\hline
Whole Cycle & $153$ & $153$ & $0$ \\
\end{tabular}
\end{center}
\begin{figure}
\includegraphics[width=14cm]{fig1323b.png}
\end{figure}
We could repeat this cycle over and over again. What is happening per
cycle? $153$ Joules of energy is added to the system, and the system
does $153$ Joules of work. This is what an engine is supposed to do.
A closed clockwise cycle will function as an engine. A closed
counter-clockwise cycle will function as a refrigerator. For our
engine cycle, a net $Q_{in} = 553 + 598 = 1151$ Joules is
transfered into the gas, and a net $Q_{out} = 998$ Joules is transfered
out of the gas. The difference $W_{by} = Q_{in} - Q_{out}$ is the work
done by the gas.
What we would like to have happen is for most of the $Q_{in}$ to be
converted to work, and little transfered out: $Q_{out}$. We define the
{\bf efficiency}, $\epsilon$, of a "heat-engine" to be:
\begin{equation}
\epsilon \equiv {{W_{by}} \over {Q_{in}}}
\end{equation}
\noindent This equation can also be written as
\begin{equation}
\epsilon \equiv 1 - {{Q_{out}} \over {Q_{in}}}
\end{equation}
\noindent since $W_{by} = Q_{in} - Q_{out}$. For our example, $\epsilon = 153/1151
\approx 0.13$, or a $13 \%$ efficiency. How can we make the
efficiency large? What kind of cycle is the best? The answers to these important
engineering questions require some thought and lead to some new ideas:
the second law of thermodynamics and entropy.
We will tackle these problems for the monatomic ideal gas,
since we have a nice working model and it is a simple system. Then
we will generalize our results.
\bigskip
\centerline{\bf Entropy}
\bigskip
When $\Delta U$ was summed up for a closed cycle the net change was zero.
This is because $U$ is a state function of the system. It only depends on
the state itself, and not how the state was formed. We represent the
sum of $\Delta U$ for a closed cycle as the integral $\oint dU$. Since $U$ is
a state function $\oint dU = 0$. Examples of other state functions are
$V$, $P$, and $T$. $Q$ and $W$ are not state functions. $\oint dQ$
is not zero as demonstrated by our last example. $\oint dW = \oint p dV$ is
also not equal to zero in general, but is the area enclosed by the path
of the cyclic process.
We would like to have a state function involving $dQ$. Equivantly, we need to
find an expression with $dQ$ such that when integrated over a complete cycle
gives zero. Then we will have something absolute
about $Q$, which will help us determine the maximum energy efficiency.
We shall show that $\oint dQ/T$ is zero for any closed cycle for
an ideal gas. First we will do a specific closed cycle example and calculate
$\int dQ/T$ to show that $\oint dQ/T = 0$. Then we will prove
it for the general case of an ideal gas.
Consider a closed cycle process for an ideal monatomic gas
that is bounded by two isothermal processes and two isometric processes.
We will calculate the following table in lecture for the $4$ legs:
\begin{center}
\begin{tabular}{c|c|c|c|c}
Process & W & Q & $\Delta U$ & $\int dQ/T$ \\
\hline
$a \rightarrow b$ & 0 & $(3/2)nR(T_H-T_C)$ & $(3/2)nR(T_H-T_C)$ & $(3/2)nR \; ln(T_H/T_C)$\\
\hline
$b \rightarrow c$ & $nRT_H \; ln(V_2/V_1)$ & $nRT_H \; ln(V_2/V_1)$ & 0 & $nR \; ln(V_2/V_1)$\\
\hline
$c \rightarrow d$ & 0 & $-(3/2)nR(T_H-T_C)$ & $-(3/2)nR(T_H-T_C)$ & $(3/2)nR \; ln(T_C/T_H)$\\
\hline
$d \rightarrow a$ & $nRT_C \; ln(V_1/V_2)$ & $nRT_C \; ln(V_1/V_2)$ & 0 & $nR \; ln(V_1/V_2)$\\
\hline
\hline
cycle & $nR(T_H-T_C)$ & $nR(T_H-T_C)$ & 0 & 0 \\
& $ln(V_2/V_1)$ & $ln(V_2/V_1)$ & & \\
\hline
\end{tabular}
\end{center}
\noindent where $a \rightarrow b$ and $c \rightarrow d$ are
isometric (constant $V$) processes, and $b \rightarrow c$ and
$d \rightarrow a$ are isothermal (constant $T$) processes.
