Download Thermodynamics

Thermodynamics
-------------------------------------
Joseph F. Alward, PhD
Department of Physics
University of the Pacific
Thermodynamics:
"thermo":
Greek therme heat
"dynamics": Greek dynamikos powerful
Physics that deals with the mechanical
action or relations between heat and work
Example 1: Heat to work
Heat Q from flame provides energy
to do work
---------------------------------------------------------Example 2: Work to heat.
Work done by person is converted
to heat energy via friction.
Internal Energy U:
(measured in joules)
Sum of random translational,
rotational, and vibrational
kinetic energies
U: change in U
U > 0 is a gain of internal
energy
U < 0 is a loss of internal
energy
----------------------------------------
Thermal Energy:
Vibrational kinetic
energy
in solids
The hotter the object,
the larger the
vibrational
kinetic energy
same as internal energy
Ideal Gases
Molecules are point-size. Gas is
low-density.
Elastic collisions are the only interactions.
----------------------------------------------------------------Average kinetic energy
per molecule = (3/2)kT
k = Boltzmann's constant
= 1.38 x 10-23 J/K
U = (3/2) NkT
N = number of molecules
In an ideal gas, internal energy is proportional
to the absolute (Kelvin) temperature.
Motions of a
diatomic
molecule in a fluiid
Heat
... is the amount of internal energy entering or leaving a system
... occurs by conduction, convection, or radiation.
... causes a substance's temperature to change
... is not the same as the internal energy of a substance
... is positive if thermal energy flows into the substance
... is negative if thermal energy flows out of the substance
... is measured in joules
------------------------------------------------------------------------------------improper: "heat flow" is redundant
Thermal Equilibrium:
Systems (or objects) are said to be in thermal equilibrium
if there is no net flow of thermal energy from one to the other.
A thermometer is in thermal equilibrium with the medium
whose temperature it measures, for example.
If two objects are in thermal equilibrium, they are at the
same temperature.
Are you now in thermal equilibrium with your environment?
Positive and Negative Work
W is positive if work is done by
system.
(Recall: Work = Fcos 
W is negative if work is done on the
system.
Air does work on the
environment: W > 0.
Environment (man) does work on
system: W < 0
(Alternative: system does negative
work because
force by air pressure on thumb is
opposite to the
direction of motion of the thumb.)
The First Law of Thermodynamics
(Conservation of Energy)
"Energy can neither be created nor
destroyed,
but only transferred from one system to
another
and transformed from one form to
another."
or
"The internal energy of an isolated
system is
constant (even though that energy may
be
transformed from one type to another)."
For thermodynamic systems, the 1st
Law is:
U = Q - W
Types of Thermodynamic
Processes
Greek
isos:
equal
baros:
weight
adiabatos:
not passable
----------------------------------------------------Isothermal:
Isobaric:
Isochoric:
Adiabatic:
Same temperature
Same pressure
Same volume
Zero heat flow (Q = 0)
First Law Example
Example: 1000 J of thermal energy flows
into a system (Q = 1000 J). At the same
time, 400 J of work is done by the system
(W = 400 J).
What is the change in the system's internal
energy U?
---------------------------------------------------------Solution:
U = Q - W
= 1000 J - 400 J
= 600 J
First Law Example
Example: 800 J of work is done on a system
(W = -800 J) as 500 J of thermal energy is
removed from the system (Q = -500 J).
What is the change in the system's internal
energy U?
-------------------------------------------------------------Solution:
U = Q - W
= -500 J - (-800 J)
= -500 J + 800 J
= 300 J
Work Done by an Expanding Gas
W = Fs = (PA) s = PV
V = Vf - Vi
W = P (Vf - Vi)
Area under pressure-volume curve is
the work done
---------------------------------------------------Isobaric Process: "same pressure"
Greek: barys, heavy
Work and the Pressure-Volume
Curve
(Related to Problems 10 and 14)
Work Done = Area Under PV curve
------------------------------------How much work is done by the
system
when the system is taken from:
(a) A to B (900 J)
(b) B to C (0 J)
(c) C to A (-1500 J)
------------------------------------Each "rectangle" has an area of
100 Pa-m3 = 100 (N/m2)-m3
= 100 N-m
= 100 Joules
Expanding Gas
(First Law)
10 grams of steam at 100 C at constant
pressure rises to 110 C:
P = 4 x 105 Pa
T =
10 C 
V = 30.0 x 10-6 m3
c = 0.48 cal/g
Example: If a gas expands at a
constant
pressure, the work done by the gas
What is the change in internal energy?
