Chapter 15
The Giancoli onLine Tutor

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The Laws of Thermodynamics.

I. First Law of Thermodynamics

The First Law of Thermodynamics states that

"The change in Internal Energy, DU, of a system is added to the total of the energy added to it by any mechanism."

The two mechanisms we consider are:
Work, W
and
Heat, Q.
Heat added to a system is positive (and negative if it is taken out).
Work done by a system is positive (and work done on a system is negative).
With these sign conventions the mathematical statement of the First Law is

DU= Q-W.

For the most part we limit systems to "containers" of a monatomic gas. A gas has the following properties:

Pressure
Volume
Temperature
mols
Internal Energy
Entropy.

The first four are related by the Ideal Gas Equation. If you know three of them you can find the fourth. (Normally the number of mols will be a constant.) The Ideal Gas Equation is called an Equation of State. An Equation of State relates two or more properties of a gas that is in a fixed state. A process equation (to be covered below) describes how related properties change as the gas is changed in a fixed way.

For an ideal gas the Internal energy only depends on the temperature. It can be calculated by using

U= 3nRT/2 .

Although generally we are only interested in the change in internal energy and use instead

DU= 3nRDT/2

The Equations above along with the equation for the work done on or by the gas

W= PDV,
But this equation is only useful in an algebra based course if the pressure is constant.

Thermodynamic Processes

In studying gas dynamics we represent the gas as a point or as a smooth series of points (called a process) on a P-V diagram. The various possible processes are outlined below.

Process	description	internal energy	heat	work
Constant Temperature aka Isothermal	T=constant Hyperbola on a P-V diagram	DU= 0	Q= W	W= nRTln(V₂/V₁)
Constant Pressure aka Isobaric	P=constant A straight horizontal line on a P-V diagram	DU= 3nRDT/2	Q=DU+W	W= PDV
Constant Volume aka Isochoric	V= constant A vertical straight line on a P-V diagram.	DU= 3nRDT/2	Q=DU	W=0
Adiabatic aka Isentropic	Q= 0 PV^1.67= constant	DU= 3nRDT/2	Q= 0	W= -DU

Interactive Example
One mol of a monatomic gas is taken at constant volume from a temperature of 200 K to 400 K, what is the work done? Enter your answer below and click outside.

To the nearest joule what is the change in Internal Energy for the gas in the above question? Enter your answer below and click outside.

To the nearest joule what is the heat added to the gas in the above question? Enter your answer below and click outside.

One mol of a monatomic gas is taken at constant pressure of 5 atm from a volume of 200 L to 400 L, what is the work done to the nearest joule? Enter your answer below and click outside.

To the nearest kelvin what is the change in temperature if the initial temperature is 200 K? Enter your answer below and click outside.

To the nearest joule what is the change in Internal Energy of the gas in the above question? Enter your answer below and click outside.

To the nearest joule what is the heat added to the gas in the above question? Enter your answer below and click outside.

Heat Engines

A Heat Engine absorbs heat from a high temperature source (reservoir) converts some of it into work and exhausts the rest to a low temperature reservoir. It does this by taking a working substance, in our case a monatomic gas, through a series of processes until it returns to its initial state. The engine is said to be operating in a cycle. Because of this DU= 0 and from First Law Equation

W= Q= Q_H-Q_L
Q_H is the heat absorbed from a high temperature reservoir, and Q_L is the heat discharged to the low temperature reservoir. The efficiency, e, is defined as what you get (W) divided by what you pay for (Q_H). From this
e= W/Q_H= 1 - Q_H/Q_L.
We deal only with reversible engines.

A reversible engine can be operated in reverse, using work to extract heat from a low temperature reservoir and delivering all the energy as heat to a high temperature reservoir. As such it can function as a refrigerator, heat pump, or air conditioner.

For reversible engines Q_H/Q_L= T_H/T_L, from which

e= 1- T_L/T_H.

Interactive Activity
The prototypical heat engine is the Carnot Engine.

The Carnot engine cycle starts with the gas at a high pressure and small volume and expands by absorbing heat from a surrounding high temperature reservoir doing work in the process (W= Q_H). At a second point the gas is insulated and allowed to expand further dropping its temperature as it continues to do work (W= -3nR(T_L-T_H) ). At a third point the gas is put in contact with a low temperature reservoir and work (W= Q_L) is done on it to compress it. At a final point it is again insulated and is further compressed raising its temperature until it returns to its starting state. The total work done in the cycle is

W= Q_H(1-T_L/T_H).

Watch the movie for awhile and then halt it by pressing the left button. Use the right buttons to step back and forth through the movie.

Note if you do not have the Quicktime plugin you can view a GIF animation version of the movie here.

Hints:

No matter what the process, if the gas is expanding it is doing work (W>0) and, conversely, if it contracts work is being done on it (W<0).
No matter what the process, the change in internal energy is always 3nRDT/2.
If a gas is carried through a series of processes and ends up with the same temperature it started with, even if it is not at its starting point, DU= 0 and Q= W.
Usually it is sufficient to know nT and not n and T separately. (nT=PV/R).
In any cycle the point closest P= 0 & V=0 has the lowest internal energy.

