Thermodynamics

Ch. 12 Thermodynamics

Internal energy – all the energy of a stationary system (chemical, nuclear, kinetic, etc.)Thermal energy – That part of the internal energy that changes T

Gases – translational, vibrational, rotational all contribute to thermal energy(monatomic gases have only translational)

Ways to change internal energy of gas system:Heat QWork WBoth!

State: the values of P, V, T, U of the system at a given time (PVTU – “state variables”)Change of State: altering P, V, T, and/or U

12.1 Work in Thermodynamic Processes

Compress gas with (approximately)constant pressure P:

W = !F"y = !PA"y = !P"V

NOTE :"y is negative, "V is negative

so W is positive work done ON gas( )Work has been done on the gas tocompress it to a smaller volume.

If compressed gas expands against thepiston, work is done BY the gas (W is thennegative).

PV Diagrams & “PV Work”In above case, P=constant - “isobaric”

The amount of work involved is equal tothe area under the curve. In compression, !V is negative and the work done on thegas, W = !P"V will be positive.

If the process went the other way, !V ispositive, and the work done ON thegas W = !P"V is negative, i.e. work isactually done BY the gas.

Other possible cases:

Decrease V at constant PThen increase P

Increase P at constant VThen decrease V

P and V both changesimultaneously

Note: the work in (b) is larger than in (a) because the change in V occurred at higher P in(b), so the force required was larger. W depends on the way the system goes from onestate to another.

How to change P or V without doing W? Add/subtract Q!

Example: Problem #6Sketch a PV diagram of the following processes:

(a) A gas expands at constant pressure P1 from volume V1 to volume V2. It is thenkept at constant volume while the pressure is reduced to P2.

(b) A gas is reduced in pressure from P1 to P2 while its volume is held constant at V1.It is then expanded at constant pressure to a final volume V2.

(c) In which process is more work done by the gas? Why?

12.2 The First law of Thermodynamics

If positive Q is energy transferred to system, positive W is work done on system, then:

!U =U

final"U

initial= Q +W

Isolated System: where Q and W are ZERO. Here U = const.! "U = 0 .

Cyclic Process: !U = 0 " Q +W = 0 i.e. Q = #W

Note that a cyclic process only requires that the SUM of Q and W is zero, not both Q andW individually. (READ Tip 12.2 regarding the sign convention here).

Isothermal Process: another special case where !U = 0 . Here !T = 0 .

Example: ideal monatomic gas in cylinder in contact with heat reservoir:

Add heat carefully, keeping T constant andallowing V to increase and P to drop.

U =3

2nRT For !T = 0 " !U = 0

But !U = Q +W " 0 = Q +W

So Q = #WSince we added heat, Q is positive and W isnegative. Work is done BY the gas on the outsideworld.

Adiabatic Process: Q = 0 .

Example: Ignition phase in internal combustion engine. Hot gas expands air againstpiston so quickly that little Q has time to be lost.

Example: Problem #14

A monatomic ideal gas undergoes the thermodynamicprocess shown in the PV diagram here. Determinewhether each of the following values of !U, Q, and Wfor the gas is positive, negative, or zero.

Example: Problem #17A thermodynamic system undergoes a process in which its internal energy decreases by500 J. If at the same time 220 J of work is done on the system. Find the energytransferred to or from it by heat.


One mole of gas is initially at a pressure of2.00 atm, volume of 0.300 L and has aninternal energy equal to 91.0 J. It its finalstate the gas is at a pressure of 1.50 atmand a volume of 0.800 L and its internalenergy equals 180 J. For the paths IAF,IBF, and IF in this figure, calculate: (a) thework done on the gas and (b) the netenergy transferred to the gas by heat in theprocess.

12.3 The First Law and Human Metabolism

What is important here is not simply the total body heat Q and work W: Q +W = !U butthe rate at which they are “performed”:

!U

!t=

Q

!t+

W

!tIn order to radiate body heat and perform work on the outside world (note that Q and Wwill be negative, in general), the body’s internal energy must be re-supplied by food andoxygen.

Average rate of oxygen used in metabolizing food – 1 liter for every 4.8 kcal (= 4.8 Cal =20 J).

!U

!tkcal / s!

= 4.8!V

O2

!tL / s!

NOTE: Extreme activity produces a lot of Watts of power, but most individuals cannotdo this for very long! You burn up almost as many Calories sleeping 8 hours as youwould in 1 hour of heavy activity!

Example: Jog 1 mile = burn 120 Calories = 2 slices of bread

Efficiency

Not everybody can produce as much workfor the same intake of energy. The more fityou are, the more efficient you will be atproducing power.

Efficiency

e =

W!t

!U!t

Smooth activity,without a lot ofstops & starts(when no W isdone but Q isstill lost) is moreefficient as well.

