Work in Thermodynamic Processes

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Work in Thermodynamic Processes


We are now ready to combine the energy transfer
mechanism of
heat

with another mechanism:
work


We will focus our discussion on work done on a
gas


Important in discussion of how system gets from one
state

to another (described by state variables
P, T, V, n, U
)


Will provide connection between microscopic and
macroscopic energy transfer mechanisms


In Chap. 5 we spoke about work done by forces on
objects undergoing a displacement


Now we will talk about the work done
on

a gas
by

the environment
W

(or
by

gas
on

environment
W
env
)


Work done
on gas
(
W
) can be positive, negative, or zero



W =
0

when no “mechanical action” taking place



W =

W
env


Work influences internal energy

Work in Thermodynamic Processes


Work done
by

a gas
on

a piston in a
cylinder:
W
env

=
F
D
y
=

PA
D
y
=

P
D
V
=
P
(
V
f



V
i
)


When the piston moves inward,
V
f

< V
i

and
W
env

< 0


When the piston moves outward,
V
f

> V
i

and
W
env

> 0


When piston doesn’t move,
W
env

=
0


Work done
by

the piston
on

the gas
=
W

=

W
env


When the piston
moves inward,
V
f

< V
i

and
W

> 0


When the piston moves outward,
V
f

> V
i

and
W

< 0



When piston doesn’t move,
W =
0

(from
University Physics
, 11
th

Ed.)

Work in Thermodynamic Processes


This expression can only be used if the pressure
remains constant during the expansion or
compression


This is called an
isobaric

(constant pressure) process (same
Greek root as “barometer”)


If the pressure changes, the
average

pressure may be
used to estimate the work done


If pressure and volume are known at each step of the
process, a
PV
diagram

helps visualize process with
curves (
paths
) connecting initial and final states


The area under the curve on a
PV
diagram is equal in
magnitude to the work done on a gas
(
W
)


True whether or not
P

remains constant



W

is positive (negative) when volume decreases (increases)

PV

Diagrams


For a gas being compressed at constant pressure:





For a gas being compressed with varying pressure:





The work done depends on the particular path


(shaded area represents work done on
the gas; here
W

> 0
)

(shaded area, the area under the curve,
represents work done on the gas)

(same initial and final states, but
work done is different in each case)

Other Thermodynamic Processes


Isovolumetric

(or
isochoric
)


Volume stays constant


For example, maintain constant piston position


Vertical line on the
PV

diagram


No work done on gas


Isothermal


Temperature stays constant


Heat flows between the system and a
reservoir

to keep
the system’s temperature constant


For example, keep oven temperature constant


Adiabatic


No energy exchanged with the surroundings via heat


For example, provide excellent insulation for system


Or, process occurs so quickly that there is no time for
heat to flow in or out of system

PV

Processes Interactive

Example Problem #12.5

Solution (details given in class):

(a)


810 J

(b)


507 J

(c)


203 J

A gas expands from
I

to
F

along the
three paths indicated in the
figure. Calculate the work done
on

the gas along paths

(a)

IAF

(b)

IF
, and

(c)

IBF
.

First Law of Thermodynamics


Both work and heat can change the internal energy
of a system


Work can be done on a rubber ball by squeezing it,
stretching it, or throwing it onto a wall


Energy can flow to the ball via heat by leaving it out in the
sun or putting it into a hot oven


These 2 methods of increasing the internal energy of
a system lead to the
first law of thermodynamics
:


The change in internal energy of a system is equal to the
heat flow into the system plus the work done on the system




D
U

=
internal energy change of system



Q
=

energy transferred
to

system by heat from the outside



W

=
work done
on

the system

(really a statement of energy conservation!)