Note that $\oint dQ/T = 0$ for any value of $V_1$, $V_2$,
$T_H$, or $T_C$. Is this true for any closed cycle?
The answer is yes. We can prove it for a monatomic ideal gas:
\begin{equation}
0 = \oint {{dV} \over V}
\end{equation}
\noindent For any closed cycle, the above integral is just
$ln{V_i/V_i} = ln(1) = 0$. Using the equation for an ideal
gas,
\begin{equation}
0 = \oint {{p dV} \over {nRT}}
\end{equation}
\noindent However, $p dV$ equals $dQ - dU$ from the
first law of thermodynamics.
\begin{equation}
0 = \oint {{dQ - dU} \over {nRT}}
\end{equation}
\noindent For a monatomic gas, $U = (3/2) nRT$. With
this substitution we have
\begin{equation}
0 = \oint {{dQ} \over {nRT}} - {2 \over 3} \oint {{dU} \over U}
\end{equation}
\noindent The second integral is zero, since $\oint dU/U = ln(U_i/U_i) = ln(1)
= 0$. Canceling out the $nR$
\begin{equation}
0 = \oint {{dQ} \over T}
\end{equation}
\noindent Although this derivation was for the monatomic gas, the same line
of reasoning will work for any ideal gas. The only difference in the
derivation will be that $U = f(T)$, that is $U$ is only a function of $T$. This
is the key property of the system. If $U$ only depends on $T$, $U(T)$, then
then the integral of $\int dU/T = \int (U'(T)/T) dT$, where $U'(T) = dU/dT$.
This integral will be zero over a closed cycle, since the integrand only
depends on T.
The above result has profound consequences. It means that if $\Delta Q/T$ is
summed up from one state $a$ to another state $b$, the integral
$\int_a^b dQ/T$ is the same for any process that changes the system from
$a \rightarrow b$. The quantity $\int_a^b dQ/T$ only depends on the
initial state $a$ and final state $b$. This integral we call the
{\bf entropy difference} between the two states. Entropy is given
the symbol $S$:
\begin{equation}
S(b) - S(a) = \int_a^b {{dQ} \over T}
\end{equation}
Entropy is a "state function" of a system, just like $U$, $V$, $T$, and $P$.
Before we try and identify what $S$ depends on for a gas, we should
remind ourselves on how it was discovered using a thermodynamics approach.
The important step in making $\oint dQ/T = 0$, was $\oint dU/T =0$. This
was made possible by our definition of temperature, which ended up being
(for an ideal gas) proportional to the average kinetic energy per molecule and
hence proportional to $U$. We can turn the argument around, and say that
{\bf it is possible to define temperature $T$ such that $\oint dQ/T = 0$ for any
cyclic process}. This definition of temperature is independent of any
thermometer, and this statement recognizes the existance of the entropy
state function.
\bigskip
\centerline{\bf Other Topics}
\bigskip
\noindent {\bf Diatomic and other ideal gases}
\bigskip
For all gases, the equation of state $PV=nRT$ is valid in the limit
of an ideal gas (i.e. in the limit as the density goes to zero).
The next most complicated gas after a monatomic gas is a diatomic gas,
whose molecule has two atoms, e.g. $H_2$, $O_2$, $N_2$. The energy
$U$ for $n$ moles of a monatomic gas was found to be $(3/2)nRT$.
We can determine how the energy of a diatomic gas depends on temperature
by measuring the heat capacity (per mole) for a constant volume process.
In a constant volume process, $\Delta U = Q = c_v \Delta T$. A graph
of $c_v$ for $H_2$ as a function of temperature will be shown in lecture.
At low temperatures, $c_v = (3/2) R$ which is the same as the
monatomic gas. However, as the temperature increases, $c_v$ jumps
up to $(5/2) R$ at around $150^\circ$K. After a further increase
in $T$, $c_v$ jumps up to $(7/2)R$ at around $3000^\circ$K.
This behavour is strange for two reasons: 1) the sudden increase (or jump) of
$c_v$ as a function of $T$, and 2) the first increase from $(3/2)R$ to $(5/2)R$.