U = Q - W
-----------------------------------------------------W = (4 x 105)(30.0 x 10-6) = 12 J
Q = mcT = (10)(.48)(10) = 48 cal
= (48 cal)(4.186 J/cal) = 201 J
U = Q - W = 201 J - 12 J = 189 J
------------------------------------------------------
is:
W = Fs = (PA)s = P(As) =
PV
What if the steam were compressed
while it was absorbing heat?
First Law Example
Key Ideas:
Example: Aluminum cube of
side L is
heated in a chamber at
atmospheric
pressure. What is the change in
the
cube's internal energy?
1. U = Q - W
2. Q = mcT
3. m = V0
4. V0 = L3
5. W = PV
6. V =
V0T
U = Q - W
Go to Solution
First Law Example
U = Q - W
W is positive
U = Q - W
Q is zero.
W = -0.03 J
U = 0 - (-0.03 J) = 0.03 J (positive)
Assuming ideal gas: U = (3 / 2) N k T
Q is zero
U is negative
Temperature drops from 41F
to - 31 F causing vapor to
condense into a cloud of
tiny droplets.
T is positive since U is positive
First Law Examples
U = Q - W
W is positive
Q is zero
U is negative
Steam temperature drops
and is cool to the touch.
U = Q - W
Q is zero. W is negative
Why does paper catch
on fire?
Work, Rubber Bands, and Internal Energy
U = Q - W
Expand rubber band: W < 0, Q = 0
U > 0
temperature increases
--------------------------------------------------------Press thick rubber band to lip and expand
it rapidly. The warming should be
obvious.
Now allow the band to contract quickly;
cooling will also be evident.
Heating a Gas at
Constant Volume
Isochoric Process
"same volume"
U = Q - W = Q
Conceptual Questions
1. 100 Joules of heat is added to a gas, and the gas expands at constant
pressure.
Is it possible that the internal energy increases by 200 J?
No.
2. A gas is compressed isothermally, and its internal energy increases. Is
the gas
an ideal gas? No. For an ideal gas, U = (3 / 2) N k T.
3. State how U changes in each of the following processes (U = Q - W)
(a)
500 J
(b)
200 J
(c)
(d)
(e)
W = -500 J
W=0
and
and
Q= 0
................. U = 0 - (- 500 J) =
Q = -200 J ................. U = - 200 J - 0 = -
W = 100 J
and
Q = 100 J
W = -100 J
and
Q = -100 J
W = 300 J
and
Q = 500 J
Adiabatic Expansion of a Ideal Gas
Key Idea #1: For an ideal gas, U depends only on the absolute temperature T.
Key Idea #2: If there's nothing to push against, an expanding gas does zero work.
Gas in Chamber A suddenly
rushes into Chamber B. How
does the final temperature of
gas compare to its initial
temperature?
---------------------------------------U = Q - W
Q
=0
W =0
U = (3 / 2) N k T = 0
T = 0
Second Law of Thermodynamics
Heat flows naturally from a region at high
temperature
to a region at low temperature. By itself, heat will not
flow from a cold to a hot body.
When an isolated system undergoes a change,
passing
from one state to another, it will do so in such a way
that
its entropy (disorder) will increase, or at best remain
the
same.
The 2d Law applied to creationism:
"The earth must have been created supernaturally,
since it is a highly-ordered system." Rebuttal: The
earth is not isolated; the sun provides the energy to
create the order.
Appendix:
Example: Aluminum cube of side L is heated in a chamber at atmospheric
pressure. What is the change in the cube's internal energy?
--------------------------------------------------------------------------U = Q - W = mcT
- PV
= ( V0 ) cT - P ( V0T )
= ( c - PV0T
--------------------------------------------------------------------------L = 20 cm
P = 1.01 x 105 Pa
 = 2.7 x 103 kg/m3
T = 100 C
c = 0.90 x 103 J/kg-C
 = 72 x 10-6 C-1 (coefficient of volume expansion)
U = ( c - PV0T
= [( 2.7 x 103 )( .48 ) - (1.01 x 105) (72 x 10-6) ] (0.20)3 (100)