Sign Conventions for First Law

Quantity

Definition

Meaning of + sign

Meaning of


sign

Q

Energy transferred
by heat flow

Heat flow into the
system

Heat flow out of
the system

W

Work

Work done on the
system

Work done by the
system

D
U

Internal energy
change

Internal energy
increase

Internal energy
decrease

In

class example


Fill in the boxes with +,

, or 0:

Situation

System

Q

W

D
U

(a) Rapidly pumping up bicycle tire

Air in the pump

(b) Pan of room
-
temperature water resting
on a hot stove

Water in the pan

(c) Air leaking quickly out of a balloon

Air rushing out of
balloon

Applications of the First Law


Isolated system
: does not interact with its
surroundings


No energy transfer takes place and no work is done


Internal energy of the isolated system remains constant



Q = W =
D
U =
0


Cyclic process
: originates and ends at the same state



D
U =
0

and
Q =

W


The net work done per cycle by the gas is equal to the area
enclosed by the path representing the process on a
PV

diagram (important for describing
heat engines
)



PV

diagram for an ideal monatomic
gas confined in a cylinder by a movable
piston undergoing a cyclic process

Applications of the First Law


Isothermal process
: constant
-

temperature process


Consider ideal monatomic gas
contained in cylinder with movable
piston


The cylinder and gas are in thermal
contact with a large source of energy (reservoir)


Allow energy to transfer into the gas by heat


The gas expands (volume increases) and pressure
falls while maintaining a constant temperature


Since

,
D
U =
0


Therefore
D
U =
0

= Q + W


So
W =


Q <
0
(system supplies work to outside world)


Applications of the First Law


Adiabatic process
: no energy transferred by heat


The work done is equal to the change in the internal energy
of the system (
W =
D
U
)


One way to accomplish a process with no heat exchange is
to have it happen very quickly (mostly true in an internal
combustion engine)


In an adiabatic expansion, the work done is negative and the
internal energy decreases


For an ideal monatomic gas (see Chap. 10):


Isovolumetric process
: no change in volume, so
W =
0


The energy added to the system goes into increasing the
internal energy of the system (
D
U = Q
)


Temperature will increase


Isobaric process
: no change in pressure



W
=


P
D
V
, so

D
U = Q



P
D
V

The First Law and Human Metabolism


The first law can be applied to living organisms


The internal energy stored in humans goes into
other forms needed by the organs and into work and
heat


When we eat, we replenish our supply of internal energy


The
metabolic rate

(
D
U

/

D
t
)

is directly proportional
to the rate of oxygen consumption by volume


Basal metabolic rate (to maintain and run organs, etc.) is
about 80 W (other rates shown in Table 12.4)


The efficiency of the body is the ratio of the
mechanical power output to the metabolic rate


What you get out divided by what you put in


Efficiencies range from about 20% (cycling) to 3%
(shoveling)

Example Problem #12.16

Solution (details given in class):

(a)
12 kJ

(b)

12 kJ

A gas is taken through the cyclic
process described by the figure.

(a)
Find the net energy transferred
to the system by heat during
one complete cycle.

(b)
If the cycle is reversed


that
is, the process follows the path
ACBA



what is the net energy
transferred by heat per cycle?

Interactive Example Problem: Triangular
Cyclic Process on a
PV
Diagram

Simulation and solution details given in class.

(
ActivPhysics Online

Exercise #8.13, copyright Addison Wesley publishing)

Heat Engines


A
heat engine

is a device that converts
internal energy to other useful forms,
such as electrical or mechanical energy


Electrical power plants


Internal combustion engine in cars


A heat engine carries some “working
substance” through a cyclical process


Energy is transferred from a source at a high
temperature (
Q
h
)


Work is done by the engine (
W
eng
)


Energy is expelled to a source at a lower
temperature (
Q
c
)


Example of piston steam engine (in
-
class movie):

Animation

Heat Engine


Since it is a cyclical process, its initial and final
internal energies are the same


So,
D
U =
0 =
Q + W
and
Q = Q
net

=


W = W
eng


The work done by the engine equals the
net

energy
absorbed by the engine


Therefore,


The work is equal to the area enclosed by the curve
of the
PV

diagram (if working substance is a gas)