The understanding of these features required "new" physics, quantum mechanics,
which we cover in our modern physics course (Phy235).
Qualitatively, we understand the sudden increase in $U$ for a diatomic
molecule as rotational energy and vibrational energy. A diatomic molecule
can translate (like a monatomic gas) in 3 different directions, rotate
in 3 different directions and vibrate (two degrees of freedom). There is
a theorum, {\bf the equipartition theorum}, which states that for every
degree of freedom that contributes quadratically to the energy, the average
energy is $(RT)/2$ per mole. If only translation is allowed, $U = (3/2) RT$,
since there are three directions, each of which contribute $(RT)/2$. Rotations
should add $3 \times (1/2)RT = (3/2)RT$, but they only seem to add $RT$ to
the energy/mole. These descrepancies are resolved using quantum mechanics.
For our applications here, we just need to know $c_v$ for the conditions of
our gas. Then,
\begin{equation}
\Delta U = n c_v \Delta T
\end{equation}
\noindent At room temperature, $300^\circ$K, $N_2$, $O_2$ and $H_2$ have
$c_v = (5/2) R$. The heat capacity at constant pressure is found in terms
of $c_v$ as:
\begin{equation}
c_p = c_v + R
\end{equation}
\noindent This can be shown as follows:
\begin{eqnarray*}
\Delta U & = & Q - p \Delta V\\
n c_v \Delta T & = & Q - p \Delta V
\end{eqnarray*}
\noindent Now for a constant pressure process, $p \Delta V = \Delta (pV)$.
So we have
\begin{eqnarray*}
n c_v \Delta T & = & Q - \Delta (pV)\\
& = & Q - \Delta (nRT)\\
n (c_v + R) \Delta T & = & Q
\end{eqnarray*}
\noindent which gives $c_p = c_v + R$.\\
The main difference between ideal gases and ideal monatomic gases (in our
treatment) is that $c_v$ and hence $c_p$ depend on energy and might be different
than the monatomic values of $(3/2)R$ and $(5/2)R$ respectively. This results
in $\gamma$ being different also, since $\gamma \equiv c_p/c_v$.
\bigskip
\noindent {\bf Efficiency and the Carnot Cycle}
\bigskip
Using $\oint dQ/T = 0$, we can determine an upper limit for
efficiency of an engine. Consider a closed cycle (for an ideal gas)
that acts like an engine. That is, the path is clockwise in the $P-V$ plane.
Divide the path into small segments. For each segment, energy will either
enter the system as heat ($\Delta Q(in)$ or leave the system as heat
$\Delta Q(out)$. $\Delta Q$ could also be zero. Label the segments
for which energy enters as $i$, and those for which energy leaves as $j$,
where both $i$ and $j$ are summed. From $\oint dQ/T = 0$ we have:
\begin{equation}
\sum_i {{Q_i(in)} \over T_i} - \sum_j {{Q_j(out)} \over T_j} = 0
\end{equation}
\noindent where we have taken $Q_j(out)$ to be the absolute value of
$Q_j(out)$. The equation can be re-written
\begin{equation}
\sum_i {{Q_i(in)} \over T_i} = \sum_j {{Q_j(out)} \over T_j}
\end{equation}
\noindent Let $T_{hottest}$ be the largest value of the temperature during
the cycle, and let $T_{coldest}$ be the lowest value of the temperature
during the cycle. Then we have:
\begin{equation}
\sum_i {{Q_i(in)} \over T_i} \geq {1 \over T_{hottest}} \sum_i Q_i(in)
\end{equation}
\noindent or
\begin{equation}
\sum_i {{Q_i(in)} \over T_i} \geq {Q_{in} \over T_{hottest}}
\end{equation}
\noindent where $Q_{in}$ is the sum of energy entering the gas via
heat. Similarly,
\begin{equation}
\sum_i {{Q_j(out)} \over T_j} \leq {1 \over T_{coldest}} \sum_i Q_j(out)
\end{equation}
\noindent or
\begin{equation}
\sum_i {{Q_j(out)} \over T_j} \leq {Q_{out} \over T_{coldest}}
\end{equation}
\noindent where $Q_{out}$ is the sum of energy leaving the gas via heat.