12.4 Heat Engines and the Second law of Thermodynamics

During heat transfer from a hot reservoir to a cold reservoir, a portion of the randomizedkinetic energy of a gas can be converted into directed non-randomized work. Example:coal-fired or gas-fired electrical power plants.

Heat Engine – A device that converts part of the internal energy into work.

Cyclic Process – Gas returned to its initial state, so that its !U = Q +W = 0 . So the workdone on the engine is W = !Q , and so the work done by the engine is We

= !W

Engine absorbs net

Q

e, net= Q

hot! Q

cold which is converted towork done by the engine

W

e= Q

e, net= Q

hot! Q

cold .

The work done by the engine for a cyclicprocess is the area enclosed by the cycliccurve in its PV diagram. Think of this asPV work.

Thermal Efficiency: the fraction of the heat flowing in that is converted to work in 1cycle:

e =W

e

Qhot

=Q

hot! Q

cold

Qhot

= 1!Q

cold

Qhot

2nd Law of Thermodynamics: It is impossible to construct a heat engine that, operatingin a cycle, produces no other effect than the absorption of energy from a reservoir andthe performance of an equal amount of work.

In essence: it is always true that e < 1 and Q

cold> 0 and Q

hot< ! . No heat

engine is 100% efficient.

Applications: power generators, etc.

Reverse process: Heat Pumps. Examples: refrigerators heat pumps for homeheating/cooling.

12.5 Reversible & Irreversible Processes

Reversible – every state along the path is in equilibrium, and can return to initial statealong same path. In practice, needs to be slow & have no “unwanted” losses (such asfriction).

Irreversible – the real world.

12.6 The Carnot Engine

An idealized engine of maximum efficiency, working in reversible cycle. (Does not reallyexist).

e = 1!T

cold

Thot

For Tcold

= Thot

e = 0

ForT

cold

Thot

" 0 e" 1

All real engines are less efficient than the Carnot Engine because they operateirreversibly (due to friction) and because they complete a cycle in a brief time period (arenever in a state of equilibrium).

Example: Problem#27

One of the most efficient engines ever built is a coal-fired steam turbine in the Ohiovalley, driving an electric generator as it operates between 1870°C and 430°C.

(a) What is its maximum theoretical efficiency?(b) Its actual efficiency is 42%. How much mechanical power does the engine deliver

if it absorbs 1.40x105 J of energy each second from the hot reservoir?

12.7 Entropy

For a reversible system, if Qris the heat absorbed or expelled by the system:

!S =

Qr

TStrictly speaking, Qr

is for a reversible path. For real (irreversible) systems, we mustmodel he process by a reversible one (with the same initial and final states, of course!).

The entropy of the Universe increases in all natural processes.

Perpetual Motion Machines – ain’t no such thing!

1st kind – violate the First Law of Thermodynamics by having !U = 0 but Q +W < 1(puts out more energy than is put in).

2nd kind – violate the Second Law of Thermodynamics by having no heat loss from thesystem Qr

= 0 (having e ! 1).

12.8 Entropy and Disorder

Observation: A disorderly arrangement is much more probable than an orderly one if thelaws of nature are allowed to operate without interference.

Isolated (“closed”) systems tend toward greater disorder and entropy is a measure ofthat disorder.

NOTE: within such an isolated system, some parts might experience a decrease inentropy, but only at the expense of an even greater increase in entropy by the rest of thesystem.

Entropy a la Boltzmann: S = kB

lnW where W represents the probability of the systemhaving that specific configuration.

Example: Drawing colored marbles from a bag with equal numbers of Red & Green:

The 2R2G result is most probable, having 6 ways to get that result. Also most disordered.

The second law of thermodynamics is really a statement of what is most probable,rather than of what must be. In terms of entropy & the bag of marbles, sometimes youreally will pull 4 red ones in a row!

Implicit with our statement that entropy increases during a process is the definition of the“arrow of time”. One often sees a dropped plate shatter on the floor. One never sees abroken plate self-assemble and jump off the floor into our hand!

Degradation of Energy

In all real processes, the energy available for doing work decreases. As time progresses,higher-grade energy (that which can do useful work) generally gets transformed intolower-grade energy (that which can do less useful work).

Example: Dropping a ball to do work. Mechanically, a ball can do mgh work. But if allthe mgh were first transformed into heat, less work could be done, because no heatengine operates with 100% efficiency.


A 70-kg log falls from a height of 25 m into a lake. If the log, the lake, and the air are allat 300 K, find the change in entropy of the universe for this process.


The surface of the Sun is at approximately 5700 K, and the temperature of the Earth’ssurface is approximately 290 K. What entropy change occurs when 1000 J of energy istransferred by heat from the Sun to Earth?

Question: If entropy continually increases, how is it that complex living organisms canform and grow out of simpler molecules on the Earth?

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Thermodynamics