Thermal efficiency

is defined as the ratio of the work
done by the engine to the energy absorbed at the
higher temperature (what you get divided by what
you put in):

Second Law of Thermodynamics


The efficiency of heat engines is addressed in the
second law of thermodynamics
:


No engine can have 100% efficiency


Other statements and implications of the second
law:


A system cannot convert heat solely into mechanical work


Heat never flows spontaneously from a colder body to a
hotter body


It is impossible for any process to have as its sole result
the transfer of heat from a cooler to a hotter body


There is no such thing as a free lunch!


Summary of 1
st

and 2
nd

laws:


1
st

law: We cannot get more energy out than we put in


2
nd

law: We cannot break even

Internal Combustion Engines






An idealized model of the thermodynamic processes of a gas
engine is called the
Otto cycle


Intake stroke
: Gas
-
air mixture drawn into cylinder (a


b)


Compression stroke
: Intake valve closes, mixture compressed (adiabatic
compression; b


c
)


Ignition
: Spark plug ignites mixture (heating at constant volume; c


d
)


Power stroke
: Hot burned mixture pushes mixture down, doing work
(adiabatic expansion; d


e
)


Cooling
: Energy released via heat at constant volume (e


b)


Exhaust stroke
: Exhaust valve opens and burned mixture is pushed out
of cylinder (b


a)

(from
University Physics
, 11
th

Ed.)

(from
College Physics
, Giambattista)

Heat Pumps: Refrigerators, Air conditioners


Refrigerators





Energy is extracted from the cold reservoir (food cabin)
and transferred to hot reservoir (kitchen)


Air conditioner works on the same principle





Both are examples of a
heat pump
(reverse heat
engines)

(from
University Physics
, 11
th

Ed.)

(from
University Physics
, 11
th

Ed.)

Heat Pumps Interactive

Carnot Engine


The
most

efficient heat engine possible is the
Carnot
engine


A
theoretical

heat engine operating in an ideal,
reversible

cycle (a cycle in which system can be returned to its initial
state along the same path) called the
Carnot cycle


Note that a truly reversible process is an idealization, and
real processes are
irreversible

(although some are close
to being reversible)


PV

diagram of the cycle:


Consists of 2 adiabatic and 2
isothermal processes, all reversible


Net work done
W

is net energy
received in one cycle:
W = Q
h


Q
c

Carnot Cycle


Assume the substance that changes
temperature is an ideal gas


Gas is contained in cylinder
with movable piston at one end


Steps of the cycle:


A


B: Gas expands isothermally
while in contact with reservoir at
T
h


B


C: Gas expands adiabatically
(
Q =
0)


C


D: Gas compressed
isothermally while in contact with reservoir
at
T
c
< T
h


D


A: Gas compressed adiabatically


Note that upward (downward) red arrows on
piston indicate removal (addition) of sand

Carnot Engine


Carnot showed that the efficiency of the engine
depends on the temperatures of the reservoirs:




Temperatures must be in Kelvin


All Carnot engines operating between the same two
temperatures will have the same efficiency


The efficiency increases as
T
c

is lowered and as
T
h

is raised


Efficiency is 0 if
T
h

= T
c


Efficiency is 100% only if
T
c

= 0 K

(not possible from
third
law of thermodynamics
)


In most practical cases
T
c

is near room temperature (300 K)
so generally
T
h

is raised to increase efficiency


All real engines are less efficient than the Carnot engine
(friction, cycle completed in brief time period)

Example Problem #12.27

Solution (details given in class):

(a)
0.672 (or 67.2%)

(b)
58.8 kW

One of the most efficient engines ever built is a coal
-
fired steam turbine engine in the Ohio River valley,
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 42.0%. How much
mechanical power does the engine deliver if it
absorbs 1.40


10
5

J of energy each second from
the hot reservoir?