Putting these equations together gives
\begin{equation}
{{Q_{in}} \over T_{hottest}} \leq
\sum {{Q_i(in)} \over T_i} =
\sum {{Q_j(out)} \over T_j} \leq
{{Q_{out}} \over T_{coldest}}
\end{equation}
\noindent This inequality can be re-written
\begin{equation}
{{Q_{out}} \over {Q_{in}}} \geq {T_{coldest} \over T_{hottest}}
\end{equation}
\noindent Since the efficiency $\epsilon = 1 - Q_{out}/Q_{in}$, we have
\begin{equation}
\epsilon \leq 1 - {T_{coldest} \over T_{hottest}}
\end{equation}
From these inequalities, one can see how to obtain the maximum efficiency.
The equal sign will hold if all the energy transfered in ($Q_{in}$) is done so
at the hottest temperature, and all the energy transfered out ($Q_{out}$) is
done at the lowest temperature. The cycle that accomplishes this is
a cycle that has two isothermal processes and two adiabatic processes.
Such a cycle is called a {\bf Carnot Cycle}. For a Carnot cycle, the
efficiency $\epsilon_{Carnot}$ is
\begin{equation}
\epsilon_{Carnot} = 1 - {{T_{cold}} \over {T_{hot}}}
\end{equation}
\noindent where $T_{hot}$ is the temperature of the hot isothermal and $T_{cold}$
is the temperature of the cold isothermal. $Q = 0$ on the adiabatic legs
of the cycle. The Carnot cycle is the most efficient engine operating
between the hottest and coldest temperatures.
\bigskip
\noindent {\bf The meaning of entropy}
\bigskip
Previously we derived an expression for the difference in entropy between two
states:
\begin{equation}
S(b) - S(a) = \int_a^b {{dQ} \over T}
\end{equation}
\noindent Note: the integral gives the same value for any path taken from
state $a$ to $b$. Although we derived this for an ideal gas, it holds for
any substance. In evaluating the right side, usually $T$ changes as
energy is added or subtracted from the system, so one usually needs to
integrate. Since $\Delta Q$ is related to $\Delta T$ via the heat capacity
$C$: $dQ = C dT$, the entropy difference equation can be written as
\begin{equation}
S(b) - S(a) = \int_a^b {{C(T) dT} \over T}
\end{equation}
\noindent where $C(T)$ is the heat capacity and may depend on the temperature
$T$. If the system changes from $a \rightarrow b$ such that $C$ is a constant,
then $C(T) = C$ factors out of the integral:
\begin{equation}
S(b) - S(a) = C \int_a^b {{dT} \over T} = C \; ln({T_b \over T_a})
\end{equation}
\noindent For solids and liquids, $C = m c_v$ where $c_v$ is the specific
heat capacity. For gases $C=n c_v$ or $C=n c_p$ depending if the process
is isometric or isobaric.
Since entropy is heat/temperature, is there a simple connection between
the direction of energy flow and entropy change? Yes there is. Consider
two systems, one at temperature $T_H$ (hot) and the other at temperature $T_C$ (cold),
where $T_H > T_C$. If a small amount of energy $\Delta Q$ flows from the
hot system to the colder one, the hot system loses entropy equal to
$\Delta Q / T_H$. The cold system gains entropy
equal to $\Delta Q / T_C$. The net entropy change of the two systems
together is
\begin{equation}
\Delta S_{net} = {{\Delta Q} \over T_C} - {{\Delta Q} \over T_H} > 0
\end{equation}
\noindent The sum is greater than zero since $T_C < T_H$. If on the
other hand, energy were to flow from the cold system to the hot one,
the net entropy change of the two systems together would be
\begin{equation}
\Delta S_{net} = {{\Delta Q} \over T_H} - {{\Delta Q} \over T_C} < 0
\end{equation}
\noindent Thus, it appears that when energy is transfered in the form of heat,
the direction of the transfer is such that the net entropy of the two
(or more) systems increases. Entropy somehow is related to the direction of
energy transfer. The second law of thermodynamics states this more precisely:
\bigskip
\noindent {\bf Second Law of Thermodynamics}: When an isolated system undergoes
a change, the entropy of the system can only increase or remain the same.
\bigskip
We know that entropy is a state function and that it is somehow related to the
direction of thermodynamic change, but specifically what is it a measure of?