II. The Second Law of Thermodynamics and Entropy

Entropy is probability. If you throw 4 dice the chances of getting two heads (HH) is 1/4; the changes of getting a head and a tail (HT) is 1/2. In very real terms, the entropy of (HT) is 1/2 and the entropy of (HH) is 1/2. If you have a system that is in two parts, one with energy U₁, volume V₁, and entropy P₁, the other with U₂, V₂, and P₂. The system as energy U=U₁+U₂, and V=V₁+V₂, but entropy P=P₁P₂. To make entropy additive we arbitrarily define it as
S= Ln(P)

now it will be additive:
S= LN(P)= Ln(P₁P₂)= LN(P₁)+LN(P₂)= S₁+S₂

Unfortunately that is not the whole story. Like temperature, people knew about entropy before they knew what it was. They knew how to calculate changes in entropy by using
DS= Q/T,
where Q is the heat added to a system at temperature T. Calculated in that way entropy has units J/K (which is really unitless since K is just a unit of energy like J). In order that the new definition of entropy match the older one entropy is now defined as
S=kLn(P),
where k is Boltzmann's constant which has the units of J/K.

Interactive Activity
To learn why nature favors disorder as the number of objects increase follow this link . You will find that with two dice even though HT is more common then HH, none-the-less HH shows up with frequency. However, with 8 dice, four of each is common, but you will have to wait until the winter Olympics is held in hell before all heads show up. Imagine how unlikely order will be with 100s, or thousands, or millions, or... of coins are thrown. Even millions is a small number when dealing with atoms. Nature favors disorder! Try it as many times as you like.

The second Law of Thermo dynamics can be stated in many ways but they all amount to Nature favors disorder! (because disorder is (much) more probable than order and what is probably going to happen is what is most probable)!

What follows are a few statements of The Second Law:

You can't buih a perfect refrigerator.
Heat can't go from cold to hot.
You can't build a perfect heat engine.
Reversible heat engines are more efficient than irreversible (read-real) heat engines.
All the King's men couldn't put Humpty Dumpty back together again.
The entropy of the universe never decreases.
In any irreversible process (read-real) the entropy of the universe always increases.
Murphy's VII Law (Mother Nature is a B****)
If you think things are mixed up now just you wait, Henry Higgins.

It is important to realize that entropy is a property. This means that it does not matter how you get from one point (state) to another, the change in entropy will be the same. In reality equation (5) is the most useful equation. For example, suppose that 20 Joules of energy conducted from a high temperature reservoir at 400K to a low temperature reservoir at 200K with nothing else happening. Your Job is to calculate the change in entropy of the universe. The significance of "nothing else happening" is to tell you that the only affected parts of the universe are the high and low temperature reservoirs. The significance of term reservoir is to tell you that, whatever it is, its temperature does not measurably change if you add or subtract a finite amount of heat. Therefore:

DS_highT= (-20)/400= -.05

DS_lowT= (20)/200= .1

DS_universe=-.05+.1= .05
Let us look at a similar problem, a heat engine operating in a cycle, extracting heat, Q_H, from a high temperature reservoir , exhausting heat, Q_L, to a low temperature reservoir and doing work equal to Q_H-Q_L. The engine is operating in a cycle so its entropy does not change (it returns to where it starts). Work is organized energy so it has zero entropy. Thus, as before, only the two reservoirs are affected so the change in entropy of the universe is
DS_universe= -Q_H/T_H+Q_L/T_L.
Looking at this you can see why:

You can't build a perfect heat engine (a engine with no exhaust): If Q_L=0 the change in entropy of the universe is negative. Impossible!

Recall that for a reversible engine
Q_L/Q_H=T_L/_TH
or
Q_L/T_L=Q_H/T_H
which shows us that for a reversible engine the change in entropy of the universe is zero! For a refrigerator heat is removed from a low temperature and delivered to a high temperature of the universe is:

DS_universe= +Q_H/T_H-Q_L/T_L.
From this you can see why: You can't build a perfect refrigerator (a device that transfers heat from cold to hot with no work being done, i.e., Q_H=Q_L): If Q_H= Q_L=Q the change in entropy of the universe is negative (because T_L<T_H). Impossible!

Interactive Example
What is the change in entropy of the universe if one kilogram of water at 280K is mixed with one kilogram of water at 300K. Answer the following questions to reach the answer.

What is the final temperature of the mixture? Enter your answer below and click outside.

To the nearest joule how much energy is lost by the 'hot' water and gained by the 'cool' water? Enter your answer below and click outside.

What is the change in entropy, to the nearest J/K, of the hot water? Enter your answer below and click outside.

What is the change in entropy, to the nearest J/K, of the cool water? Enter your answer below and click outside.

What is the change in entropy, to the nearest J/K, of the 'universe'? Enter your answer below and click outside.

© 1997-1999 Carl Adler mailto:Carl@Image-ination.com

For a thermodynamic process to be reversible all change must occur in small microscopic steps effectively taking an infinite time to do anything. Thus a reversible heat engine is an idealization serving as a upper limit on what we can do in reality. Since power is Energy/Time a reversible engine can be thought of as an engine which delivers energy very efficiently with zero power.