Since we have a working model of a monatomic gas, lets determine how the
entropy depends on its macroscopic quantities. Suppose we have $n$ moles of
a monatomic gas. Let state $a$ have a pressure $P_a$, a volume $V_a$, and a
temperature $T_a = (P_a V_a)/(nR)$. Let state $b$ have a pressure $P_b$, a
volume $V_b$, and a temperature $T_b = (P_b V_b)/(nR)$. Let the process from
$a \rightarrow b$ be in two parts: an isothermal followed by an isometric.
For the isothermal, the temperature is constant at $T_a$:
\begin{equation}
\Delta S_{isothermal} = \int {{dQ} \over T} = {Q \over T_a}
\end{equation}
\noindent since $T_a$ factors out of the integral. $Q$ is the net heat,
and is equal to the work $W$ done since $\Delta U = 0$ for an isothermal
process. So we have
\begin{eqnarray*}
\Delta S_{isothermal} & = & {W \over T_a}\\
& = & {1 \over T_a} \int P dV\\
& = & {1 \over T_a} \int {{nRT_a dV} \over V}\\
\Delta S_{isothermal} & = & nR \; ln{V_b \over V_a}
\end{eqnarray*}
\noindent For the isometric leg, $\Delta Q = (3/2) nR \Delta T$.
The entropy change for this section is
\begin{eqnarray*}
\Delta S_{isometric} & = & \int {{3nR} \over 2} {{dT} \over T}\\
& = & {{3nR} \over 2} ln{T_b \over T_a}
\end{eqnarray*}
\noindent Adding the two pieces together, we have for the entropy difference
between the states $a$ and $b$ for an ideal monatomic gas:
\begin{equation}
S(b) - S(a) = nR(ln{V_b \over V_a} + {3 \over 2} ln{T_b \over T_a})
\end{equation}
\noindent Since the temperature is proportional to the total internal energy
$U$, $(T_b / T_a) = (U_b / U_a)$:
\begin{equation}
S(b) - S(a) = nR(ln{V_b \over V_a} + {3 \over 2} ln{U_b \over U_a})
\end{equation}
\noindent in terms of the energy $U$. We can examine this equation to see
what the {\bf entropy} depends on for a {\bf monatomic gas}:\\
\noindent {\it Energy dependence}: The entropy is proportional to the
log of the energy $U$ of the gas (if $n$ and $V$ are constant). Entropy
increases with energy, via the log.\\
\noindent {\it Volume dependence}: The entropy is proportional to the
log of the volume $V$ of the gas (if $n$ and $U$ are constant).\\
\noindent The energy dependence gives us a relationship between entropy,
energy and temperature. If we differentiate $S$ with respect to $U$,
\begin{eqnarray*}
{{\partial S} \over {\partial U}} & = & {{3nR} \over {2U}}\\
{{\partial S} \over {\partial U}} & = & {1 \over T}
\end{eqnarray*}
\noindent since $U = (3/2) nR T$ for a monatomic gas. As we mention
in the last paragraph (without proof), this relationship can be derived
using statistical physics and can be used to define temperature.\\
The volume dependence is interesting. If the volume of the gas gets
larger, with no change in the energy, the entropy increases. Consider a
container insulated from the "outside" with two chambers inside. Suppose
the gas can travel from one chamber to the other only through a valve.
Suppose the gas starts in one chamber with the valve closed, and the other
chamber is a vacuum. If the valve is opened, the gas can move to the other
chamber and after a while both chambers will be occupied by the gas.
The energy of the gas did not change $\Delta U = 0$. The temperature
didn't change. However, the entropy increased by an amount $nR \; ln2$.
For a monatomic gas, entropy has something to do with the volume that
the gas can move around in. The true meaning of entropy was discovered by
Ludwig Boltzmann (1844-1906).
He realized that entropy is related to the number of possible configurations
of a system. He started a new field of physics called {\bf Statistical Physics},
and revolutionized our treatment of thermodynamics. Statistical physics is a
course of its own and usually taught in the senior year (Phy407).
Boltzmann defined a function $\Omega (U)$, which equals the number of
different ways that a system can arrange itself with total energy
$U$. Entropy $S$ is related to $\Omega$ as
\begin{equation}
S = k \; ln(\Omega)
\end{equation}
\noindent where $k$ is Boltzmann's constant. In statistical mechanics,
the temperature of a system is defined by
\begin{equation}
{1 \over T} \equiv {{\partial S} \over {\partial U}}
\end{equation}
\noindent This definition of temperature is universal. It works for
any system. For an ideal gas, this definition for temperature results
in temperature being proportional to $U$. Serendipitously one obtains the
same temperature using a constant-volume ideal-gas thermometer, for which
temperature was initially defined proportional to pressure and later
(using the kinetic theory) found to also be proportional to $U$.
\bigskip
\centerline{\bf Summary of Some Processes}
\bigskip
The simplist way to find $\Delta U$, $Q$, $W_{by}$, and $\Delta S$ for a reversible
process is to do the following. {\it First} determine $\Delta U$, which equals
$\Delta U = (3/2)nR \Delta T$ or $\Delta U = (3/2) \Delta (PV)$ for a monatomic gas.
For a diatomic gas at room temperature just replace the $(3/2)$ with $(5/2)$.
{\it Second} determine $W_{by}$ by finding the {\bf area under the curve in the
P-V plane}. $W_{by}$ is the same for any type of gas. {\it Third} find $Q$,
which is just $Q = \Delta U + W_{by}$. This is true for any type of gas.
{\it Finally} determine $\Delta S$ by carrying out the integral
$\Delta S = \int (dQ)/T$, where $T$ is the temperature (in Kelvin) at which the
incremental heat energy $dQ$ is transfered.
\bigskip
\noindent {\bf Isothermal Process}\\
For an isothermal process, $\Delta T=0$ and hence $\Delta U = 0$. This is true for both a
monatomic and diatomic gas. If the process is reversible, then $PV = nR T_0$ during the process,
where $T_0$ is the constant temperature. The work done by the gas is then $W_{by} = \int P dV = nRT_0 \int (dV)/V$.
The integral of $\int dV/V = ln(V)$, and taking the limits from the initial volume $V_i$ to the
final volume $V_f$ gives $W_{by} = nR T_0 ln(V_f/V_i)$. This is the work for both a monatomic and a
diatomic gas. Since $\Delta U = Q - W_{by}$, we have $Q = W_{by} = nR T_0 ln(V_f/V_i)$.
To find the change in entropy for an isothermal process, we need to do the integral
$\Delta S = \int (dQ)/T = (1/T_0) \int dQ = Q/T_0$, since the temperature is constant.
Using the expression for $Q$, we have $\Delta S = nR ln(V_f/V_i)$. We can summarize our results
in the following chart:
\begin{center}
\begin{tabular}{c|c|c|c|c}
Process & $W_{by}$ & $Q$ & $\Delta U$ & $\Delta S$ \\
\hline
Isothermal & $nRT_0 \; ln(V_f/V_i)$ & $nRT_0 \; ln(V_f/V_i)$ & $0$ & $nR \; ln(V_f/V_i)$ \\
\end{tabular}
\end{center}
\noindent for both monatomic and diatomic gases.
\bigskip
\noindent {\bf Isometric Process}\\
For an isometric process, $\Delta V = 0$, and hence $W_{by} = 0$. Since $W_{by} = 0$,
$\Delta U = Q$. The change in internal energy $\Delta U$ will depend on $\Delta T$.
For a monatomic gas, $\Delta U = (3/2)nR \Delta T$ and for a diatomic gas (at room temperature)
$\Delta U = (5/2) nR \Delta T$. We can also express $\Delta U$ in terms of the change in
the product of $PV$, since $nRT = PV$. In terms of $PV$, we have $\Delta U = (3/2) \Delta (PV)$
for a monatomic gas, and $\Delta U = (5/2) \Delta (PV)$ for a diatomic gas at room temperature.
To find the change in entropy for an isometric process, we need to carry out the integral
$\Delta S = \int (dQ)/T$. For a monatomic gas, $\Delta Q = (3/2)nR \Delta T$. Thus,
$\Delta S = (3/2)nR \int (dT)/T = (3/2)nR \; ln(T_f/T_i)$. For a diatomic gas,
$\Delta Q = (5/2) nR \Delta T$. In this case, the change in entropy is
$\Delta S = (5/2) nR \; ln(T_f/T_i)$. We can summarize these results in the following
charts:
\begin{center}
\begin{tabular}{c|c|c|c|c}
Process & $W_{by}$ & $Q$ & $\Delta U$ & $\Delta S$ \\
\hline
Isometric & $0$ & $ (3/2) \Delta (PV)$ & $(3/2) \Delta (PV)$ & $ (3/2)nR \; ln(T_f/T_i)$ \\
\end{tabular}
\end{center}
\noindent for a monatomic gas. For a diatomic gas, the results are
\begin{center}
\begin{tabular}{c|c|c|c|c}
Process & $W_{by}$ & $Q$ & $\Delta U$ & $\Delta S$ \\
\hline
Isometric & $0$ & $ (5/2) \Delta (PV)$ & $(5/2) \Delta (PV)$ & $ (5/2)nR \; ln(T_f/T_i)$ \\
\end{tabular}
\end{center}
\bigskip
\noindent {\bf Adiabatic Processes}\\
For an adiabatic process there is no "heat" exchange, $Q = 0$. Thus, $\Delta S = 0$, and
$W_{by} = - \Delta U$. For a monatomic gas, $\Delta U = (3/2) nR \Delta T = (3/2) \Delta (PV)$
For a diatomic gas, $\Delta U = (5/2) nR \Delta T = (5/2) \Delta (PV)$. These results can
be summarized in following charts:
\begin{center}
\begin{tabular}{c|c|c|c|c}
Process & $W_{by}$ & $Q$ & $\Delta U$ & $\Delta S$ \\
\hline
Adiabatic & $-(3/2) \Delta (PV)$ & $0$ & $(3/2) \Delta (PV)$ & $0$ \\
\end{tabular}
\end{center}
\noindent for a monatomic gas. For a diatomic gas, the results are
\begin{center}
\begin{tabular}{c|c|c|c|c}
Process & $W_{by}$ & $Q$ & $\Delta U$ & $\Delta S$ \\
\hline
Adiabatic & $-(5/2) \Delta (PV)$ & $0$ & $(5/2) \Delta (PV)$ & $0$ \\
\end{tabular}
\end{center}
\bigskip
\noindent {\bf Isobaric Processes}\\
For an isobaric process, all the quantities $\Delta U$, $Q$, $W_{by}$, and $\Delta S$
are non-zero. Let the pressure be $P_0$, and the initial and final volumes be
$V_i$ and $V_f$. Then $\Delta U = (3/2)nR \Delta T = (3/2) \Delta (PV) =
(3/2) P_0 (V_f - V_i)$. $W_{by} = P_0 (V_f - V_i)$. The "heat" energy transfer
is $Q = \Delta U + W_{by} = (5/2) P_0 (V_f - V_i)$, since $(3/2) + 1 = (5/2)$.
The change in entropy is $\Delta S = \int (dQ)/T = \int (5/2) nR (dT)/T$.
Evaluating the integral gives $\Delta S = (5/2) nR \; ln(T_f/T_i)$, which
equals $\Delta S = (5/2) nR \; ln(V_f/V_i)$, since $(T_f/T_i) = (V_f/V_i)$ for
an isobaric process. These results can be summarized in the following charts:
\begin{center}
\begin{tabular}{c|c|c|c|c}
Process & $W_{by}$ & $Q$ & $\Delta U$ & $\Delta S$ \\
\hline
Isobaric & $P_0(V_f - V_i)$ & $ (5/2)P_0(V_f - V_i)$ & $(3/2) P_0(V_f - V_i)$ & $(5/2)nR \; ln(V_f/V_i)$ \\
\end{tabular}
\end{center}
\noindent for a monatomic gas. For a diatomic gas, the results are
\begin{center}
\begin{tabular}{c|c|c|c|c}
Process & $W_{by}$ & $Q$ & $\Delta U$ & $\Delta S$ \\
\hline
Isobaric & $P_0(V_f - V_i)$ & $ (7/2)P_0(V_f - V_i)$ & $(5/2) P_0(V_f - V_i)$ & $(7/2)nR \; ln(V_f/V_i)$ \\
\end{tabular}
\end{center}
%\begin{figure}
%\includegraphics[width=14cm]{fig1b.png}
%\end{figure}
\end{document}